PHYS 305 - INTRODUCTION TO MODERN PHYSICS Rotational kinematics Question Bank Set 6 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Rotational

kinematics

Question Bank - Set 6

Liberty University

Question 1

Question

A solid sphere of radius Ris released from rest on a ramp that is inclined at

an angle θto the horizontal. The sphere rolls without slipping down the ramp.

Find an expression for the acceleration of the center of mass of the sphere as a

function of the angle θ.

Solution

Step 1: First, let’s consider the forces acting on the sphere. The weight mg

acts downwards, the normal force Nacts perpendicular to the surface of the

ramp, and the frictional force fsacts parallel to the surface of the ramp. The

component of the weight acting along the ramp is mg sin θ, and the component

perpendicular to the ramp is mg cos θ.

Step 2: The net force causing acceleration down the ramp is the component

of the weight along the ramp, so we have Fnet =mg sin θ.

Step 3: The acceleration of the center of mass of the sphere is given by

a=Fnet

m, where mis the mass of the sphere.

Step 4: We can express the mass min terms of the radius Rand density ρ

of the sphere: m=4

3πR3ρ.

Step 5: Substituting the expression for mass into the acceleration formula,

we get:

a=mg sin θ

3πR3ρ

Step 6: Simplifying the expression, we find:

a=3

gsin θ

Therefore, the acceleration of the center of mass of the sphere as a function

of the angle θis 3

gsin θ

Question 2

Question

A disk with a radius of 0.1 m starts from rest and accelerates with a constant

angular acceleration of 5 rad/s2. What is the angular velocity of the disk after

3 seconds?

Solution

Step 1: Find the angular velocity at time tusing the equation ω=ω0+αt,

where ωis the final angular velocity, ω0is the initial angular velocity, αis the

angular acceleration, and tis the time.

ω= 0 + 5 ×3 = 15 rad/s

Therefore, the angular velocity of the disk after 3 seconds is 15 rad/s.

Question 3

Question

A disk with a radius of 0.2 m starts from rest and accelerates with a constant

angular acceleration of 3 rad/s2for 4 seconds. After this time, the disk reaches

an angular velocity of 12 rad/s. What is the angular displacement of a point on

the rim of the disk during this time?

Solution

Step 1: Calculate the angular velocity at the end of the 4-second period using

the equation:

ωf=ωi+α·t

where ωfis the final angular velocity, ωiis the initial angular velocity (0 rad/s),

αis the angular acceleration, and tis the time.

ωf= 0 + 3 ·4 = 12 rad/s

Step 2: Calculate the average angular velocity during the 4-second period

using:

¯ω=ωi+ωf

¯ω=0 + 12

2= 6 rad/s

Step 3: Calculate the angular displacement using the equation:

θ=ωi·t+1

2α·t2

where θis the angular displacement, ωiis the initial angular velocity, αis the

angular acceleration, and tis the time.

θ= 0 ·4 + 1

2·3·42= 24 rad

Therefore, the angular displacement of a point on the rim of the disk during

the 4-second period is 24 radians.

Question 4

Question

A disk of radius 0.2 m is rotating about a fixed axis passing through its center

with an initial angular velocity of 5 rad/s. The disk is subjected to a constant

angular acceleration of −0.1 rad/s2. Compute the time it takes for the disk to

come to a stop.

Solution

Step 1: The angular acceleration can be expressed in terms of the final angular

velocity using the equation α=ωf−ωi

t, where αis the angular acceleration, ωf

is the final angular velocity, ωiis the initial angular velocity, and tis time. Step

2: Plugging in α=−0.1 rad/s2,ωi= 5 rad/s, and ωf= 0 into the equation, we

get −0.1 = 0−5

t. Step 3: Solving for t, we find t=5

0.1= 50 seconds. Therefore,

it will take 50 seconds for the disk to come to a stop.

Question 5

Question

A disk with a radius of 0.5 m is rotating with an angular velocity of 4 rad/s. The

rotational inertia of the disk is 0.2 kg ·m2. At a certain time, a constant torque

is applied to the disk, causing it to decelerate at a rate of 2 rad/s2. Calculate

the angular position of a point on the edge of the disk after 3 seconds.

Solution

Step 1: Calculate the initial angular acceleration using the torque equation

τ=Iα. Given: r= 0.5 m, ω0= 4 rad/s, α=−2 rad/s2,I= 0.2 kg ·m2.

From the torque equation:

τ=Iα

τ=I·(−2)

τ=−0.4 Nm

Step 2: Calculate the angular velocity after 3 seconds using the kinematic

equation ω=ω0+αt.

ω= 4 + (−2) ·3

ω= 4 −6

ω=−2 rad/s

Step 3: Calculate the angular displacement after 3 seconds using the kine-

matic equation θ=θ0+ω0t+1

2αt2. Given: θ0= 0 (assuming the initial position

is at 0 radians).

θ= 0 + 4 ·3 + 1

2·(−2) ·32

θ= 12 −9

θ= 3 rad

Therefore, the angular position of a point on the edge of the disk after 3

seconds is 3 radians.

Question 6

Question

A thin, uniform rod of length Lis rotating about an axis which is perpendicular

to the rod and passes through one end of the rod. The rod is initially at rest

and then a constant force Fis applied perpendicular to the rod at the other

end. Find the angular acceleration of the rod in terms of F,L, and the moment

of inertia of the rod.

Solution

Step 1: The moment of inertia of the rod about the axis of rotation is I=1

3mL2,

where mis the mass of the rod.

Step 2: The torque applied to the rod is given by τ=F r, where ris the

distance from the force to the axis of rotation. In this case, r=L.

Step 3: The torque applied causes an angular acceleration α, related to the

torque and moment of inertia by the equation τ=Iα.

Step 4: Substituting the values for torque and moment of inertia into the

equation above, we get F r =1

3mL2α.

Step 5: Solving for α, we find α=3F

mL .

Therefore, the angular acceleration of the rod in terms of F,L, and the

moment of inertia is α=3F

mL .

Question 7

Question

A disk of radius Ris initially at rest. It starts rotating with a constant angular

acceleration α. At some time t, the angular velocity of the disk is ω. What is

the magnitude of the tangential acceleration of a point on the rim of the disk

at this time?

Solution

Step 1: The angular velocity ωis related to the angular acceleration αand time

tusing the equation

ω=αt

Step 2: The tangential velocity vof a point on the rim of the disk is related to

the angular velocity ωusing the equation

v=Rω

Step 3: Taking the derivative of vwith respect to time tgives the tangential

acceleration atof the point on the rim of the disk:

at=Rdω

Step 4: Substituting ω=αt into the equation above gives

at=Rd(αt)

Step 5: Taking the derivative of αt with respect to tgives

at=Rα

Step 6: Therefore, the magnitude of the tangential acceleration of a point on

the rim of the disk at time tis Rα.

Question 8

Question

A thin rod of length Land mass Mis pivoted at one end and released from rest

in a horizontal position. Find the angular velocity of the rod when it makes an

angle of θwith the vertical.

Solution

Step 1: We will start by calculating the moment of inertia of the rod about the

pivot point. The moment of inertia of a rod rotating about one end is given

by I=1

3mL2, where mis the mass per unit length of the rod. Since the total

mass of the rod is Mand its length is L,m=M

L. Thus, the moment of inertia

of the rod about the pivot point is:

I=1

3M

LL2=1

3ML2

Step 2: Next, we will apply the conservation of mechanical energy to find

the angular velocity of the rod at an angle θ. The initial mechanical energy is

purely potential energy: Ei=mgh, where his the vertical height the center

of mass has dropped to (which is L(1 −cos θ)) and gis the acceleration due to

gravity. The final mechanical energy consists of both potential and rotational

kinetic energy: Ef=mgh′+1

2Iω2, where h′is the vertical height of the center

of mass when the rod makes an angle θ, and ωis the angular velocity. Setting

Ei=Efand solving for ωgives:

mgh =mgh′+1

2Iω2

mgh =mgL(1 −cos θ) + 1

21

3ML2ω2

Step 3: Simplifying the energy equation gives us:

21

3ML2ω2= 2mgL(1 −cos θ)

6ML2ω2= 2mgL(1 −cos θ)

ω=r12g(1 −cos θ)

Therefore, the angular velocity of the rod when it makes an angle θwith the

vertical is ω=q12g(1−cos θ)

Question 9

Question

A solid disk with a radius of 0.5 m and a mass of 2 kg is rotating about its

central axis with an angular acceleration of 4 rad/s2. At a certain instant, the

angular velocity of the disk is 5 rad/s. What is the total kinetic energy of the

rotating disk at this instant?

Solution

Step 1: We first find the moment of inertia of the disk using the formula for a

solid disk:

I=1

2mr2

where m= 2 kg (mass of the disk) and r= 0.5 m (radius of the disk). Substitute

these values to get

I=1

2×2 kg ×(0.5 m)2= 0.25 kg m2

Step 2: Now, we can find the kinetic energy of the disk at the given instant

using the formula for rotational kinetic energy:

KE =1

2Iω2

where ω= 5 rad/s (angular velocity of the disk). Substitute the values of Iand

ωto find

KE =1

2×0.25 kg m2×(5 rad/s)2= 6.25 J

Step 3: Therefore, the total kinetic energy of the rotating disk at the given

instant is 6.25 Joules.

Question 10

Question

A thin, uniform rod of length Land mass Mis free to rotate about a horizontal

axis passing through one end. Initially, the rod is at rest. A small object of

mass mtraveling horizontally with speed vcollides with the free end of the rod

and sticks to it. Find the final angular velocity of the system.

Solution

Step 1: Find the initial angular momentum of the system. The initial angular

momentum of the system is given by the equation:

Li=Iω

where Iis the moment of inertia and ωis the angular velocity. The moment of

inertia Iof the rod about the end where it can rotate is 1

3ML2. So,

Li=1

3ML2×0 = 0

Step 2: Find the final moment of inertia of the system. After the collision,

the system consists of a rod with mass M+mand length Lrotating about one

end. The moment of inertia of the combined system is 1

3(M+m)L2.

Step 3: Apply conservation of angular momentum. By conservation of an-

gular momentum, the initial angular momentum equals the final angular mo-

mentum.

Lf=Ifωf

0 = 1

3(M+m)L2×ωf

Step 4: Find the final angular velocity of the system. Solving for ωf, we get:

ωf= 0

Therefore, the final angular velocity of the system is 0.

Question 11

Question

A disk with radius Rstarts from rest and accelerates uniformly for 5 seconds

until it reaches an angular speed of 100 rad/s. After that, the disk decelerates

uniformly until it comes to a stop after rotating through one full revolution.

Calculate the angular acceleration when the disk is decelerating.

Solution

Step 1: Find the initial angular speed ωiof the disk when it starts accelerating.

Given that the disk starts from rest, ωi= 0.

Step 2: Find the angular acceleration αduring acceleration phase.

Using the equation of motion for rotational accelerated motion:

ωf=ωi+αt

Substitute the known values: ωf= 100 rad/s, ωi= 0, t= 5 s.

100 = 0 + α×5

α= 20 rad/s2

Step 3: Find the final angular speed ωfafter deceleration phase.

When the disk completes one full revolution, its final angular speed is 0 rad/s.

Step 4: Find the angular deceleration αdduring deceleration phase.

Using the equation of motion for rotational decelerated motion:

ωf=ωi+αdt

Substitute the known values: ωf= 0 rad/s, ωi= 100 rad/s, t= total time

taken for one revolution.

αd=ωf−ωi

αd=0−100

Step 5: Find the total time Ttaken for the disk to complete one revolution.

The total time for completing one revolution is the sum of time taken for accel-

eration and time taken for deceleration.

T= Time for acceleration + Time for deceleration

T= 5 + 100

αd

Step 6: Solve for αd.

Substitute the expression for T,

αd=0−100

5 + 100

αd

Step 7: Rearrange the equation to solve for αd.

αd=−100

5 + 100

αd

Thus, the angular acceleration when the disk is decelerating is −100

5+ 100

αd

rad/s

Question 12

Question

A disk with radius Rand moment of inertia Iabout its center is initially at

rest. A constant force 

Fis applied tangentially at the edge of the disk. After

the force has been applied for a time t, the disk acquires an angular velocity ω.

Find the angular acceleration of the disk in terms of I,R,F, and ω.

Solution

1. Let’s start by writing down the torque equation for the given situation. The

torque applied by the force 

Fat the edge of the disk is given by:

τ=RF sin θ

Here, θis the angle between 

Fand the radius Rof the disk. Since the force is

tangential, θ= 90◦, and sin θ= 1. So the torque simplifies to:

τ=RF

2. The torque applied to an object is related to its angular acceleration α

by the equation:

τ=Iα

Substitute the expression for torque into this equation:

RF =Iα

3. We also know that the angular acceleration αis related to the angular

velocity ωand the time tby:

ω=αt

Solving for α, we get:

α=ω

4. Substitute the expression for αinto our equation for torque:

RF =Iω

t

5. Finally, solve for the angular acceleration α:

α=RF t

Therefore, the angular acceleration of the disk in terms of I,R,F, and ωis

RF t

Question 13

Question

A wheel starts from rest and rotates with a constant angular acceleration of 2

rad/s2. After 4 seconds, what is the angular velocity of the wheel?

Solution

Step 1: Use the equation for angular velocity under constant angular accelera-

tion:

ω=ω0+αt

where



ωis the final angular velocity,



ω0is the initial angular velocity (in this case, 0 since the wheel starts from

rest),



αis the angular acceleration (given as 2 rad/s2), and



tis the time (given as 4 seconds).

Step 2: Substitute the given values into the equation:

ω= 0 + 2 ×4 = 8 rad/s

Therefore, after 4 seconds, the angular velocity of the wheel is 8 rad/s.

Question 14

Question

A solid sphere of radius Rand mass Mrolls without slipping down a ramp

inclined at an angle θ. The sphere is released from rest at the top of the ramp.

What is the linear speed of the center of the sphere when it reaches the bottom

of the ramp?

Solution

Let’s denote the linear speed of the center of the sphere as v, the radius as R,

the mass as M, and the angle of inclination as θ. The moment of inertia of a

solid sphere about its center is 2

5MR2.

Step 1: Find the acceleration of the sphere down the ramp using energy con-

siderations. The initial potential energy is converted into translational kinetic

energy and rotational kinetic energy at the bottom of the ramp.

Initial potential energy = Final kinetic energy

Mgh =1

2Mv2+1

2Iω2

Since the sphere rolls without slipping, ω=v

MgR sin θ=1

2Mv2+1

22

5MR2v

R2

gR sin θ=1

2v2+1

5v2

v=r10

7gR sin θ

Therefore, the linear speed of the center of the sphere when it reaches the

bottom of the ramp is r10

7gR sin θ.

Question 15

Question

A spherical object of radius Rstarts from rest at the top of a frictionless incline

that makes an angle of θwith the horizontal. The object rolls without slipping

down the incline. What is the linear acceleration of the center of mass of the

object in terms of gand θ?

Solution

Let’s denote the linear acceleration of the center of mass as a.

Step 1: The acceleration of a rolling object involves both translational and

rotational components. Considering the object is rolling without slipping, the

relationship between the linear acceleration and angular acceleration is given by

a=Rα, where αis the angular acceleration.

Step 2: To find the angular acceleration α, we can use the torque equation.

The torque about the center of mass due to gravity is τ=−mgR sin(θ), where

the negative sign indicates the direction opposite to the rotation.

Step 3: The net torque acting on the object is equal to Iα, where Iis the

moment of inertia of the object. For a solid sphere rolling down an incline, the

moment of inertia about an axis through its center and perpendicular to the

incline is I=2

5mR2.

Step 4: Equating the torque equation with the net torque equation, we

have −mgR sin(θ) = 2

5mR2α. Solving for α, we get α=−5gsin(θ)

2R.

Step 5: Now, substituting αback into the relationship between linear and

angular acceleration, we find a=R−5gsin(θ)

2R=−5gsin(θ)

Step 6: Therefore, the linear acceleration of the center of mass of the object

rolling down the incline is a=−5gsin(θ)

Question 16

Question

A thin rod of length Land mass Mis rotating about a fixed axis passing through

one of its ends with an angular velocity ω0. At a certain moment, a point mass

mis attached to the free end of the rod. What is the new angular velocity of

the system if the linear speed of the point mass at that moment is v0?

Solution

1. To solve this problem, we will use the principle of conservation of angular

momentum. Initially, the angular momentum of the system is given by:

Linitial =Iω0

where Iis the moment of inertia of the rod about the fixed axis.

2. When the mass mis attached to the free end of the rod, the new angular

momentum is given by:

Lfinal = (I+mL2)ωf

where ωfis the final angular velocity of the system.

3. Since angular momentum is conserved, we have:

Iω0= (I+mL2)ωf

4. The moment of inertia of the rod about the fixed axis is I=1

3ML2.

Substituting this into the conservation equation, we get:

3ML2ω0=1

3ML2+mL2ωf

5. Simplifying the expression, we find the new angular velocity ωfto be:

ωf=

3ML2ω0

3ML2+mL2

6. Since the linear velocity v0of the mass mis related to the angular velocity

by v=ωfL, we have:

v0=ωfL

7. Substituting the expression for ωfinto the linear velocity equation, we

can solve for the final angular velocity ωf.

Question 17

Question

A disc of radius Ris initially at rest when a constant torque τis applied to it.

The torque is then removed when the disc has rotated through an angle θ. If

the moment of inertia of the disc about its center of mass is I, determine the

angular velocity of the disc as a function of the angle rotated, ω(θ).

Solution

Step 1: By Newton’s second law for rotation, the torque applied can be related

to the angular acceleration αas τ=Iα.

Step 2: We know that α=dω

dt , and ω=dθ

dt . So, we have τ=Id2θ

dt2.

Step 3: Since τis a constant, we can write τ=Id2θ

dt2as τ=Id2θ

dθ2

dθ

dt .

Step 4: Rearranging the terms and solving the differential equation gives

dω

dθ =τ

Step 5: Integrating both sides with respect to θgives ω=τ

Iθ+C, where C

is a constant of integration.

Therefore, the angular velocity of the disc as a function of rotated angle is

given by ω(θ) = τ

Iθ+C.

Question 18

Question

A wheel initially at rest starts rotating with a constant angular acceleration of

2.0 rad/s2. Through what angle does the wheel turn in 5.0 s?

Solution

Step 1: The formula for angular displacement under constant angular accelera-

tion is given by

θ=θ0+ω0t+1

2αt2

where θis the final angular position, θ0is the initial angular position, ω0is the

initial angular velocity, αis the angular acceleration, and tis the time taken.

Step 2: Since the wheel is initially at rest, ω0= 0 and θ0= 0. We can

substitute the given values α= 2.0 rad/s2and t= 5.0 s into the formula to find

the angular displacement.

Step 3: Substituting the values into the formula gives

θ=0+0+ 1

2×2.0×(5.0)2

Step 4: Calculating the expression gives

θ=0+0+ 1

2×2.0×25

Step 5: Therefore, the angular displacement of the wheel in 5.0 s is

θ= 25 rad

Step 6: So, the wheel turns through 25 radians in 5.0 seconds under a

constant angular acceleration of 2.0 rad/s2.

Question 19

Question

A wheel initially at rest accelerates uniformly for 4 seconds before coming to a

stop in 6 seconds. If the wheel makes 6 rotations during this time, determine

the angular acceleration of the wheel.

Solution

Step 1: Find the initial angular velocity of the wheel.

Given that the wheel initially at rest, we have ωi= 0.

Step 2: Find the final angular velocity of the wheel.

Using the kinematic equation for rotational motion:

ωf=ωi+αt

where ωfis the final angular velocity, ωiis the initial angular velocity, αis

the angular acceleration, and tis the time.

Since the wheel comes to a stop, ωf= 0, and ωi= 0.

Step 3: Find the angular acceleration of the wheel.

Using ωf=ωi+αt again, but this time for the second part when the wheel

accelerates, we have:

ωf=ωi+αt

Substitute the values we know: ωf= 0, ωi= 0, t= 4 s.

0 = 0 + α×4

α= 0 rad/s2

Step 4: Calculate the total angular displacement of the wheel.

Since the wheel makes 6 rotations, it completes 6 ×2πrad of angular dis-

placement.

Step 5: Find the average angular speed of the wheel.

The average angular speed is given by:

¯ω=∆θ

∆t

where ∆θis the total angular displacement and ∆tis the total time taken.

¯ω=6×2π

6+4

¯ω=12π

10 =6π

5rad/s

Step 6: Find the angular acceleration.

Using the relationship between angular acceleration, angular velocity, and

time:

¯ω=1

2(ωi+ωf)

ωf= 2¯ω−ωi

Since ωi= 0, we have:

ωf= 2¯ω

ωf= 2 ×6π

5=12π

5rad/s

The angular acceleration αcan be found using the formula:

α=ωf−ωi

Substitute the values: ωf=12π

5rad/s, ωi= 0 rad/s, t= 6 s.

α=

12π

5−0

α=12π

α=2π

5rad/s2

Therefore, the angular acceleration of the wheel is 2π

5rad/s2.

Question 20

Question

A thin rod of length Land mass Mis rotating about an axis passing through

one end with an angular velocity ω. The rod is abruptly stopped in t0seconds.

Find the angular acceleration of the rod.

Solution

Step 1: Let’s denote the angular acceleration of the rod as α. Since the rod

is initially rotating with an angular velocity ωand comes to a stop in time t0,

we can relate the angular acceleration, angular velocity, and time through the

equation:

ω=αt0

Step 2: Next, we need to consider the torque acting on the rod to find the

angular acceleration. The only torque acting on the rod is due to the stopping

force, which exerts a torque about the axis of rotation. The torque τdue to

this force is given by:

τ=Iα

where Iis the moment of inertia of the rod about the axis of rotation.

Step 3: The moment of inertia of a thin rod rotating about one end is given

by:

I=1

3ML2

Step 4: Since the torque τis also equal to the rate of change of angular

momentum, we have:

τ=dL

Step 5: Here, the initial angular momentum of the rod is Li=Iω =1

3ML2ω,

and the final angular momentum is Lf= 0 (since the rod comes to a stop). So,

the change in angular momentum is ∆L=Lf−Li=−1

3ML2ω.

Step 6: Setting the torque τequal to the change in angular momentum ∆L

and substituting the expressions for τand I, we get:

3ML2α=−1

3ML2ω

Step 7: Simplifying the above equation, we find the angular acceleration α:

α=−ω

Therefore, the angular acceleration of the rod when abruptly stopped is −ω.

Question 21

Question

A disk of radius Rand mass Mis rotating about its central axis with an angular

velocity ω0(initially). A torque τis applied to the disk until it comes to a stop

in a time interval T. Determine the angular acceleration of the disk as a function

of time during this deceleration.

Solution

1. The torque applied to an object is related to its angular acceleration according

to the equation τ=Iα, where τis the torque, Iis the rotational inertia, and α

is the angular acceleration.

2. The rotational inertia of a solid disk rotating about its central axis is

I=1

2MR2.

3. Since the disk is coming to a stop, the final angular velocity will be 0,

and the initial angular velocity is ωi=ω0.

4. The angular acceleration αcan be determined using the equation of

rotational kinematics: ωf=ωi+αt, where ωfis the final angular velocity.

5. Since ωf= 0, we have 0 = ω0+αT .

6. Solving for α, we find α=−ω0

7. Therefore, the angular acceleration of the disk as a function of time during

deceleration is α(t) = −ω0

Tt.

Question 22

Question

A solid cylinder of radius Rand mass Mis initially at rest on a horizontal

surface. A constant horizontal force Fis applied to the cylinder, causing it to

roll without slipping. If the force is applied at a distance rfrom the center of

the cylinder, determine the speed of the cylinder once it has rolled a distance d.

Solution

Step 1: Find the moment of inertia of the cylinder. The moment of inertia of a

solid cylinder of radius Rand mass Mabout its central axis is given by:

I=1

2MR2

Step 2: Write the equation for the work done on the cylinder. The work

done on the cylinder is equal to the change in kinetic energy. The work done

by the force Fthat acts at a distance rcan be written as:

W=F·d

Step 3: Calculate the work done by the force F. The work done by the force

Fcan also be expressed in terms of the torque produced by the force about the

center of the cylinder:

W=τ·θ

where the angle θthrough which the force acts is related to the distance dby

d=Rθ.

Step 4: Determine the torque produced by the force F. The torque produced

by the force Fabout the center of the cylinder is given by:

τ=F·r

Step 5: Set the work done by the force equal to the change in kinetic energy.

Equating the work done by the force to the change in kinetic energy gives:

F·d=1

2Iω2

where ωis the angular velocity of the cylinder.

Step 6: Write the relationship between linear and angular velocity. Since

the cylinder is rolling without slipping, the linear velocity vis related to the

angular velocity ωby:

v=Rω

Step 7: Solve for the angular velocity of the cylinder. Substitute I,d, and

vinto the equation from step 5, and use the relationship between linear and

angular velocity to solve for ω:

F d =1

21

2MR2v

R2

Step 8: Calculate the speed of the cylinder. Finally, substitute the value of

ωinto the relationship between linear and angular velocity to find the speed of

the cylinder when it has rolled a distance d:

v= 2 F d

M1/2

Question 23

Question

A wheel initially at rest starts rotating with a constant angular acceleration

α= 0.5 rad/s2. At the same time, a bug located at the edge of the wheel starts

walking along the edge with a constant speed of 0.1 m/s. How long will it take

for the bug to reach the opposite side of the wheel, assuming the wheel has a

radius of 0.5 m?

Solution

Step 1: Let’s first find the angular velocity of the wheel at any time tusing the

equation for angular acceleration:

ω=ω0+αt

where ωis the final angular velocity, ω0is the initial angular velocity (which is

zero in this case), αis the angular acceleration, and tis the time. So, for our

case:

ω= 0 + 0.5t= 0.5trad/s

Step 2: Next, we need to find the angle through which the wheel has rotated

at time tusing the equation:

θ=θ0+ω0t+1

2αt2

where θis the final angle, θ0is the initial angle (which is zero in this case), ω0

is the initial angular velocity (which is zero), αis the angular acceleration, and

tis the time. Substitute the values into the equation:

θ=0+0+ 1

2×0.5t2= 0.25t2rad

Step 3: Since the bug is moving with a linear speed of 0.1 m/s along the

edge of the wheel, the bug will traverse a distance equal to the circumference

of the wheel. The circumference of the wheel is given by 2πr, where ris the

radius of the wheel. So, the time taken for the bug to cross to the other side is:

t=2πr

Substitute the given values:

t=2π×0.5

0.1= 10πs≈31.4 s

Therefore, it will take the bug approximately 31.4 seconds to reach the op-

posite side of the wheel.

Question 24

Question

A disc of radius Rstarts from rest and rolls without slipping down a frictionless

incline. Determine the linear speed of the center of mass of the disc when it has

descended a vertical distance h.

Solution

Let’s denote the linear speed of the center of mass of the disc as vcm.

Step 1: Calculate the angular speed of the disc when it has descended a

vertical distance h. Since the disc rolls without slipping, the linear speed of the

outer edge of the disc is equal to the radius times the angular speed: vedge =Rω,

where ωis the angular speed of the disc. The total mechanical energy of the

disc is conserved, so we have:

2Iω2+mgh =1

2mv2

edge

where Iis the moment of inertia of the disc, mis the mass of the disc, and g

is the acceleration due to gravity. Since the disc is a solid cylinder, I=1

2mR2.

Substituting these values in, we get:

4mR2ω2+mgh =1

2mR2ω2

Solving for ωgives:

ω=r8gh

Step 2: Calculate the linear speed of the center of mass. The linear speed

of the center of mass vcm can be related to the angular speed ωby:

vcm =Rω

Substitute in the expression for ωwe found in Step 1:

vcm =Rr8gh

3R=r8ghR

Therefore, the linear speed of the center of mass of the disc when it has

descended a vertical distance his q8ghR

Question 25

Question

A thin rod of length Land mass Mis rotating about one end with an angular

velocity ω. The rod is then rotated about its center at the same angular velocity.

Determine the kinetic energy of the rod after it is rotated about its center

compared to when it was rotated about one end.

Solution

Step 1: Let’s first calculate the initial kinetic energy of the rod when it is

rotating about one end.

The moment of inertia of a thin rod rotating about one end is given by I=

3ML2. The kinetic energy is given by KE =1

2Iω2. Substituting I=1

3ML2

into the expression for kinetic energy, we have:

KEinitial =1

2×1

3ML2×ω2=1

6Mω2L2

Step 2: Next, let’s calculate the kinetic energy of the rod after it is rotated

about its center.

When the rod is rotated about its center, the moment of inertia changes to

I=1

12 ML2. Therefore, the kinetic energy is given by KE =1

2Iω2. Substituting

I=1

12 ML2into the expression for kinetic energy, we have:

KEfinal =1

2×1

12ML2×ω2=1

24Mω2L2

Step 3: To determine the ratio of the final kinetic energy to the initial kinetic

energy, we divide the final kinetic energy by the initial kinetic energy:

KEfinal

KEinitial

24 Mω2L2

6Mω2L2=1

Therefore, the kinetic energy of the rod after it is rotated about its center is

4times the kinetic energy when it was rotated about one end.

Question 26

Question

A disk of radius 0.5 m is rotating about an axis perpendicular to the disk and

passing through its center. The angular position of a point on the rim of the

disk is given by θ= 2t2+ 3t, where θis in radians and tis in seconds. Calculate

the velocity of the point on the rim at t= 2 s.

Solution

Step 1: Find the angular velocity ωat t= 2 s. Given that θ= 2t2+ 3t, we first

need to find the angular velocity, ω, which is the derivative of θwith respect to

ω=dθ

dt =d

dt(2t2+ 3t) = 4t+ 3

Step 2: Calculate the angular velocity ωat t= 2 s. Substitute t= 2 into

the equation for ωto find

ω= 4(2) + 3 = 8 + 3 = 11 rad/s

Step 3: Calculate the linear velocity vat t= 2 s. The linear velocity vof

a point on the rim of the disk is related to the angular velocity ωby v=rω,

where ris the radius of the disk (0.5 m).

v=rω = 0.5×11 = 5.5 m/s

Therefore, the velocity of the point on the rim of the disk at t= 2 s is 5.5

m/s.

Question 27

Question

A disc of radius 0.25 m rotates about a fixed axis perpendicular to the disc

and through its center. The disc accelerates from rest and reaches an angular

velocity of 12 rad/s in 3 seconds. What is the angular acceleration of the disc?

Solution

Step 1: Identify the known quantities.

The initial angular velocity ω0= 0, the final angular velocity ω= 12 rad/s, and

the time t= 3 s.

Step 2: Calculate the angular acceleration using the formula for angular

acceleration.

The angular acceleration αcan be found using the equation:

α=∆ω

∆t

where ∆ω=ω−ω0and ∆t=t. Substitute the known values:

α=12 −0

3= 4 rad/s2

Step 3: State the final answer.

The angular acceleration of the disc is 4 rad/s2.

Question 28

Question

An object starts from rest and undergoes angular acceleration in a straight line.

The object reaches an angular velocity of 60 rad/s in a time of 4 seconds. If the

angular acceleration is constant, what is the angular acceleration of the object?

Solution

Step 1: Identify the given values The initial angular velocity, ω0, is 0 rad/s. The

final angular velocity, ω, is 60 rad/s. The time interval, t, is 4 seconds.

Step 2: Use the equation of rotational kinematics The equation of rotational

kinematics relating angular acceleration, initial angular velocity, final angular

velocity, and time interval is:

ω=ω0+α·t

where ωis the final angular velocity, ω0is the initial angular velocity, αis the

angular acceleration, and tis the time interval.

Step 3: Substitute the given values into the equation

60 = 0 + α·4

Step 4: Solve for the angular acceleration, α

α=60

4= 15 rad/s2

Therefore, the angular acceleration of the object is 15 rad/s2.

Question 29

Question

A thin uniform rod of length Land mass Mis free to rotate about a frictionless

pivot at one end. The rod is held horizontally and then released. What is the

angular acceleration of the rod when it makes an angle θwith the vertical?

Solution

To solve this problem, we will apply the principles of rotational motion and use

torque to calculate the angular acceleration of the rod.

Step 1: First, let’s identify the forces acting on the rod when it is at an

angle θwith the vertical. The forces acting on the rod are the gravitational

force mg acting at the center of mass, and the normal force acting at the pivot

point.

Step 2: Next, we need to calculate the torque produced by the gravitational

force. The torque τproduced by a force Fat a distance rfrom the pivot point

is given by τ=F·r·sin(ϕ), where ϕis the angle between the force vector and

the position vector.

For the gravitational force mg, the torque about the pivot point is τ=

(mg)·L

2·sin(θ).

Step 3: Now, we can apply Newton’s second law for rotational motion,

which states that the net torque is equal to the moment of inertia Itimes the

angular acceleration α:Pτ=I·α.

Substitute the expression for torque and moment of inertia I=1

3ML2into

the equation. We have (mg)·L

2·sin(θ) = 1

3ML2·α.

Step 4: Solve for the angular acceleration, α. We find α=3g

2Lsin(θ).

Thus, the angular acceleration of the rod when it makes an angle θwith the

vertical is α=3g

2Lsin(θ).

Question 30

Question

A solid disk of radius Rand mass Mis rotating about its axis with an angular

velocity ω0. A small block with mass mis placed on the disk at a distance r

from the center. The coefficient of static friction between the block and the disk

is µ. What is the maximum value of rfor which the block stays on the disk as

it slows down due to friction?

Solution

1. Let’s first determine the condition for the block to not slip off the disk. The

maximum static frictional force fmax is given by

fmax =µN,

where Nis the normal force acting on the block. The normal force can be

resolved into components perpendicular and parallel to the disk’s surface. The

perpendicular component balances the block’s weight mg, giving N=mg.

2. Therefore, the maximum frictional force fmax =µmg must provide the

required centripetal force when ris at its maximum value in order for the block

to stay on the disk.

3. The centripetal force required for the block to stay on the disk is provided

by the frictional force:

fmax =mv2

where vis the tangential velocity of the block.

4. The tangential velocity of the block can be expressed in terms of the

angular velocity of the disk. As the disk slows down due to friction, the angular

velocity decreases, and the block’s velocity along the disk decreases accordingly:

v=r(ω−αt),

where αis the angular acceleration of the disk.

5. The frictional force when the block is at its maximum distance from the

center is

µmg =m(r(ω−αt)2)/r.

6. Simplifying the above equation, we get

µg = (m/r)(ω2−2ωαt).

7. At the moment when the block slips off the disk, the frictional force

reaches its maximum value µmg. Substituting fmax =µmg into the equation

above, we have

µg = (m/r)(ω2

0−2αt).

8. Before the block slips off the disk, the tangential velocity of the block is

zero. At this moment, the disk has slowed down to angular velocity ωf, which

can be calculated using the kinematic equation

ωf=ω0−αt.

9. Substituting ωf= 0 and ω0−αt into the equation found in step 7 yields

the maximum value of r:

r=1

µg m

ω2

0.

Therefore, the maximum value of rfor which the block stays on the disk as

it slows down due to friction is m

µgω2

Question 31

Question

A thin rod of length Land mass Mis pivoted at one end and is allowed to swing

like a pendulum. Initially, the rod is at rest in a horizontal position. What is

the angular velocity of the rod when it is vertical?

Solution

Given: Length of the rod, LMass of the rod, M

We want to find the angular velocity of the rod when it is vertical.

Step 1: Let’s calculate the potential energy of the rod when it is horizontal.

When the rod is horizontal, the center of mass is located at a distance L/2

from the pivot point. The potential energy can be calculated using the formula

U=mgh, where mis the mass of the portion of the rod above the pivot point,

gis the acceleration due to gravity, and his the vertical height of the center of

mass above the lowest point. The potential energy when the rod is horizontal

is given by:

Uhorizontal =Mg L

2

Step 2: Now, let’s calculate the potential energy of the rod when it is

vertical. When the rod is vertical, the full length Lof the rod is below the pivot

point so the potential energy can be calculated using the formula U=mgh.

The potential energy when the rod is vertical is given by:

Uvertical =MgL

Step 3: The change in potential energy is equal to the kinetic energy ac-

quired by the rod when it reaches the vertical position. So, we can equate the

two expressions for potential energy and solve for the rotational kinetic energy.

Uhorizontal −Uvertical =1

2Iω2

where I=1

3ML2is the moment of inertia of the rod about the pivot.

Substitute the expressions for potential energy and moment of inertia into the

equation above.

Step 4: Simplify the equation and solve for ω, the angular velocity of the

rod when it reaches the vertical position.

Mg L

2−MgL =1

21

3ML2ω2

Solving for ωgives:

ω=r3g

So, the angular velocity of the rod when it is vertical is q3g

2L.

Question 32

Question

A disk of radius 0.2 m starts from rest and accelerates with a constant angular

acceleration of 0.5rad/s2for 5 seconds.

1. What is the angular velocity of the disk after 5 seconds?

2. How many revolutions has the disk made during this time?

3. What is the linear velocity of a point on the rim of the disk after 5 seconds?

Solution

1. To find the angular velocity of the disk after 5 seconds, we can use the

equation for angular velocity:

ω=ω0+αt

where: - ωis the final angular velocity, - ω0is the initial angular velocity (0

since the disk starts from rest), - αis the angular acceleration, - tis the time.

Step 1: Calculate the final angular velocity ω.

ω= 0 + 0.5×5=2.5rad/s

2. To find how many revolutions the disk has made during this time, we

need to first calculate the total angle rotated by the disk in 5 seconds. This can

be done using the equation:

θ=ω0t+1

2αt2

where: - θis the total angular displacement, - ω0is the initial angular velocity

(0), - αis the angular acceleration, - tis the time.

Step 2: Calculate the total angular displacement θ.

θ= 0 + 0.5×1

2×52= 6.25 rad

Since one revolution is equivalent to 2πradians, the number of revolutions

is:

Number of revolutions = 6.25 rad

2π rad/rev ≈0.995 rev

3. Lastly, to find the linear velocity of a point on the rim of the disk after 5

seconds, we use the equation:

v=rω

where: - vis the linear velocity, - ris the radius of the disk, - ωis the final

angular velocity.

Step 3: Calculate the linear velocity v.

v= 0.2×2.5 = 0.5m/s

So, after 5 seconds, the angular velocity of the disk is 2.5rad/s, the disk has

completed approximately 0.995 revolutions, and a point on the rim of the disk

has a linear velocity of 0.5m/s.

Question 33

Question

A disk with a radius of 0.5 m is spinning counterclockwise with an angular speed

of 4 rad/s. A small bug starts at the center of the disk and walks outwards along

the rim of the disk at a constant speed of 0.1 m/s. How far has the bug traveled

along the rim of the disk when it reaches the edge?

Solution

Step 1: Find the angular speed of the bug. The bug is moving tangentially

along the edge of the disk, so its linear speed is equal to the angular speed times

the radius of the disk. Given: Angular speed of disk, ωdisk = 4 rad/s Radius of

disk, r= 0.5 m

The bug’s angular speed, ωbug, is given by:

ωbug =linear speed of bug

radius of disk =0.1

0.5= 0.2 rad/s

Step 2: Find the time taken for the bug to reach the edge. The bug will

reach the edge when it has traveled an angle equivalent to the angle subtended

by the radius of the disk:

θ=distance traveled

radius of disk =2πr

r= 2πrad

The time taken for the bug to reach the edge is given by:

t=θ

ωbug

=2π

0.2= 10πs

Step 3: Calculate the distance traveled by the bug. The distance traveled

by the bug is given by the linear speed of the bug multiplied by the time taken:

Distance = linear speed ×time taken = 0.1×10π= 10πm

Therefore, the bug has traveled 10πmeters along the rim of the disk when

it reaches the edge.

Question 34

Question

A disk of radius 0.2 m is rotating with an angular velocity of 5 rad/s. A bug is

initially at rest on the rim of the disk. What is the bug’s tangential velocity 2

seconds after the bug starts crawling towards the center of the disk?

Solution

Step 1: The bug’s initial radial distance from the center of the disk, rinitial, is

equal to the radius of the disk, which is 0.2 m.

Step 2: The bug moves radially towards the center of the disk with a constant

radial velocity. The bug’s final radial distance from the center of the disk, rfinal,

decreases as it moves inwards. The time taken by the bug to reach the center

of the disk is 2 seconds.

Step 3: The bug’s tangential velocity vtis given by the equation:

vt=r·ω

where ris the radial distance from the center (in this case, rfinal), and ωis the

angular velocity of the disk.

Step 4: To find rfinal, we can use the fact that the bug moves at a constant

radial velocity. The bug’s radial velocity can be calculated using the formula:

vr=rinitial −rfinal

time

Step 5: Substituting the known values into the formula for vr:

vr=0.2 m −rfinal

2 s

Step 6: Since vris constant, we can also write:

vr=rfinal ·α

where αis the angular velocity of the bug.

Step 7: Equating the expressions for vr:

0.2 m −rfinal

2 s =rfinal ·α

Step 8: Solving for αgives:

α=0.2 m

2 s + 2 ·rfinal

Step 9: Since α=ω, we know that:

vt=rfinal ·ω=rfinal ·α=rfinal ·0.2 m

2 s + 2 ·rfinal

Step 10: Therefore, the bug’s tangential velocity 2 seconds after it starts

crawling towards the center of the disk is 0.2·rfinal

2+2·rfinal .

Question 35

Question

A disk of radius Ris rotating about its axis with a constant angular acceleration

of α. At time t= 0, the disk has an angular velocity of ω0and is at an angular

position of θ0. Find the angular position of a point on the rim of the disk at

time t.

Solution

Step 1: To find the angular position of a point on the rim of the disk at time t,

we can use the rotational kinematics equation:

θ=θ0+ω0t+1

2αt2

Therefore, the acceleration of the center of mass of the sphere as a function

of the angle θis 3

gsin θ

Question 2

Question

A disk with a radius of 0.1 m starts from rest and accelerates with a constant

angular acceleration of 5 rad/s2. What is the angular velocity of the disk after

3 seconds?

Solution

Step 1: Find the angular velocity at time tusing the equation ω=ω0+αt,

where ωis the final angular velocity, ω0is the initial angular velocity, αis the

angular acceleration, and tis the time.

ω= 0 + 5 ×3 = 15 rad/s

Therefore, the angular velocity of the disk after 3 seconds is 15 rad/s.

Question 3

Question

A disk with a radius of 0.2 m starts from rest and accelerates with a constant

angular acceleration of 3 rad/s2for 4 seconds. After this time, the disk reaches

an angular velocity of 12 rad/s. What is the angular displacement of a point on

the rim of the disk during this time?

Solution

Step 1: Calculate the angular velocity at the end of the 4-second period using

the equation:

ωf=ωi+α·t

where ωfis the final angular velocity, ωiis the initial angular velocity (0 rad/s),

αis the angular acceleration, and tis the time.

ωf= 0 + 3 ·4 = 12 rad/s

Step 2: Calculate the average angular velocity during the 4-second period

using:

¯ω=ωi+ωf

¯ω=0 + 12

2= 6 rad/s

Step 3: Calculate the angular displacement using the equation:

θ=ωi·t+1

2α·t2

where θis the angular displacement, ωiis the initial angular velocity, αis the

angular acceleration, and tis the time.

θ= 0 ·4 + 1

2·3·42= 24 rad

Therefore, the angular displacement of a point on the rim of the disk during

the 4-second period is 24 radians.

Question 4

Question

A disk of radius 0.2 m is rotating about a fixed axis passing through its center

with an initial angular velocity of 5 rad/s. The disk is subjected to a constant

angular acceleration of −0.1 rad/s2. Compute the time it takes for the disk to

come to a stop.

Solution

Step 1: The angular acceleration can be expressed in terms of the final angular

velocity using the equation α=ωf−ωi

t, where αis the angular acceleration, ωf

is the final angular velocity, ωiis the initial angular velocity, and tis time. Step

2: Plugging in α=−0.1 rad/s2,ωi= 5 rad/s, and ωf= 0 into the equation, we

get −0.1 = 0−5

t. Step 3: Solving for t, we find t=5

0.1= 50 seconds. Therefore,

it will take 50 seconds for the disk to come to a stop.

Question 5

Question

A disk with a radius of 0.5 m is rotating with an angular velocity of 4 rad/s. The

rotational inertia of the disk is 0.2 kg ·m2. At a certain time, a constant torque

is applied to the disk, causing it to decelerate at a rate of 2 rad/s2. Calculate

the angular position of a point on the edge of the disk after 3 seconds.

Solution

Step 1: Calculate the initial angular acceleration using the torque equation

τ=Iα. Given: r= 0.5 m, ω0= 4 rad/s, α=−2 rad/s2,I= 0.2 kg ·m2.

From the torque equation:

τ=Iα

τ=I·(−2)

τ=−0.4 Nm

Step 2: Calculate the angular velocity after 3 seconds using the kinematic

equation ω=ω0+αt.

ω= 4 + (−2) ·3

ω= 4 −6

ω=−2 rad/s

Step 3: Calculate the angular displacement after 3 seconds using the kine-

matic equation θ=θ0+ω0t+1

2αt2. Given: θ0= 0 (assuming the initial position

is at 0 radians).

θ= 0 + 4 ·3 + 1

2·(−2) ·32

θ= 12 −9

θ= 3 rad

Therefore, the angular position of a point on the edge of the disk after 3

seconds is 3 radians.

Question 6

Question

A thin, uniform rod of length Lis rotating about an axis which is perpendicular

to the rod and passes through one end of the rod. The rod is initially at rest

and then a constant force Fis applied perpendicular to the rod at the other

end. Find the angular acceleration of the rod in terms of F,L, and the moment

of inertia of the rod.

Solution

Step 1: The moment of inertia of the rod about the axis of rotation is I=1

3mL2,

where mis the mass of the rod.

Step 2: The torque applied to the rod is given by τ=F r, where ris the

distance from the force to the axis of rotation. In this case, r=L.

Step 3: The torque applied causes an angular acceleration α, related to the

torque and moment of inertia by the equation τ=Iα.

Step 4: Substituting the values for torque and moment of inertia into the

equation above, we get F r =1

3mL2α.

Step 5: Solving for α, we find α=3F

mL .

Therefore, the angular acceleration of the rod in terms of F,L, and the

moment of inertia is α=3F

mL .

Question 7

Question

A disk of radius Ris initially at rest. It starts rotating with a constant angular

acceleration α. At some time t, the angular velocity of the disk is ω. What is

the magnitude of the tangential acceleration of a point on the rim of the disk

at this time?

Solution

Step 1: The angular velocity ωis related to the angular acceleration αand time

tusing the equation

ω=αt

Step 2: The tangential velocity vof a point on the rim of the disk is related to

the angular velocity ωusing the equation

v=Rω

Step 3: Taking the derivative of vwith respect to time tgives the tangential

acceleration atof the point on the rim of the disk:

at=Rdω

Step 4: Substituting ω=αt into the equation above gives

at=Rd(αt)

Step 5: Taking the derivative of αt with respect to tgives

at=Rα

Step 6: Therefore, the magnitude of the tangential acceleration of a point on

the rim of the disk at time tis Rα.

Question 8

Question

A thin rod of length Land mass Mis pivoted at one end and released from rest

in a horizontal position. Find the angular velocity of the rod when it makes an

angle of θwith the vertical.

Solution

Step 1: We will start by calculating the moment of inertia of the rod about the

pivot point. The moment of inertia of a rod rotating about one end is given

by I=1

3mL2, where mis the mass per unit length of the rod. Since the total

mass of the rod is Mand its length is L,m=M

L. Thus, the moment of inertia

of the rod about the pivot point is:

I=1

3M

LL2=1

3ML2

Step 2: Next, we will apply the conservation of mechanical energy to find

the angular velocity of the rod at an angle θ. The initial mechanical energy is

purely potential energy: Ei=mgh, where his the vertical height the center

of mass has dropped to (which is L(1 −cos θ)) and gis the acceleration due to

gravity. The final mechanical energy consists of both potential and rotational

kinetic energy: Ef=mgh′+1

2Iω2, where h′is the vertical height of the center

of mass when the rod makes an angle θ, and ωis the angular velocity. Setting

Ei=Efand solving for ωgives:

mgh =mgh′+1

2Iω2

mgh =mgL(1 −cos θ) + 1

21

3ML2ω2

Step 3: Simplifying the energy equation gives us:

21

3ML2ω2= 2mgL(1 −cos θ)

6ML2ω2= 2mgL(1 −cos θ)

ω=r12g(1 −cos θ)

Therefore, the angular velocity of the rod when it makes an angle θwith the

vertical is ω=q12g(1−cos θ)

Question 9

Question

A solid disk with a radius of 0.5 m and a mass of 2 kg is rotating about its

central axis with an angular acceleration of 4 rad/s2. At a certain instant, the

angular velocity of the disk is 5 rad/s. What is the total kinetic energy of the

rotating disk at this instant?

Solution

Step 1: We first find the moment of inertia of the disk using the formula for a

solid disk:

I=1

2mr2

where m= 2 kg (mass of the disk) and r= 0.5 m (radius of the disk). Substitute

these values to get

I=1

2×2 kg ×(0.5 m)2= 0.25 kg m2

Step 2: Now, we can find the kinetic energy of the disk at the given instant

using the formula for rotational kinetic energy:

KE =1

2Iω2

where ω= 5 rad/s (angular velocity of the disk). Substitute the values of Iand

ωto find

KE =1

2×0.25 kg m2×(5 rad/s)2= 6.25 J

Step 3: Therefore, the total kinetic energy of the rotating disk at the given

instant is 6.25 Joules.

Question 10

Question

A thin, uniform rod of length Land mass Mis free to rotate about a horizontal

axis passing through one end. Initially, the rod is at rest. A small object of

mass mtraveling horizontally with speed vcollides with the free end of the rod

and sticks to it. Find the final angular velocity of the system.

Solution

Step 1: Find the initial angular momentum of the system. The initial angular

momentum of the system is given by the equation:

Li=Iω

where Iis the moment of inertia and ωis the angular velocity. The moment of

inertia Iof the rod about the end where it can rotate is 1

3ML2. So,

Li=1

3ML2×0 = 0

Step 2: Find the final moment of inertia of the system. After the collision,

the system consists of a rod with mass M+mand length Lrotating about one

end. The moment of inertia of the combined system is 1

3(M+m)L2.

Step 3: Apply conservation of angular momentum. By conservation of an-

gular momentum, the initial angular momentum equals the final angular mo-

mentum.

Lf=Ifωf

0 = 1

3(M+m)L2×ωf

Step 4: Find the final angular velocity of the system. Solving for ωf, we get:

ωf= 0

Therefore, the final angular velocity of the system is 0.

Question 11

Question

A disk with radius Rstarts from rest and accelerates uniformly for 5 seconds

until it reaches an angular speed of 100 rad/s. After that, the disk decelerates

uniformly until it comes to a stop after rotating through one full revolution.

Calculate the angular acceleration when the disk is decelerating.

Solution

Step 1: Find the initial angular speed ωiof the disk when it starts accelerating.

Given that the disk starts from rest, ωi= 0.

Step 2: Find the angular acceleration αduring acceleration phase.

Using the equation of motion for rotational accelerated motion:

ωf=ωi+αt

Substitute the known values: ωf= 100 rad/s, ωi= 0, t= 5 s.

100 = 0 + α×5

α= 20 rad/s2

Step 3: Find the final angular speed ωfafter deceleration phase.

When the disk completes one full revolution, its final angular speed is 0 rad/s.

Step 4: Find the angular deceleration αdduring deceleration phase.

Using the equation of motion for rotational decelerated motion:

ωf=ωi+αdt

Substitute the known values: ωf= 0 rad/s, ωi= 100 rad/s, t= total time

taken for one revolution.

αd=ωf−ωi

αd=0−100

Step 5: Find the total time Ttaken for the disk to complete one revolution.

The total time for completing one revolution is the sum of time taken for accel-

eration and time taken for deceleration.

T= Time for acceleration + Time for deceleration

T= 5 + 100

αd

Step 6: Solve for αd.

Substitute the expression for T,

αd=0−100

5 + 100

αd

Step 7: Rearrange the equation to solve for αd.

αd=−100

5 + 100

αd

Thus, the angular acceleration when the disk is decelerating is −100

5+ 100

αd

rad/s

Question 12

Question

A disk with radius Rand moment of inertia Iabout its center is initially at

rest. A constant force 

Fis applied tangentially at the edge of the disk. After

the force has been applied for a time t, the disk acquires an angular velocity ω.

Find the angular acceleration of the disk in terms of I,R,F, and ω.

Solution

1. Let’s start by writing down the torque equation for the given situation. The

torque applied by the force 

Fat the edge of the disk is given by:

τ=RF sin θ

Here, θis the angle between 

Fand the radius Rof the disk. Since the force is

tangential, θ= 90◦, and sin θ= 1. So the torque simplifies to:

τ=RF

2. The torque applied to an object is related to its angular acceleration α

by the equation:

τ=Iα

Substitute the expression for torque into this equation:

RF =Iα

3. We also know that the angular acceleration αis related to the angular

velocity ωand the time tby:

ω=αt

Solving for α, we get:

α=ω

4. Substitute the expression for αinto our equation for torque:

RF =Iω

t

5. Finally, solve for the angular acceleration α:

α=RF t

Therefore, the angular acceleration of the disk in terms of I,R,F, and ωis

RF t

Question 13

Question

A wheel starts from rest and rotates with a constant angular acceleration of 2

rad/s2. After 4 seconds, what is the angular velocity of the wheel?

Solution

Step 1: Use the equation for angular velocity under constant angular accelera-

tion:

ω=ω0+αt

where



ωis the final angular velocity,



ω0is the initial angular velocity (in this case, 0 since the wheel starts from

rest),



αis the angular acceleration (given as 2 rad/s2), and



tis the time (given as 4 seconds).

Step 2: Substitute the given values into the equation:

ω= 0 + 2 ×4 = 8 rad/s

Therefore, after 4 seconds, the angular velocity of the wheel is 8 rad/s.

Question 14

Question

A solid sphere of radius Rand mass Mrolls without slipping down a ramp

inclined at an angle θ. The sphere is released from rest at the top of the ramp.

What is the linear speed of the center of the sphere when it reaches the bottom

of the ramp?

Solution

Let’s denote the linear speed of the center of the sphere as v, the radius as R,

the mass as M, and the angle of inclination as θ. The moment of inertia of a

solid sphere about its center is 2

5MR2.

Step 1: Find the acceleration of the sphere down the ramp using energy con-

siderations. The initial potential energy is converted into translational kinetic

energy and rotational kinetic energy at the bottom of the ramp.

Initial potential energy = Final kinetic energy

Mgh =1

2Mv2+1

2Iω2

Since the sphere rolls without slipping, ω=v

MgR sin θ=1

2Mv2+1

22

5MR2v

R2

gR sin θ=1

2v2+1

5v2

v=r10

7gR sin θ

Therefore, the linear speed of the center of the sphere when it reaches the

bottom of the ramp is r10

7gR sin θ.

Question 15

Question

A spherical object of radius Rstarts from rest at the top of a frictionless incline

that makes an angle of θwith the horizontal. The object rolls without slipping

down the incline. What is the linear acceleration of the center of mass of the

object in terms of gand θ?

Solution

Let’s denote the linear acceleration of the center of mass as a.

Step 1: The acceleration of a rolling object involves both translational and

rotational components. Considering the object is rolling without slipping, the

relationship between the linear acceleration and angular acceleration is given by

a=Rα, where αis the angular acceleration.

Step 2: To find the angular acceleration α, we can use the torque equation.

The torque about the center of mass due to gravity is τ=−mgR sin(θ), where

the negative sign indicates the direction opposite to the rotation.

Step 3: The net torque acting on the object is equal to Iα, where Iis the

moment of inertia of the object. For a solid sphere rolling down an incline, the

moment of inertia about an axis through its center and perpendicular to the

incline is I=2

5mR2.

Step 4: Equating the torque equation with the net torque equation, we

have −mgR sin(θ) = 2

5mR2α. Solving for α, we get α=−5gsin(θ)

2R.

Step 5: Now, substituting αback into the relationship between linear and

angular acceleration, we find a=R−5gsin(θ)

2R=−5gsin(θ)

Step 6: Therefore, the linear acceleration of the center of mass of the object

rolling down the incline is a=−5gsin(θ)

Question 16

Question

A thin rod of length Land mass Mis rotating about a fixed axis passing through

one of its ends with an angular velocity ω0. At a certain moment, a point mass

mis attached to the free end of the rod. What is the new angular velocity of

the system if the linear speed of the point mass at that moment is v0?

Solution

1. To solve this problem, we will use the principle of conservation of angular

momentum. Initially, the angular momentum of the system is given by:

Linitial =Iω0

where Iis the moment of inertia of the rod about the fixed axis.

2. When the mass mis attached to the free end of the rod, the new angular

momentum is given by:

Lfinal = (I+mL2)ωf

where ωfis the final angular velocity of the system.

3. Since angular momentum is conserved, we have:

Iω0= (I+mL2)ωf

4. The moment of inertia of the rod about the fixed axis is I=1

3ML2.

Substituting this into the conservation equation, we get:

3ML2ω0=1

3ML2+mL2ωf

5. Simplifying the expression, we find the new angular velocity ωfto be:

ωf=

3ML2ω0

3ML2+mL2

6. Since the linear velocity v0of the mass mis related to the angular velocity

by v=ωfL, we have:

v0=ωfL

7. Substituting the expression for ωfinto the linear velocity equation, we

can solve for the final angular velocity ωf.

Question 17

Question

A disc of radius Ris initially at rest when a constant torque τis applied to it.

The torque is then removed when the disc has rotated through an angle θ. If

the moment of inertia of the disc about its center of mass is I, determine the

angular velocity of the disc as a function of the angle rotated, ω(θ).

Solution

Step 1: By Newton’s second law for rotation, the torque applied can be related

to the angular acceleration αas τ=Iα.

Step 2: We know that α=dω

dt , and ω=dθ

dt . So, we have τ=Id2θ

dt2.

Step 3: Since τis a constant, we can write τ=Id2θ

dt2as τ=Id2θ

dθ2

dθ

dt .

Step 4: Rearranging the terms and solving the differential equation gives

dω

dθ =τ

Step 5: Integrating both sides with respect to θgives ω=τ

Iθ+C, where C

is a constant of integration.

Therefore, the angular velocity of the disc as a function of rotated angle is

given by ω(θ) = τ

Iθ+C.

Question 18

Question

A wheel initially at rest starts rotating with a constant angular acceleration of

2.0 rad/s2. Through what angle does the wheel turn in 5.0 s?

Solution

Step 1: The formula for angular displacement under constant angular accelera-

tion is given by

θ=θ0+ω0t+1

2αt2

where θis the final angular position, θ0is the initial angular position, ω0is the

initial angular velocity, αis the angular acceleration, and tis the time taken.

Step 2: Since the wheel is initially at rest, ω0= 0 and θ0= 0. We can

substitute the given values α= 2.0 rad/s2and t= 5.0 s into the formula to find

the angular displacement.

Step 3: Substituting the values into the formula gives

θ=0+0+ 1

2×2.0×(5.0)2

Step 4: Calculating the expression gives

θ=0+0+ 1

2×2.0×25

Step 5: Therefore, the angular displacement of the wheel in 5.0 s is

θ= 25 rad

Step 6: So, the wheel turns through 25 radians in 5.0 seconds under a

constant angular acceleration of 2.0 rad/s2.

Question 19

Question

A wheel initially at rest accelerates uniformly for 4 seconds before coming to a

stop in 6 seconds. If the wheel makes 6 rotations during this time, determine

the angular acceleration of the wheel.

Solution

Step 1: Find the initial angular velocity of the wheel.

Given that the wheel initially at rest, we have ωi= 0.

Step 2: Find the final angular velocity of the wheel.

Using the kinematic equation for rotational motion:

ωf=ωi+αt

where ωfis the final angular velocity, ωiis the initial angular velocity, αis

the angular acceleration, and tis the time.

Since the wheel comes to a stop, ωf= 0, and ωi= 0.

Step 3: Find the angular acceleration of the wheel.

Using ωf=ωi+αt again, but this time for the second part when the wheel

accelerates, we have:

ωf=ωi+αt

Substitute the values we know: ωf= 0, ωi= 0, t= 4 s.

0 = 0 + α×4

α= 0 rad/s2

Step 4: Calculate the total angular displacement of the wheel.

Since the wheel makes 6 rotations, it completes 6 ×2πrad of angular dis-

placement.

Step 5: Find the average angular speed of the wheel.

The average angular speed is given by:

¯ω=∆θ

∆t

where ∆θis the total angular displacement and ∆tis the total time taken.

¯ω=6×2π

6+4

¯ω=12π

10 =6π

5rad/s

Step 6: Find the angular acceleration.

Using the relationship between angular acceleration, angular velocity, and

time:

¯ω=1

2(ωi+ωf)

ωf= 2¯ω−ωi

Since ωi= 0, we have:

ωf= 2¯ω

ωf= 2 ×6π

5=12π

5rad/s

The angular acceleration αcan be found using the formula:

α=ωf−ωi

Substitute the values: ωf=12π

5rad/s, ωi= 0 rad/s, t= 6 s.

α=

12π

5−0

α=12π

α=2π

5rad/s2

Therefore, the angular acceleration of the wheel is 2π

5rad/s2.

Question 20

Question

A thin rod of length Land mass Mis rotating about an axis passing through

one end with an angular velocity ω. The rod is abruptly stopped in t0seconds.

Find the angular acceleration of the rod.

Solution

Step 1: Let’s denote the angular acceleration of the rod as α. Since the rod

is initially rotating with an angular velocity ωand comes to a stop in time t0,

we can relate the angular acceleration, angular velocity, and time through the

equation:

ω=αt0

Step 2: Next, we need to consider the torque acting on the rod to find the

angular acceleration. The only torque acting on the rod is due to the stopping

force, which exerts a torque about the axis of rotation. The torque τdue to

this force is given by:

τ=Iα

where Iis the moment of inertia of the rod about the axis of rotation.

Step 3: The moment of inertia of a thin rod rotating about one end is given

by:

I=1

3ML2

Step 4: Since the torque τis also equal to the rate of change of angular

momentum, we have:

τ=dL

Step 5: Here, the initial angular momentum of the rod is Li=Iω =1

3ML2ω,

and the final angular momentum is Lf= 0 (since the rod comes to a stop). So,

the change in angular momentum is ∆L=Lf−Li=−1

3ML2ω.

Step 6: Setting the torque τequal to the change in angular momentum ∆L

and substituting the expressions for τand I, we get:

3ML2α=−1

3ML2ω

Step 7: Simplifying the above equation, we find the angular acceleration α:

α=−ω

Therefore, the angular acceleration of the rod when abruptly stopped is −ω.

Question 21

Question

A disk of radius Rand mass Mis rotating about its central axis with an angular

velocity ω0(initially). A torque τis applied to the disk until it comes to a stop

in a time interval T. Determine the angular acceleration of the disk as a function

of time during this deceleration.

Solution

1. The torque applied to an object is related to its angular acceleration according

to the equation τ=Iα, where τis the torque, Iis the rotational inertia, and α

is the angular acceleration.

2. The rotational inertia of a solid disk rotating about its central axis is

I=1

2MR2.

3. Since the disk is coming to a stop, the final angular velocity will be 0,

and the initial angular velocity is ωi=ω0.

4. The angular acceleration αcan be determined using the equation of

rotational kinematics: ωf=ωi+αt, where ωfis the final angular velocity.

5. Since ωf= 0, we have 0 = ω0+αT .

6. Solving for α, we find α=−ω0

7. Therefore, the angular acceleration of the disk as a function of time during

deceleration is α(t) = −ω0

Tt.

Question 22

Question

A solid cylinder of radius Rand mass Mis initially at rest on a horizontal

surface. A constant horizontal force Fis applied to the cylinder, causing it to

roll without slipping. If the force is applied at a distance rfrom the center of

the cylinder, determine the speed of the cylinder once it has rolled a distance d.

Solution

Step 1: Find the moment of inertia of the cylinder. The moment of inertia of a

solid cylinder of radius Rand mass Mabout its central axis is given by:

I=1

2MR2

Step 2: Write the equation for the work done on the cylinder. The work

done on the cylinder is equal to the change in kinetic energy. The work done

by the force Fthat acts at a distance rcan be written as:

W=F·d

Step 3: Calculate the work done by the force F. The work done by the force

Fcan also be expressed in terms of the torque produced by the force about the

center of the cylinder:

W=τ·θ

where the angle θthrough which the force acts is related to the distance dby

d=Rθ.

Step 4: Determine the torque produced by the force F. The torque produced

by the force Fabout the center of the cylinder is given by:

τ=F·r

Step 5: Set the work done by the force equal to the change in kinetic energy.

Equating the work done by the force to the change in kinetic energy gives:

F·d=1

2Iω2

where ωis the angular velocity of the cylinder.

Step 6: Write the relationship between linear and angular velocity. Since

the cylinder is rolling without slipping, the linear velocity vis related to the

angular velocity ωby:

v=Rω

Step 7: Solve for the angular velocity of the cylinder. Substitute I,d, and

vinto the equation from step 5, and use the relationship between linear and

angular velocity to solve for ω:

F d =1

21

2MR2v

R2

Step 8: Calculate the speed of the cylinder. Finally, substitute the value of

ωinto the relationship between linear and angular velocity to find the speed of

the cylinder when it has rolled a distance d:

v= 2 F d

M1/2

Question 23

Question

A wheel initially at rest starts rotating with a constant angular acceleration

α= 0.5 rad/s2. At the same time, a bug located at the edge of the wheel starts

walking along the edge with a constant speed of 0.1 m/s. How long will it take

for the bug to reach the opposite side of the wheel, assuming the wheel has a

radius of 0.5 m?

Solution

Step 1: Let’s first find the angular velocity of the wheel at any time tusing the

equation for angular acceleration:

ω=ω0+αt

where ωis the final angular velocity, ω0is the initial angular velocity (which is

zero in this case), αis the angular acceleration, and tis the time. So, for our

case:

ω= 0 + 0.5t= 0.5trad/s

Step 2: Next, we need to find the angle through which the wheel has rotated

at time tusing the equation:

θ=θ0+ω0t+1

2αt2

where θis the final angle, θ0is the initial angle (which is zero in this case), ω0

is the initial angular velocity (which is zero), αis the angular acceleration, and

tis the time. Substitute the values into the equation:

θ=0+0+ 1

2×0.5t2= 0.25t2rad

Step 3: Since the bug is moving with a linear speed of 0.1 m/s along the

edge of the wheel, the bug will traverse a distance equal to the circumference

of the wheel. The circumference of the wheel is given by 2πr, where ris the

radius of the wheel. So, the time taken for the bug to cross to the other side is:

t=2πr

Substitute the given values:

t=2π×0.5

0.1= 10πs≈31.4 s

Therefore, it will take the bug approximately 31.4 seconds to reach the op-

posite side of the wheel.

Question 24

Question

A disc of radius Rstarts from rest and rolls without slipping down a frictionless

incline. Determine the linear speed of the center of mass of the disc when it has

descended a vertical distance h.

Solution

Let’s denote the linear speed of the center of mass of the disc as vcm.

Step 1: Calculate the angular speed of the disc when it has descended a

vertical distance h. Since the disc rolls without slipping, the linear speed of the

outer edge of the disc is equal to the radius times the angular speed: vedge =Rω,

where ωis the angular speed of the disc. The total mechanical energy of the

disc is conserved, so we have:

2Iω2+mgh =1

2mv2

edge

where Iis the moment of inertia of the disc, mis the mass of the disc, and g

is the acceleration due to gravity. Since the disc is a solid cylinder, I=1

2mR2.

Substituting these values in, we get:

4mR2ω2+mgh =1

2mR2ω2

Solving for ωgives:

ω=r8gh

Step 2: Calculate the linear speed of the center of mass. The linear speed

of the center of mass vcm can be related to the angular speed ωby:

vcm =Rω

Substitute in the expression for ωwe found in Step 1:

vcm =Rr8gh

3R=r8ghR

Therefore, the linear speed of the center of mass of the disc when it has

descended a vertical distance his q8ghR

Question 25

Question

A thin rod of length Land mass Mis rotating about one end with an angular

velocity ω. The rod is then rotated about its center at the same angular velocity.

Determine the kinetic energy of the rod after it is rotated about its center

compared to when it was rotated about one end.

Solution

Step 1: Let’s first calculate the initial kinetic energy of the rod when it is

rotating about one end.

The moment of inertia of a thin rod rotating about one end is given by I=

3ML2. The kinetic energy is given by KE =1

2Iω2. Substituting I=1

3ML2

into the expression for kinetic energy, we have:

KEinitial =1

2×1

3ML2×ω2=1

6Mω2L2

Step 2: Next, let’s calculate the kinetic energy of the rod after it is rotated

about its center.

When the rod is rotated about its center, the moment of inertia changes to

I=1

12 ML2. Therefore, the kinetic energy is given by KE =1

2Iω2. Substituting

I=1

12 ML2into the expression for kinetic energy, we have:

KEfinal =1

2×1

12ML2×ω2=1

24Mω2L2

Step 3: To determine the ratio of the final kinetic energy to the initial kinetic

energy, we divide the final kinetic energy by the initial kinetic energy:

KEfinal

KEinitial

24 Mω2L2

6Mω2L2=1

Therefore, the kinetic energy of the rod after it is rotated about its center is

4times the kinetic energy when it was rotated about one end.

Question 26

Question

A disk of radius 0.5 m is rotating about an axis perpendicular to the disk and

passing through its center. The angular position of a point on the rim of the

disk is given by θ= 2t2+ 3t, where θis in radians and tis in seconds. Calculate

the velocity of the point on the rim at t= 2 s.

Solution

Step 1: Find the angular velocity ωat t= 2 s. Given that θ= 2t2+ 3t, we first

need to find the angular velocity, ω, which is the derivative of θwith respect to

ω=dθ

dt =d

dt(2t2+ 3t) = 4t+ 3

Step 2: Calculate the angular velocity ωat t= 2 s. Substitute t= 2 into

the equation for ωto find

ω= 4(2) + 3 = 8 + 3 = 11 rad/s

Step 3: Calculate the linear velocity vat t= 2 s. The linear velocity vof

a point on the rim of the disk is related to the angular velocity ωby v=rω,

where ris the radius of the disk (0.5 m).

v=rω = 0.5×11 = 5.5 m/s

Therefore, the velocity of the point on the rim of the disk at t= 2 s is 5.5

m/s.

Question 27

Question

A disc of radius 0.25 m rotates about a fixed axis perpendicular to the disc

and through its center. The disc accelerates from rest and reaches an angular

velocity of 12 rad/s in 3 seconds. What is the angular acceleration of the disc?

Solution

Step 1: Identify the known quantities.

The initial angular velocity ω0= 0, the final angular velocity ω= 12 rad/s, and

the time t= 3 s.

Step 2: Calculate the angular acceleration using the formula for angular

acceleration.

The angular acceleration αcan be found using the equation:

α=∆ω

∆t

where ∆ω=ω−ω0and ∆t=t. Substitute the known values:

α=12 −0

3= 4 rad/s2

Step 3: State the final answer.

The angular acceleration of the disc is 4 rad/s2.

Question 28

Question

An object starts from rest and undergoes angular acceleration in a straight line.

The object reaches an angular velocity of 60 rad/s in a time of 4 seconds. If the

angular acceleration is constant, what is the angular acceleration of the object?

Solution

Step 1: Identify the given values The initial angular velocity, ω0, is 0 rad/s. The

final angular velocity, ω, is 60 rad/s. The time interval, t, is 4 seconds.

Step 2: Use the equation of rotational kinematics The equation of rotational

kinematics relating angular acceleration, initial angular velocity, final angular

velocity, and time interval is:

ω=ω0+α·t

where ωis the final angular velocity, ω0is the initial angular velocity, αis the

angular acceleration, and tis the time interval.

Step 3: Substitute the given values into the equation

60 = 0 + α·4

Step 4: Solve for the angular acceleration, α

α=60

4= 15 rad/s2

Therefore, the angular acceleration of the object is 15 rad/s2.

Question 29

Question

A thin uniform rod of length Land mass Mis free to rotate about a frictionless

pivot at one end. The rod is held horizontally and then released. What is the

angular acceleration of the rod when it makes an angle θwith the vertical?

Solution

To solve this problem, we will apply the principles of rotational motion and use

torque to calculate the angular acceleration of the rod.

Step 1: First, let’s identify the forces acting on the rod when it is at an

angle θwith the vertical. The forces acting on the rod are the gravitational

force mg acting at the center of mass, and the normal force acting at the pivot

point.

Step 2: Next, we need to calculate the torque produced by the gravitational

force. The torque τproduced by a force Fat a distance rfrom the pivot point

is given by τ=F·r·sin(ϕ), where ϕis the angle between the force vector and

the position vector.

For the gravitational force mg, the torque about the pivot point is τ=

(mg)·L

2·sin(θ).

Step 3: Now, we can apply Newton’s second law for rotational motion,

which states that the net torque is equal to the moment of inertia Itimes the

angular acceleration α:Pτ=I·α.

Substitute the expression for torque and moment of inertia I=1

3ML2into

the equation. We have (mg)·L

2·sin(θ) = 1

3ML2·α.

Step 4: Solve for the angular acceleration, α. We find α=3g

2Lsin(θ).

Thus, the angular acceleration of the rod when it makes an angle θwith the

vertical is α=3g

2Lsin(θ).

Question 30

Question

A solid disk of radius Rand mass Mis rotating about its axis with an angular

velocity ω0. A small block with mass mis placed on the disk at a distance r

from the center. The coefficient of static friction between the block and the disk

is µ. What is the maximum value of rfor which the block stays on the disk as

it slows down due to friction?

Solution

1. Let’s first determine the condition for the block to not slip off the disk. The

maximum static frictional force fmax is given by

fmax =µN,

where Nis the normal force acting on the block. The normal force can be

resolved into components perpendicular and parallel to the disk’s surface. The

perpendicular component balances the block’s weight mg, giving N=mg.

2. Therefore, the maximum frictional force fmax =µmg must provide the

required centripetal force when ris at its maximum value in order for the block

to stay on the disk.

3. The centripetal force required for the block to stay on the disk is provided

by the frictional force:

fmax =mv2

where vis the tangential velocity of the block.

4. The tangential velocity of the block can be expressed in terms of the

angular velocity of the disk. As the disk slows down due to friction, the angular

velocity decreases, and the block’s velocity along the disk decreases accordingly:

v=r(ω−αt),

where αis the angular acceleration of the disk.

5. The frictional force when the block is at its maximum distance from the

center is

µmg =m(r(ω−αt)2)/r.

6. Simplifying the above equation, we get

µg = (m/r)(ω2−2ωαt).

7. At the moment when the block slips off the disk, the frictional force

reaches its maximum value µmg. Substituting fmax =µmg into the equation

above, we have

µg = (m/r)(ω2

0−2αt).

8. Before the block slips off the disk, the tangential velocity of the block is

zero. At this moment, the disk has slowed down to angular velocity ωf, which

can be calculated using the kinematic equation

ωf=ω0−αt.

9. Substituting ωf= 0 and ω0−αt into the equation found in step 7 yields

the maximum value of r:

r=1

µg m

ω2

0.

Therefore, the maximum value of rfor which the block stays on the disk as

it slows down due to friction is m

µgω2

Question 31

Question

A thin rod of length Land mass Mis pivoted at one end and is allowed to swing

like a pendulum. Initially, the rod is at rest in a horizontal position. What is

the angular velocity of the rod when it is vertical?

Solution

Given: Length of the rod, LMass of the rod, M

We want to find the angular velocity of the rod when it is vertical.

Step 1: Let’s calculate the potential energy of the rod when it is horizontal.

When the rod is horizontal, the center of mass is located at a distance L/2

from the pivot point. The potential energy can be calculated using the formula

U=mgh, where mis the mass of the portion of the rod above the pivot point,

gis the acceleration due to gravity, and his the vertical height of the center of

mass above the lowest point. The potential energy when the rod is horizontal

is given by:

Uhorizontal =Mg L

2

Step 2: Now, let’s calculate the potential energy of the rod when it is

vertical. When the rod is vertical, the full length Lof the rod is below the pivot

point so the potential energy can be calculated using the formula U=mgh.

The potential energy when the rod is vertical is given by:

Uvertical =MgL

Step 3: The change in potential energy is equal to the kinetic energy ac-

quired by the rod when it reaches the vertical position. So, we can equate the

two expressions for potential energy and solve for the rotational kinetic energy.

Uhorizontal −Uvertical =1

2Iω2

where I=1

3ML2is the moment of inertia of the rod about the pivot.

Substitute the expressions for potential energy and moment of inertia into the

equation above.

Step 4: Simplify the equation and solve for ω, the angular velocity of the

rod when it reaches the vertical position.

Mg L

2−MgL =1

21

3ML2ω2

Solving for ωgives:

ω=r3g

So, the angular velocity of the rod when it is vertical is q3g

2L.

Question 32

Question

A disk of radius 0.2 m starts from rest and accelerates with a constant angular

acceleration of 0.5rad/s2for 5 seconds.

1. What is the angular velocity of the disk after 5 seconds?

2. How many revolutions has the disk made during this time?

3. What is the linear velocity of a point on the rim of the disk after 5 seconds?

Solution

1. To find the angular velocity of the disk after 5 seconds, we can use the

equation for angular velocity:

ω=ω0+αt

where: - ωis the final angular velocity, - ω0is the initial angular velocity (0

since the disk starts from rest), - αis the angular acceleration, - tis the time.

Step 1: Calculate the final angular velocity ω.

ω= 0 + 0.5×5=2.5rad/s

2. To find how many revolutions the disk has made during this time, we

need to first calculate the total angle rotated by the disk in 5 seconds. This can

be done using the equation:

θ=ω0t+1

2αt2

where: - θis the total angular displacement, - ω0is the initial angular velocity

(0), - αis the angular acceleration, - tis the time.

Step 2: Calculate the total angular displacement θ.

θ= 0 + 0.5×1

2×52= 6.25 rad

Since one revolution is equivalent to 2πradians, the number of revolutions

is:

Number of revolutions = 6.25 rad

2π rad/rev ≈0.995 rev

3. Lastly, to find the linear velocity of a point on the rim of the disk after 5

seconds, we use the equation:

v=rω

where: - vis the linear velocity, - ris the radius of the disk, - ωis the final

angular velocity.

Step 3: Calculate the linear velocity v.

v= 0.2×2.5 = 0.5m/s

So, after 5 seconds, the angular velocity of the disk is 2.5rad/s, the disk has

completed approximately 0.995 revolutions, and a point on the rim of the disk

has a linear velocity of 0.5m/s.

Question 33

Question

A disk with a radius of 0.5 m is spinning counterclockwise with an angular speed

of 4 rad/s. A small bug starts at the center of the disk and walks outwards along

the rim of the disk at a constant speed of 0.1 m/s. How far has the bug traveled

along the rim of the disk when it reaches the edge?

Solution

Step 1: Find the angular speed of the bug. The bug is moving tangentially

along the edge of the disk, so its linear speed is equal to the angular speed times

the radius of the disk. Given: Angular speed of disk, ωdisk = 4 rad/s Radius of

disk, r= 0.5 m

The bug’s angular speed, ωbug, is given by:

ωbug =linear speed of bug

radius of disk =0.1

0.5= 0.2 rad/s

Step 2: Find the time taken for the bug to reach the edge. The bug will

reach the edge when it has traveled an angle equivalent to the angle subtended

by the radius of the disk:

θ=distance traveled

radius of disk =2πr

r= 2πrad

The time taken for the bug to reach the edge is given by:

t=θ

ωbug

=2π

0.2= 10πs

Step 3: Calculate the distance traveled by the bug. The distance traveled

by the bug is given by the linear speed of the bug multiplied by the time taken:

Distance = linear speed ×time taken = 0.1×10π= 10πm

Therefore, the bug has traveled 10πmeters along the rim of the disk when

it reaches the edge.

Question 34

Question

A disk of radius 0.2 m is rotating with an angular velocity of 5 rad/s. A bug is

initially at rest on the rim of the disk. What is the bug’s tangential velocity 2

seconds after the bug starts crawling towards the center of the disk?

Solution

Step 1: The bug’s initial radial distance from the center of the disk, rinitial, is

equal to the radius of the disk, which is 0.2 m.

Step 2: The bug moves radially towards the center of the disk with a constant

radial velocity. The bug’s final radial distance from the center of the disk, rfinal,

decreases as it moves inwards. The time taken by the bug to reach the center

of the disk is 2 seconds.

Step 3: The bug’s tangential velocity vtis given by the equation:

vt=r·ω

where ris the radial distance from the center (in this case, rfinal), and ωis the

angular velocity of the disk.

Step 4: To find rfinal, we can use the fact that the bug moves at a constant

radial velocity. The bug’s radial velocity can be calculated using the formula:

vr=rinitial −rfinal

time

Step 5: Substituting the known values into the formula for vr:

vr=0.2 m −rfinal

2 s

Step 6: Since vris constant, we can also write:

vr=rfinal ·α

where αis the angular velocity of the bug.

Step 7: Equating the expressions for vr:

0.2 m −rfinal

2 s =rfinal ·α

Step 8: Solving for αgives:

α=0.2 m

2 s + 2 ·rfinal

Step 9: Since α=ω, we know that:

vt=rfinal ·ω=rfinal ·α=rfinal ·0.2 m

2 s + 2 ·rfinal

Step 10: Therefore, the bug’s tangential velocity 2 seconds after it starts

crawling towards the center of the disk is 0.2·rfinal

2+2·rfinal .

Question 35

Question

A disk of radius Ris rotating about its axis with a constant angular acceleration

of α. At time t= 0, the disk has an angular velocity of ω0and is at an angular

position of θ0. Find the angular position of a point on the rim of the disk at

time t.

Solution

Step 1: To find the angular position of a point on the rim of the disk at time t,

we can use the rotational kinematics equation:

θ=θ0+ω0t+1

2αt2

Step 2: Substituting the given values into the equation, we have:

θ=θ0+ω0t+1

2αt2

Step 3: Since we are looking for the angular position of a point on the rim

of the disk, the angular position will be θ= 2π. This is because the point on

the rim travels around a full circle.

2π=θ0+ω0t+1

2αt2

Step 4: Rearranging the equation to solve for t, we have:

αt2+ 2ω0t+ 2π−θ0= 0

Step 5: We can solve this quadratic equation for time tusing the quadratic

formula:

t=−b±√b2−4ac

where a=α,b= 2ω0, and c= 2π−θ0.

Step 6: Once we find the values of t, we can substitute them back into the

equation to determine the angular position of the point on the rim of the disk

at time t.

PHYS 305 - INTRODUCTION TO MODERN PHYSICS Rotational kinematics Question Bank Set 6

Students also viewed