PHYS 305 - INTRODUCTION TO MODERN PHYSICS Rotational kinematics Question Bank Set 9 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Rotational

kinematics

Question Bank - Set 9

Liberty University

Question 1

Question

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

passing through one end. Initially at rest, a constant force Fis applied perpen-

dicular to the other end of the rod for a time t. Find the angular speed of the

rod immediately after the force is removed.

Solution

Step 1: Determine the torque exerted by the force on the rod. The torque

exerted on the rod is given by τ=F r, where ris the distance from the axis

of rotation to the point where the force is applied. Since the force is applied

perpendicular to the rod at a distance L, we have r=L. Thus, τ=F L.

Step 2: Determine the angular acceleration of the rod. From Newton’s

second law for rotation, we have τ=Iα, where Iis the moment of inertia of

the rod and αis the angular acceleration. For a thin rod rotating about one

end, the moment of inertia is I=1

3ML2. Substituting these values into the

equation, we get F L =1

3ML2α. Solving for α, we find α=3F

ML .

Step 3: Determine the angular speed of the rod. The final angular speed

ωfof the rod can be found using the kinematic equation ωf=ω0+αt, where

ω0is the initial angular speed. Since the rod is initially at rest, ω0= 0. Thus,

ωf=αt =3F t

ML .

Question 2

Question

A solid sphere of radius Rand mass Mrolls without slipping down an inclined

plane that makes an angle θwith the horizontal. If the sphere starts from rest

at the top of the incline, determine its speed when it reaches the bottom of the

incline.

Solution

Step 1: The total kinetic energy of the system is the sum of the translational

kinetic energy and the rotational kinetic energy:

2mv2+1

2Iω2

where mis the mass of the sphere, vis its linear velocity, Iis the moment of

inertia of the sphere, and ωis its angular velocity.

Step 2: The linear velocity of the sphere can be related to its angular velocity

by v=Rω, where Ris the radius of the sphere.

Step 3: The moment of inertia of a solid sphere rotating about an axis

through its center and perpendicular to its surface is I=2

5MR2.

Step 4: We use the constraint of rolling without slipping, which means that

the linear velocity is related to the angular velocity by v=Rω.

Step 5: The potential energy at the top of the incline is P E =M gR sin θ,

where Mis the mass of the sphere, gis the acceleration due to gravity, and θ

is the angle of the incline.

Step 6: The total energy at the top is the sum of the potential energy and

the initial kinetic energy (which is zero since the sphere starts from rest).

Step 7: The total energy at the bottom is the sum of the kinetic energy

(translational and rotational).

Step 8: By conservation of energy, we equate the total energy at the top to

the total energy at the bottom:

MgR sin θ=1

2mv2+1

2Iω2

Step 9: Substitute v=Rω and I=2

5MR2into the equation and solve for

MgR sin θ=1

2mR2ω2+1

22

5MR2ω2

Step 10: Simplifying the equation gives:

MgR sin θ=7

10mR2+1

2mR2ω2

Step 11: The linear velocity vat the bottom of the incline is v=Rω. Solving

for vgives:

v=R7

10gsin θ1/2

Therefore, the speed of the sphere at the bottom of the incline is v=

R7

10 gsin θ1/2.

Question 3

Question

A particle is located on the rim of a wheel of radius 0.5 meters. The wheel starts

from rest and accelerates with a constant angular acceleration of 0.8 rad/s2for

4 seconds. At t= 4 seconds, what is the magnitude of the particle’s acceleration

and the angle it has turned through?

Solution

Step 1: To find the magnitude of the particle’s acceleration at t= 4 seconds,

we use the equation for angular acceleration α:

α= 0.8 rad/s2

Step 2: To find the linear acceleration aof the particle located on the rim

of the wheel, we use the equation:

a=r·α

a= 0.5 m ×0.8 rad/s2

a= 0.4 m/s2

Step 3: Therefore, the magnitude of the particle’s acceleration at t= 4

seconds is 0.4 m/s2.

Step 4: To find the angle the particle has turned through at t= 4 seconds,

we need to use the kinematic equation for angular displacement θ:

θ=θ0+ω0t+1

2αt2

Step 5: Since the wheel starts from rest, the initial angular velocity ω0= 0

and the initial angle θ0= 0. Plugging in the values, we get:

θ= 0 + 0 ×4 + 1

2×0.8×42

θ= 0 + 0 + 0.8×8

θ= 6.4 radians

Step 6: Therefore, at t= 4 seconds, the particle has turned through an angle

of 6.4 radians.

Question 4

Question

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

one end with an angular velocity ω. The rod is released at an angle θ0below

the horizontal. What is the angular velocity of the rod when it is vertical?

Solution

1. Let’s first find the initial angular momentum of the rod when it is released.

The initial angular momentum, Linitial, is given by:

Linitial =Iω

where Iis the moment of inertia of the rod and ωis the initial angular velocity.

For a rod rotating about one end, the moment of inertia is I=1

3ML2. So, we

have:

Linitial =1

3ML2·ω

2. Next, let’s find the final angular momentum of the rod when it is vertical.

When the rod is vertical, the moment of inertia is I=ML2/3. Let the final

angular velocity be ωfinal. Then, the final angular momentum Lfinal is given by:

Lfinal =1

3ML2·ωfinal

3. Conservation of angular momentum states that initial angular momentum

equals final angular momentum. Therefore, we have:

Linitial =Lfinal

3ML2·ω=1

3ML2·ωfinal

4. Solving for ωfinal, we find:

ωfinal =ω

5. Therefore, the angular velocity of the rod when it is vertical is the same

as the initial angular velocity, ω.

Question 5

Question

A solid cylinder of radius Rand mass Mrolls without slipping down an incline

of angle θfrom rest. Calculate the time it takes for the cylinder to reach

the bottom of the incline. Assume the cylinder starts from rest, the incline is

frictionless, and neglect air resistance.

Solution

Step 1: Firstly, we need to determine the acceleration of the cylinder as it rolls

down the incline. The net torque acting on the cylinder is due to the force

of gravity causing a torque M gR sin(θ) and the friction force causing a torque

−fR, where fis the frictional force and is zero in this case. The net torque is

equal to the moment of inertia of the cylinder times its angular acceleration α.

Xτ=Iα

MgR sin(θ) = 1

2MR2α

a=Rα

a=2

3gsin(θ)

Step 2: Now we can find the time it takes for the cylinder to reach the

bottom of the incline using the kinematic equation for rotational motion. The

final velocity of the cylinder when it reaches the bottom of the incline is given

by the equation:

v2=u2+ 2as

Where u= 0 (initial velocity), ais the acceleration found above, and sis the

distance traveled down the incline. Since the incline makes an angle θwith the

horizontal, the distance traveled is s=h

sin(θ)where his the height of the incline.

Given that h=Rcos(θ), we get s=Rcot(θ). So,

v2= 0 + 22

3gsin(θ)Rcot(θ)

v=r4

3gR sin(θ)

Step 3: Finally, we can find the time tit takes for the cylinder to reach the

bottom of the incline using the formula:

v=u+at

t=v

t=q4

3gR sin(θ)

3gsin(θ)

t= 2sR

Question 6

Question

A thin rod of length Land mass Mis free to rotate about one end. A force F

is applied perpendicular to the rod at a distance afrom the axis of rotation. If

the rod starts from rest, determine the angular acceleration of the rod.

Solution

Step 1: We will start by considering the torque acting on the rod. The torque

due to the force Fis given by τ=F·a.

Step 2: The moment of inertia of a rod rotating about one end is I=1

3ML2.

Step 3: Using Newton’s second law for rotational motion, τ=I·α, where

αis the angular acceleration.

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

the rotational analog of Newton’s second law, we get F·a=1

3ML2·α.

Step 5: Solving for α, we find α=3F·a

ML2.

Therefore, the angular acceleration of the rod is α=3F·a

ML2.

Question 7

Question

A rigid body starts from rest and accelerates uniformly for 5 seconds to reach

an angular velocity of 10 rad/s. If the body continues to rotate at this con-

stant angular velocity for an additional 10 seconds, what is the total angular

displacement of the body during this time period?

Solution

Step 1: Find the angular acceleration of the body during the first 5 seconds.

Given that the body starts from rest and reaches an angular velocity of 10 rad/s

in 5 seconds, we can use the equation for angular acceleration:

angular acceleration, α=change in angular velocity

time interval =10 rad/s −0 rad/s

5 s = 2 rad/s2

Step 2: Calculate the angular displacement during the first 5 seconds. Using

the equation for angular displacement under uniform acceleration:

θ1=1

2αt2=1

2×2 rad/s2×(5 s)2= 25 rad

Step 3: Determine the total angular displacement for the entire period of

rotation. Since the body rotates at a constant angular velocity of 10 rad/s for

the next 10 seconds, the angular displacement during this time is given by:

θ2= angular velocity ×time = 10 rad/s ×10 s = 100 rad

Step 4: Calculate the total angular displacement. The total angular displace-

ment is the sum of the angular displacements during the two time intervals:

Total angular displacement = θ1+θ2= 25 rad + 100 rad = 125 rad

Therefore, the total angular displacement of the body during the given time

period is 125 radians.

Question 8

Question

A disc of radius 0.2 m rotates about a fixed axis through its center with an

initial angular velocity of 5 rad/s. The angular acceleration of the disc is given

by α=−0.2trad/s2. Find the angular velocity of the disc after 4 seconds.

Solution

Step 1: Write the equation for angular velocity as a function of time.

The angular acceleration can be expressed as the derivative of angular ve-

locity with respect to time:

α=dω

Integrating both sides with respect to time, we get:

Zαdt =Zdω

dt dt

Z(−0.2t)dt =Zdω

−0.1t2+C1=ω

where C1is the constant of integration.

Step 2: Determine the value of the constant of integration.

Given that the initial angular velocity is 5 rad/s when t= 0, we can substi-

tute these values into the equation:

−0.1(0)2+C1= 5

C1= 5

Step 3: Substitute the value of C1back into the equation to find the angular

velocity at t= 4 s.

Substitute C1= 5 into the equation −0.1t2+C1=ω:

−0.1(4)2+ 5 = ω

−0.1(16) + 5 = ω

−1.6 + 5 = ω

ω= 3.4 rad/s

Therefore, the angular velocity of the disc after 4 seconds is 3.4 rad/s.

Question 9

Question

A solid cylinder of mass Mand radius Ris initially at rest on a horizontal

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

to roll without slipping. Calculate the time it takes for the cylinder to reach a

linear speed of v.

Solution

Step 1: First, we need to find the acceleration of the cylinder. The friction force

on the cylinder will provide the torque needed for the rotation and the force

needed for the linear acceleration. Since the cylinder rolls without slipping, we

have the relationship a=αR, where ais the linear acceleration and αis the

angular acceleration.

Step 2: The net torque on the cylinder is equal to F R −ffr, where ffis the

friction force. This torque is also equal to Iα, where Iis the moment of inertia

of the cylinder. For a solid cylinder rotating about its axis, I=1

2MR2.

Step 3: The friction force is equal to µN, where µis the coefficient of friction

and Nis the normal force. For a cylinder, N=Mg. Substituting these values

into our torque equation and equating it to Iα, we get F R −µM gR =1

2MR2α.

Step 4: Since the cylinder rolls without slipping, the linear acceleration a

is equal to the angular acceleration αR. Therefore, we can rewrite the above

equation in terms of a:F R −µMgR =1

2MRa.

Step 5: Given that F−ff=M a for the linear motion, we can substitute

the expression for ffto get F−µMg = (M+1

2M)a.

Step 6: Solving for the linear acceleration a, we find a=F−µMg

2M. Now that

we have the acceleration, we can use the kinematic equation v=u+at to find

the time tit takes for the cylinder to reach a linear speed of v. Since the cylinder

starts from rest, u= 0, so v=at implies t=v

a. Substituting the expression for

agives us the final answer.

Question 10

Question

A wheel initially at rest has an angular acceleration of 3.0 rad/s2. If the wheel

turns through 45 rad, what is its final angular velocity?

Solution

Step 1: Given parameters: The initial angular velocity, ωi, is 0 rad/s. The

angular acceleration, α, is 3.0 rad/s2. The final angle rotated, θ, is 45 rad.

Step 2: Using the equation for angular motion: We can use the relationship

between angular displacement, initial angular velocity, angular acceleration, and

final angular velocity:

ω2

f=ω2

i+ 2αθ

Step 3: Substituting the known values: Plugging in the values we have:

ω2

f= (0)2+ 2(3.0)(45)

ω2

f= 270

Step 4: Finding the final angular velocity: Taking the square root of both

sides:

ωf=√270 ≈16.43 rad/s

Step 5: Therefore, the final angular velocity of the wheel is approximately

16.43 rad/s.

Question 11

Question

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

2.0 rad/s2. If the wheel makes 10 complete revolutions, calculate the time it

takes for the wheel to make these revolutions.

Solution

Step 1: Identify the given values and the unknowns. Let abe the angular

acceleration of the wheel, which is 2.0 rad/s2, and let Nbe the number of

complete revolutions made by the wheel, which is 10. We are asked to find the

time tit takes for the wheel to make these revolutions.

Step 2: Convert the number of revolutions into radians. One complete revo-

lution is equivalent to 2πradians. Therefore, 10 complete revolutions correspond

to 10 ×2π= 20πradians.

Step 3: Use the equation for angular displacement with constant angular

acceleration. The equation to relate angular displacement, angular acceleration,

initial angular velocity, and time is:

θ=ωit+1

2at2

where θis the angular displacement, ωiis the initial angular velocity (which is

0 since the wheel starts at rest), ais the angular acceleration, and tis the time.

We can simplify this equation to:

θ=1

2at2

Substitute in the known values:

20π=1

2×2.0×t2

Step 4: Solve for the time t. Solve for tin the equation:

20π=t2

t=√20π≈6.32 s

Therefore, it takes approximately 6.32 seconds for the wheel to make 10

complete revolutions.

Question 12

Question

A thin hoop with radius Rand mass Mis rolling without slipping on a horizontal

surface. Initially, the hoop is at rest. A particle of mass mis moving horizontally

at speed vtoward the hoop. The particle collides with the hoop and sticks to

it. Find the angular velocity of the system just after the collision.

Solution

Let the initial speed of the particle be vand the angular velocity of the system

just after the collision be ω.

Step 1: Let’s first determine the initial angular momentum of the system.

Since the hoop is initially at rest, the initial angular momentum of the system

is solely due to the particle and is given by

Linitial =m·(R+ 0) ·v=mRv

Step 2: After the collision, the mass of the system is M+mand the new

angular velocity of the hoop-particle system is ω. The final angular momentum

of the system is given by

Lfinal = (M+m)R2·ω

Step 3: Since angular momentum is conserved in the absence of external

torques,

Linitial =Lfinal

mRv = (M+m)R2·ω

Step 4: Solving for ωgives

ω=mRv

(M+m)R2=mv

M+m

Question 13

Question

A solid sphere with radius Rand mass Mrolls without slipping down an inclined

plane making an angle θwith the horizontal. The sphere starts from rest at the

top of the incline. What is its linear acceleration after it has rolled a distance

hdown the incline?

Solution

Step 1: Finding the angular acceleration

The net torque on the sphere about its center of mass can be calculated as:

τ=Iα

where Iis the moment of inertia of the sphere about its center of mass, αis the

angular acceleration, and τis the net torque acting on the sphere. Since the

sphere is rolling without slipping, we have:

α=a

where ais the linear acceleration of the sphere and Ris its radius.

The net torque acting on the sphere is due to the force of gravity and the

normal force. The component of the gravitational force parallel to the incline

causes a torque:

τgravity =MgR sin θ

The torque due to the normal force is perpendicular to the radius of the sphere

and therefore does not contribute to the torque about the center of mass.

Setting the net torque equal to Iα, we have:

MgR sin θ=2

5MR2a

Which simplifies to:

a=5

2gsin θ

Step 2: Finding the linear acceleration

The linear acceleration of the sphere down the incline is the same as the

acceleration due to gravity along the incline since there is no slipping. Using

trigonometry, we can find the acceleration down the incline:

adown incline =gsin θ

Therefore, the linear acceleration of the sphere after it has rolled a distance

hdown the incline is a=5

2gsin θ.

Question 14

Question

A disk of radius 0.2 m starts from rest and undergoes an angular acceleration

of 4.0 rad/s2. What is the angular velocity of the disk after 3 seconds?

Solution

Step 1: We can use the equation for angular motion to find the angular ve-

locity. The equation relating angular velocity, initial angular velocity, angular

acceleration, and time is:

ωf=ωi+αt

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

angular acceleration, and tis the time.

Step 2: Given that the disk starts from rest, the initial angular velocity is

0. Thus, we have:

ωf= 0 + (4.0 rad/s2)(3 s)

Step 3: Calculating the final angular velocity:

ωf= 12 rad/s

Step 4: Therefore, the angular velocity of the disk after 3 seconds is 12 rad/s .

Question 15

Question

A disk with a radius of 0.5 meters starts from rest and accelerates with a constant

angular acceleration of 2 rad/s2. Find the angular velocity of the disk after it

has rotated through an angle of π/3 radians.

Solution

Step 1: Determine the angular velocity using the kinematic equation for rota-

tional motion:

ω2

f=ω2

i+ 2αθ

where - ωiis the initial angular velocity (0 rad/s, since the disk starts from

rest), - αis the angular acceleration (2 rad/s2), and - θis the angle rotated

through (π/3 radians). Thus,

ω2

f= 0 + 2(2)(π/3)

ω2

f= 8π/3

ωf=p8π/3

ωf=2√6π

3rad/s

Therefore, the angular velocity of the disk after rotating through π/3 radians

is 2√6π

3rad/s.

Question 16

Question

A uniform disk of radius Rand mass Mstarts from rest and rolls down a

frictionless incline that makes an angle θwith the horizontal. What is the

linear speed of the center of the disk when it reaches the bottom of the incline?

Solution

Step 1: Start by calculating the moment of inertia of the disk rolling down

the incline. The moment of inertia of a disk about its center is I=1

2MR2.

Using the parallel axis theorem, the moment of inertia about the center of

mass is Icm =I+Md2, where dis the perpendicular distance between the

center of mass and the rotation axis. For a disk, this distance is d=R

2. So,

Icm =1

2MR2+M(R

2)2=3

4MR2.

Step 2: Find the acceleration of the disk down the incline. The gravita-

tional force component down the incline is mg sin θ, which provides a torque,

causing angular acceleration. The torque is τ=Icmα, where αis the angular

acceleration. The angular acceleration is related to the linear acceleration aby

a=αR. So, τ=Icm a

R. The torque is also equal to Rmg sin θ, so we have

Rmg sin θ=3

4MR2a

R. Solving for a, we get a=4

3gsin θ.

Step 3: Calculate the linear speed at the bottom of the incline. At the

bottom, the linear speed vis related to the angular speed ωby v=Rω. Using

the kinematic equation v2=u2+ 2as, where uis the initial velocity (which is 0

in this case), ais the acceleration, and sis the linear distance traveled, we get

v=√2as. Substituting a=4

3gsin θand s=R(height of the incline), we find

v=q2·4

3gsin θ·R=q8

3gR sin θ.

Question 17

Question

A thin rod of length Land mass Mis rotating about an axis perpendicular to the

rod and passing through one end with an angular velocity ω. How much work

is required to stop the rod from rotating? Assume the rod’s mass is uniformly

distributed along its length.

Solution

1. The rotational kinetic energy of the rotating rod is given by

K=1

2Iω2

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

thin rod rotating about an axis perpendicular to the rod and passing through

one end, the moment of inertia is I=1

3ML2.

2. Substituting the moment of inertia into the equation for kinetic energy,

we have

K=1

21

3ML2ω2=1

6Mω2L2

3. To stop the rotating rod, we need to remove all of the kinetic energy.

Hence, the work required is equal to the initial kinetic energy of the system.

Therefore, the work done to stop the rod is

W=1

6Mω2L2

Question 18

Question

A disc of radius 0.2 m rotates with a constant angular acceleration of 2 rad/s2.

At t= 0, the angular velocity of the disc is 1 rad/s and the disc is rotating

counterclockwise. Find the angular velocity of the disc after 4 seconds.

Solution

Step 1: Determine the angular displacement of the disc after 4 seconds. Given

that the disc is rotating with a constant angular acceleration, the angular dis-

placement θcan be calculated using the equation:

θ=ωit+1

2αt2,

where ωiis the initial angular velocity, αis the angular acceleration, and tis

the time. Plugging in the values: ωi= 1 rad/s, α= 2 rad/s2, and t= 4 s, we

get:

θ= 1(4) + 1

2·2·42= 4 + 16 = 20 rad.

Step 2: Find the final angular velocity of the disc. The final angular velocity

ωfcan be calculated using the equation:

ωf=ωi+αt,

where ωiis the initial angular velocity, αis the angular acceleration, and tis

the time. Plugging in the values: ωi= 1 rad/s, α= 2 rad/s2, and t= 4 s, we

get:

ωf= 1 + 2 ·4 = 1 + 8 = 9 rad/s.

Therefore, the angular velocity of the disc after 4 seconds is 9 rad/s.

Question 19

Question

A wheel with an initial angular velocity of 10 rad/s is acted upon by a constant

angular acceleration of −2 rad/s2. How long will it take for the wheel to come

to a complete stop?

Solution

Step 1: Find the angular velocity at which the wheel stops.

Given: Initial angular velocity, ω0= 10 rad/s Angular acceleration, α=

−2 rad/s2

The final angular velocity, ωf, at which the wheel stops can be found using

the equation of rotational motion:

ωf=ω0+αt

Substitute the known values:

ωf= 10 rad/s −2t

Step 2: Solve for the time tat which the wheel stops.

When the wheel comes to a complete stop, ωf= 0. Therefore, we have:

0 = 10 rad/s −2t

Solving for t:

2t= 10 rad/s

t=10

2= 5 s

Therefore, it will take 5 seconds for the wheel to come to a complete stop.

Question 20

Question

A solid sphere of radius Rand mass Mstarts from rest and rolls without

slipping down an inclined plane making an angle θwith the horizontal. Find

the acceleration of the center of mass of the sphere.

Solution

Step 1: The forces acting on the sphere are the gravitational force Mg sin(θ)

down the plane, the normal force Nperpendicular to the plane (which provides

the necessary centripetal force for rolling without slipping), and the friction

force fup the plane.

Step 2: The net force acting on the sphere is:

f−Mg sin(θ) = Macm

Step 3: The torque about the center of mass causing the rotational acceler-

ation is due to the friction force. The torque equation is:

fR =Iα

Step 4: For a solid sphere of radius Rand mass M, the moment of inertia

I=2

5MR2. The rotational acceleration α=acm

Step 5: Substituting the moment of inertia Iand rotational acceleration α

into the torque equation gives:

fR =2

5MR2·acm

Step 6: Simplifying the equation above, we have:

f=2

5Macm

Step 7: Now we can substitute the expression for finto the net force equa-

tion: 2

5Macm −Mg sin(θ) = Macm

Step 8: Solving for the acceleration of the center of mass acm, we find:

acm =5

7gsin(θ)

Question 21

Question

A solid sphere of mass Mand radius Ris initially at rest on a horizontal

surface. It starts rolling without slipping due to an applied constant force 

acting horizontally on it from a certain distance above the ground. The sphere

reaches the ground with an angular velocity ω. What is the speed of the sphere’s

center of mass just before it hits the ground, in terms of M,R,g(acceleration

due to gravity), ω, and any other necessary constants?

Solution

Step 1: Find the acceleration of the center of mass of the sphere

The net torque acting on the sphere is due to the applied force 

F, which

only acts horizontally. The moment of inertia of a solid sphere about its center

is 2

5MR2. Thus, we can use the equation for rotational motion to find the

resultant torque:

τ=Iα

where

τ=F R =2

5MR2·α

Since the sphere is rolling without slipping, we have the relationship α=a

where ais the linear acceleration of the center of mass of the sphere. Therefore,

F R =2

5MR2·a

F=2

5Ma

The net force acting on the sphere is responsible for its acceleration. Therefore,

the net force is given by

F−mg =Ma

Substitute the expression for Fin terms of ainto the above equation:

5Ma −mg =Ma

5a−g=a

a=5

Step 2: Find the speed of the center of mass just before hitting the ground

The final speed vof the center of mass just before it hits the ground is given

by the equation of motion:

v2=u2+ 2as

where uis the initial speed of the center of mass (which is 0 in this case), sis

the distance traveled by the center of mass, and ais the acceleration we found

in Step 1. Rearranging the equation, we get:

v=√2as =r2·5

7g·R

Therefore, the speed of the sphere’s center of mass just before hitting the

ground is r10

7gR .

Question 22

Question

A solid sphere of radius Rrolls without slipping down an incline of angle θ. If

the sphere starts from rest, determine the linear acceleration of the center of

mass of the sphere after it has rolled a distance dalong the incline.

Solution

Step 1: The moment of inertia for a solid sphere rotating about an axis through

its center is given by I=2

5MR2, where Mis the mass of the sphere. Step 2:

The torque on the sphere is due to the gravitational force mg sin(θ) acting at

a distance Rfrom the center. The torque equation is τ=Iα =Rmg sin(θ).

Step 3: The linear acceleration aof the center of mass can be related to the

angular acceleration αusing the equation a=Rα since the sphere is rolling

without slipping. Step 4: We can relate aand αusing the kinematic equation

of rotational motion α=a

R. Step 5: Substitute τ=Iα into the torque equation

Rmg sin(θ) = 2

5MR2·a

R. Step 6: Simplifying the equation Rmg sin(θ) = 2

5Ma,

we can solve for ato find a=5

2gsin(θ), which is the linear acceleration of the

center of mass of the sphere after it has rolled a distance dalong the incline.

Question 23

Question

A wheel starts from rest and rotates with a constant angular acceleration of 2.5

rad/s2. If the diameter of the wheel is 60 cm, determine the time it takes for

the wheel to reach an angular speed of 10 rad/s.

Solution

Step 1: First, we need to find the angular acceleration in terms of the radius of

the wheel. Given that the diameter is 60 cm, the radius rcan be calculated as

r=60

2= 30 cm = 0.3 m.

Step 2: We know that the angular acceleration αis equal to at

r, where atis

the tangential acceleration and ris the radius of the wheel. So, we can rewrite

the given angular acceleration as: α= 2.5 rad/s2=at

0.3.

Step 3: Next, we need to find the tangential acceleration at. Since at=r·α,

we have at= 0.3×2.5 = 0.75 m/s2.

Step 4: To find the time it takes for the wheel to reach an angular speed

of 10 rad/s, we can use the equation ω=ω0+αt, where ωis the final angular

speed, ω0is the initial angular speed, αis the angular acceleration, and tis the

time.

Step 5: Setting ω= 10 rad/s, ω0= 0 rad/s, and α= 2.5 rad/s2, we get:

10 = 0 + 2.5t. Solving for t, we find that t=10

2.5= 4 seconds.

Therefore, it takes 4 seconds for the wheel to reach an angular speed of 10

rad/s.

Question 24

Question

A wheel of radius 2.5 m starts from rest and accelerates with a constant angular

acceleration of 0.5 rad/s2.

1. What is the angular speed of the wheel after 3 seconds?

2. How many revolutions has the wheel completed after 6 seconds?

Solution

1. Let’s first find the angular speed of the wheel after 3 seconds using the

equation:

ω=ω0+αt

where ωis the final angular speed, ω0is the initial angular speed (which is 0 in

this case), αis the angular acceleration, and tis the time.

ω= 0 + (0.5 rad/s2)(3 s)

ω= 1.5 rad/s

So, after 3 seconds, the angular speed of the wheel is 1.5 rad/s.

2. Next, we need to find the number of revolutions completed by the wheel

after 6 seconds. We can calculate this by first finding the total angle rotated by

the wheel in 6 seconds using the equation:

θ=ω0t+1

2αt2

where θis the total angle rotated, ω0is the initial angular speed, αis the angular

acceleration, and tis the time.

θ= 0 ×6 + 1

2×0.5×(6)2

θ= 9 rad

Now, we can find the number of revolutions completed by dividing the total

angle by 2π:

Number of revolutions = 9 rad

2π

Number of revolutions ≈1.43 rev

Therefore, after 6 seconds, the wheel has completed approximately 1.43 revolu-

tions.

Question 25

Question

A thin hoop of radius Rand mass Mrolls without slipping down an inclined

plane that makes an angle θwith the horizontal. The hoop starts from rest at

the top of the incline. What is the linear acceleration of the center of the hoop

after it has rolled a distance ddown the incline?

Solution

Step 1: First, we’ll find the angular acceleration of the hoop as it rolls down the

incline. Given that the hoop is rolling without slipping, the linear acceleration

aof the center of mass is related to the angular acceleration αby a=Rα.

Step 2: The torque about the center of mass of the hoop is caused by the

force of gravity. The torque τis given by τ=R·sin(θ)·mg =Iα, where Iis

the moment of inertia for a hoop of mass Mand radius Rabout its center of

mass. For a hoop, I=MR2.

Step 3: Solving for αin the torque equation, we have R·sin(θ)·mg =MR2α,

which simplifies to α=R·sin(θ)·g

R= sin(θ)·g.

Step 4: Now, we can find the linear acceleration of the center of the hoop

using a=Rα. Substituting α= sin(θ)·g, we get a=R·sin(θ)·g.

Step 5: Finally, the linear acceleration of the center of the hoop after it has

rolled a distance ddown the incline is the component of aparallel to the incline.

This acceleration is aparallel =a·sin(θ) = R·sin2(θ)·g.

Question 26

Question

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

bug starts at the edge of the disk and crawls at a constant speed of 0.1 m/s

towards the center. How far has the bug traveled when it reaches the center of

the disk?

Solution

Step 1: The bug travels along a radius of the disk while the disk is rotating, so

its radial distance r(t) from the center as a function of time tis changing. We

can relate the angular and linear velocities using the equation v=ωr, where v

is the linear velocity, ωis the angular velocity, and ris the radius. Therefore,

the bug’s radial velocity at a distance rfrom the center is given by vradial =ωr.

Step 2: Since the bug crawls towards the center at a constant speed of

0.1 m/s, we have vradial = 0.1 m/s. Substituting in the given angular velocity

ω= 5 rad/s, we can find the bug’s radial distance at any point in time t.

Step 3: The bug’s radial velocity is constant, so we can write r(t) = vradialt.

Step 4: When the bug reaches the center of the disk, its radial distance r

will be equal to the initial radius r0= 0.2 m. So, we have r0=vradialt.

Step 5: Substituting in the given values, we have 0.2=0.1·t. Solving for t,

we find t= 2 s.

Step 6: Finally, we can find the distance the bug has traveled by the time it

reaches the center of the disk. The distance traveled is given by d=vradial ·t.

Step 7: Substituting in the known values, we get d= 0.1 m/s ·2 s = 0.2 m.

Therefore, the bug has traveled 0.2 meters when it reaches the center of the

disk.

Question 27

Question

A disk of radius Ris rotating with an angular velocity ω0. A bug is at rest on

the edge of the disk when it starts crawling towards the center with a constant

acceleration a. At what angular velocity will the bug be when it reaches the

center of the disk?

Solution

Step 1: The bug’s motion towards the center of the disk can be described in

terms of the conservation of angular momentum. Initially, the bug has zero

angular momentum, and finally, at the center of the disk, it will have some

angular momentum due to its linear velocity.

Step 2: Initially, the disk is rotating with angular velocity ω0and angular

momentum Iω0, where Iis the moment of inertia of the disk.

Step 3: The bug is crawling towards the center of the disk with a linear

acceleration a. At any point in time during its motion, the bug’s radial position

rfrom the center of the disk and linear velocity vcan be related to the angular

velocity ωas v=rω.

Step 4: Using conservation of angular momentum, we have Iω0=Iω +mvr,

where mis the mass of the bug.

Step 5: At the edge of the disk, r=Rand ω=ω0. At the center of the

disk, r= 0 and ω=ωf(final angular velocity of the bug).

Step 6: Substituting the initial and final conditions into the conservation of

angular momentum equation, we get Iω0=mRω0+ 0. Solving for ωf, we find

ωf=Iω0

mR .

Therefore, when the bug reaches the center of the disk, its angular velocity

will be ωf=Iω0

mR .

Question 28

Question

A solid sphere of radius Rand mass Mstarts from rest and rolls without slipping

down an incline plane that makes an angle θwith the horizontal. What is the

linear acceleration of the center of the sphere when it reaches the bottom of the

incline?

Solution

Step 1: Determine the moment of inertia of the solid sphere. The moment of

inertia of a solid sphere about its center is given by I=2

5MR2.

Step 2: Calculate the gravitational torque on the sphere. The torque due to

gravity about the center of the sphere is τ=mgR sin θ.

Step 3: Apply the rotational analogue of Newton’s second law. The net

torque acting on the sphere is equal to the moment of inertia times the angular

acceleration: τ=Iα, where αis the angular acceleration.

Step 4: Find the angular acceleration of the sphere. Substitute in the values:

mgR sin θ=2

5MR2α. Simplify to get α=5gsin θ

2R.

Step 5: Relate the linear and angular accelerations. For a rolling object, the

linear and angular accelerations are related by a=αR.

Step 6: Calculate the linear acceleration of the center of the sphere. Substi-

tute in the values: a=5gsin θ

2. Therefore, the linear acceleration of the center

of the sphere when it reaches the bottom of the incline is 5gsin θ

Question 29

Question

A disk rotates with a constant angular acceleration of 2.0 rad/s2. At time t= 0,

its angular velocity is 3.0 rad/s and its radius is 0.50 m. What is the angular

velocity of the disk after it has rotated through an angle θ= 4.0 rad?

Solution

Step 1: We can use the following kinematic equation relating angular velocity,

angular acceleration, and displacement:

ω2

f=ω2

i+ 2αθ

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

the angular acceleration, and - θis the angular displacement.

Step 2: Plugging in the given values, we have:

ω2

f= (3.0 rad/s)2+ 2(2.0 rad/s2)(4.0 rad)

Step 3: Calculating the right-hand side of the equation, we get:

ω2

f= 9.0 rad2/s2+ 16.0 rad2/s2

ω2

f= 25.0 rad2/s2

Step 4: Taking the square root of both sides to solve for ωf, we find:

ωf=√25.0 rad/s

ωf= 5.0 rad/s

Step 5: Therefore, the angular velocity of the disk after rotating through an

angle of 4.0 rad is 5.0 rad/s .

Question 30

Question

A wheel of radius 0.5 m starts from rest and has a constant angular acceleration

of 2 rad/s2.

What is the angular velocity of the wheel after 3 seconds?

Solution

Step 1: Write down the relevant equation for angular motion. The angular

velocity of an object rotating with a constant angular acceleration can be cal-

culated using the equation:

ωf=ωi+αt

where: ωf= final angular velocity, ωi= initial angular velocity (in this case,

the wheel starts from rest so ωi= 0), α= angular acceleration, t= time.

Step 2: Substitute the given values into the equation. Plugging in the values

given: ωi= 0, α= 2 rad/s2,t= 3 s, we get:

ωf= 0 + 2 ×3 = 6 rad/s

Step 3: Calculate the final angular velocity. Therefore, the angular velocity

of the wheel after 3 seconds is 6 rad/s.

Question 31

Question

A disk with a radius of 0.1 m starts rotating from rest with a constant angular

acceleration of 2 rad/s2. What is the magnitude of the tangential acceleration

of a point on the edge of the disk after 3 seconds?

Solution

Step 1: Find the angular velocity (ω) after 3 seconds using the kinematic equa-

tion ω=ω0+αt, where ω0is the initial angular velocity, αis the angular

acceleration, and tis the time.

Initial angular velocity ω0= 0 rad/s

Angular acceleration α= 2 rad/s2

t= 3 s

ω= 0 + 2 ×3 = 6 rad/s

Step 2: Find the tangential speed (v) of a point on the edge of the disk using

v=r×ω, where ris the radius of the disk and ωis the angular velocity.

r= 0.1 m

ω= 6 rad/s

v= 0.1×6=0.6 m/s

Step 3: Find the tangential acceleration of a point on the edge of the disk

using at=r×α, where αis the angular acceleration.

r= 0.1 m

α= 2 rad/s2

at= 0.1×2=0.2 m/s2

Therefore, the magnitude of the tangential acceleration of a point on the

edge of the disk after 3 seconds is 0.2 m/s2.

Question 32

Question

A disc of mass 0.5 kg and radius 0.2 m is rotating about its central axis with an

angular velocity of 4 rad/s. A small mass of 0.1 kg is attached to the rim of the

disc by a string wound around the disc. Initially, the string is held stationary

as the disc continues to rotate. When the string is released, the mass falls

vertically. Assuming no energy losses to friction or air resistance, calculate the

velocity of the mass just before it hits the ground. (Take g= 9.81 m/s2)

Solution

Step 1: Find the moment of inertia of the disc. The moment of inertia Iof a

disc rotating around its central axis is given by:

I=1

2mR2

where mis the mass of the disc and Ris the radius of the disc. Substitute the

given values: m= 0.5 kg, R= 0.2 m

I=1

2×0.5×(0.2)2= 0.01 kg m2

Step 2: Determine the final angular velocity of the system. When the mass

falls, it exerts a torque on the disc causing it to rotate in the opposite direction.

From conservation of angular momentum, the change in angular momentum

of the system must be zero. Initially, the angular momentum is given by Iω,

where ω= 4 rad/s. When the mass falls, the system will rotate in the opposite

direction with a final angular velocity ωf. The angular momentum after the

mass falls is Iωf. Therefore, Iω =−Iωf

0.01 ×4 = −0.01 ×ωf

ωf=−4 rad/s

Step 3: Calculate the linear speed of the mass just before it hits the ground.

The linear speed of the mass can be calculated using the equation:

v=ωfR

Substitute the value of ωfand R= 0.2 m:

v=−4×0.2 = −0.8 m/s

Since the speed of the mass is in the opposite direction of its fall, we take

the magnitude of the velocity:

|v|= 0.8 m/s

Therefore, the velocity of the mass just before it hits the ground is 0.8 m/s.

Question 33

Question

A thin, uniform rod of length Lis pivoted at one end and released from rest in

a vertical position. What is the speed of the free end when the rod makes an

angle θwith the vertical?

Solution

Let xbe the distance between the pivot point and the free end of the rod, and

let ωbe the angular speed of the rod. The instantaneous speed of the free end

is given by v=ωx.

Step 1: Find an expression for the angular position of the rod in terms of

the angle θ. The arc length of the circular path traced out by the free end of

the rod is equal to the length of the rod. Therefore, we have:

x=Lθ

Step 2: Find an expression for the angular speed ωin terms of θ. From

the conservation of energy, the initial gravitational potential energy is converted

into rotational kinetic energy at this point:

mgh =1

2Iω2

Since the rod is rotating about one end and is pivoted at that end, the moment

of inertia Iis 1

3mL2. Substituting this in, we get:

mgh =1

21

3mL2ω2

ω=r3gh

Step 3: Find an expression for the speed of the free end in terms of θ.

Substitute the expression for ωinto the equation v=ωx:

v=r3gh

2L·Lθ

v=r3gh

2·θ

Question 34

Question

A disk with a radius of 0.5 m starts from rest and accelerates uniformly for 5

seconds to reach an angular velocity of 10 rad/s.

1. Find the angular acceleration of the disk.

2. Determine the total angle rotated by the disk during this time interval.

Solution

1. To find the angular acceleration of the disk, we can use the formula:

α=ωf−ωi

where: - αis the angular acceleration, - ωfis the final angular velocity (10

rad/s), - ωiis the initial angular velocity (0 rad/s), - tis the time (5 s).

Step 1: Calculate the angular acceleration:

α=10 rad/s −0 rad/s

5 s = 2 rad/s2

2. To determine the total angle rotated by the disk during this time interval,

we can use the formula:

θ=ωit+1

2αt2

where: - θis the total angle rotated, - ωiis the initial angular velocity (0 rad/s),

-αis the angular acceleration (2 rad/s2), - tis the time (5 s).

Step 2: Calculate the total angle rotated:

θ= 0 rad/s ×5 s + 1

2×2 rad/s2×(5 s)2= 25 rad

Therefore, the angular acceleration of the disk is 2 rad/s2and the total angle

rotated by the disk during this time interval is 25 radians.

Question 35

Question

A solid disk of mass Mand radius Ris initially at rest on a frictionless horizontal

surface. A constant force 

Fis applied tangentially to the edge of the disk. It

takes the disk t1seconds to roll x1meters. If the force is doubled and applied

for t2seconds, how far will the disk have rolled?

Solution

Step 1: Find the acceleration of the disk when the force 

Fis applied. The

net acceleration of the disk is the result of two components: the tangential

acceleration due to the force applied and the angular acceleration due to the

torque caused by the force. The magnitude of the torque is given by τ=R×F.

The moment of inertia for a solid disk rotating around its center is I=1

2MR2.

Question 2

Question

A solid sphere of radius Rand mass Mrolls without slipping down an inclined

plane that makes an angle θwith the horizontal. If the sphere starts from rest

at the top of the incline, determine its speed when it reaches the bottom of the

incline.

Solution

Step 1: The total kinetic energy of the system is the sum of the translational

kinetic energy and the rotational kinetic energy:

2mv2+1

2Iω2

where mis the mass of the sphere, vis its linear velocity, Iis the moment of

inertia of the sphere, and ωis its angular velocity.

Step 2: The linear velocity of the sphere can be related to its angular velocity

by v=Rω, where Ris the radius of the sphere.

Step 3: The moment of inertia of a solid sphere rotating about an axis

through its center and perpendicular to its surface is I=2

5MR2.

Step 4: We use the constraint of rolling without slipping, which means that

the linear velocity is related to the angular velocity by v=Rω.

Step 5: The potential energy at the top of the incline is P E =M gR sin θ,

where Mis the mass of the sphere, gis the acceleration due to gravity, and θ

is the angle of the incline.

Step 6: The total energy at the top is the sum of the potential energy and

the initial kinetic energy (which is zero since the sphere starts from rest).

Step 7: The total energy at the bottom is the sum of the kinetic energy

(translational and rotational).

Step 8: By conservation of energy, we equate the total energy at the top to

the total energy at the bottom:

MgR sin θ=1

2mv2+1

2Iω2

Step 9: Substitute v=Rω and I=2

5MR2into the equation and solve for

MgR sin θ=1

2mR2ω2+1

22

5MR2ω2

Step 10: Simplifying the equation gives:

MgR sin θ=7

10mR2+1

2mR2ω2

Step 11: The linear velocity vat the bottom of the incline is v=Rω. Solving

for vgives:

v=R7

10gsin θ1/2

Therefore, the speed of the sphere at the bottom of the incline is v=

R7

10 gsin θ1/2.

Question 3

Question

A particle is located on the rim of a wheel of radius 0.5 meters. The wheel starts

from rest and accelerates with a constant angular acceleration of 0.8 rad/s2for

4 seconds. At t= 4 seconds, what is the magnitude of the particle’s acceleration

and the angle it has turned through?

Solution

Step 1: To find the magnitude of the particle’s acceleration at t= 4 seconds,

we use the equation for angular acceleration α:

α= 0.8 rad/s2

Step 2: To find the linear acceleration aof the particle located on the rim

of the wheel, we use the equation:

a=r·α

a= 0.5 m ×0.8 rad/s2

a= 0.4 m/s2

Step 3: Therefore, the magnitude of the particle’s acceleration at t= 4

seconds is 0.4 m/s2.

Step 4: To find the angle the particle has turned through at t= 4 seconds,

we need to use the kinematic equation for angular displacement θ:

θ=θ0+ω0t+1

2αt2

Step 5: Since the wheel starts from rest, the initial angular velocity ω0= 0

and the initial angle θ0= 0. Plugging in the values, we get:

θ= 0 + 0 ×4 + 1

2×0.8×42

θ= 0 + 0 + 0.8×8

θ= 6.4 radians

Step 6: Therefore, at t= 4 seconds, the particle has turned through an angle

of 6.4 radians.

Question 4

Question

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

one end with an angular velocity ω. The rod is released at an angle θ0below

the horizontal. What is the angular velocity of the rod when it is vertical?

Solution

1. Let’s first find the initial angular momentum of the rod when it is released.

The initial angular momentum, Linitial, is given by:

Linitial =Iω

where Iis the moment of inertia of the rod and ωis the initial angular velocity.

For a rod rotating about one end, the moment of inertia is I=1

3ML2. So, we

have:

Linitial =1

3ML2·ω

2. Next, let’s find the final angular momentum of the rod when it is vertical.

When the rod is vertical, the moment of inertia is I=ML2/3. Let the final

angular velocity be ωfinal. Then, the final angular momentum Lfinal is given by:

Lfinal =1

3ML2·ωfinal

3. Conservation of angular momentum states that initial angular momentum

equals final angular momentum. Therefore, we have:

Linitial =Lfinal

3ML2·ω=1

3ML2·ωfinal

4. Solving for ωfinal, we find:

ωfinal =ω

5. Therefore, the angular velocity of the rod when it is vertical is the same

as the initial angular velocity, ω.

Question 5

Question

A solid cylinder of radius Rand mass Mrolls without slipping down an incline

of angle θfrom rest. Calculate the time it takes for the cylinder to reach

the bottom of the incline. Assume the cylinder starts from rest, the incline is

frictionless, and neglect air resistance.

Solution

Step 1: Firstly, we need to determine the acceleration of the cylinder as it rolls

down the incline. The net torque acting on the cylinder is due to the force

of gravity causing a torque M gR sin(θ) and the friction force causing a torque

−fR, where fis the frictional force and is zero in this case. The net torque is

equal to the moment of inertia of the cylinder times its angular acceleration α.

Xτ=Iα

MgR sin(θ) = 1

2MR2α

a=Rα

a=2

3gsin(θ)

Step 2: Now we can find the time it takes for the cylinder to reach the

bottom of the incline using the kinematic equation for rotational motion. The

final velocity of the cylinder when it reaches the bottom of the incline is given

by the equation:

v2=u2+ 2as

Where u= 0 (initial velocity), ais the acceleration found above, and sis the

distance traveled down the incline. Since the incline makes an angle θwith the

horizontal, the distance traveled is s=h

sin(θ)where his the height of the incline.

Given that h=Rcos(θ), we get s=Rcot(θ). So,

v2= 0 + 22

3gsin(θ)Rcot(θ)

v=r4

3gR sin(θ)

Step 3: Finally, we can find the time tit takes for the cylinder to reach the

bottom of the incline using the formula:

v=u+at

t=v

t=q4

3gR sin(θ)

3gsin(θ)

t= 2sR

Question 6

Question

A thin rod of length Land mass Mis free to rotate about one end. A force F

is applied perpendicular to the rod at a distance afrom the axis of rotation. If

the rod starts from rest, determine the angular acceleration of the rod.

Solution

Step 1: We will start by considering the torque acting on the rod. The torque

due to the force Fis given by τ=F·a.

Step 2: The moment of inertia of a rod rotating about one end is I=1

3ML2.

Step 3: Using Newton’s second law for rotational motion, τ=I·α, where

αis the angular acceleration.

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

the rotational analog of Newton’s second law, we get F·a=1

3ML2·α.

Step 5: Solving for α, we find α=3F·a

ML2.

Therefore, the angular acceleration of the rod is α=3F·a

ML2.

Question 7

Question

A rigid body starts from rest and accelerates uniformly for 5 seconds to reach

an angular velocity of 10 rad/s. If the body continues to rotate at this con-

stant angular velocity for an additional 10 seconds, what is the total angular

displacement of the body during this time period?

Solution

Step 1: Find the angular acceleration of the body during the first 5 seconds.

Given that the body starts from rest and reaches an angular velocity of 10 rad/s

in 5 seconds, we can use the equation for angular acceleration:

angular acceleration, α=change in angular velocity

time interval =10 rad/s −0 rad/s

5 s = 2 rad/s2

Step 2: Calculate the angular displacement during the first 5 seconds. Using

the equation for angular displacement under uniform acceleration:

θ1=1

2αt2=1

2×2 rad/s2×(5 s)2= 25 rad

Step 3: Determine the total angular displacement for the entire period of

rotation. Since the body rotates at a constant angular velocity of 10 rad/s for

the next 10 seconds, the angular displacement during this time is given by:

θ2= angular velocity ×time = 10 rad/s ×10 s = 100 rad

Step 4: Calculate the total angular displacement. The total angular displace-

ment is the sum of the angular displacements during the two time intervals:

Total angular displacement = θ1+θ2= 25 rad + 100 rad = 125 rad

Therefore, the total angular displacement of the body during the given time

period is 125 radians.

Question 8

Question

A disc of radius 0.2 m rotates about a fixed axis through its center with an

initial angular velocity of 5 rad/s. The angular acceleration of the disc is given

by α=−0.2trad/s2. Find the angular velocity of the disc after 4 seconds.

Solution

Step 1: Write the equation for angular velocity as a function of time.

The angular acceleration can be expressed as the derivative of angular ve-

locity with respect to time:

α=dω

Integrating both sides with respect to time, we get:

Zαdt =Zdω

dt dt

Z(−0.2t)dt =Zdω

−0.1t2+C1=ω

where C1is the constant of integration.

Step 2: Determine the value of the constant of integration.

Given that the initial angular velocity is 5 rad/s when t= 0, we can substi-

tute these values into the equation:

−0.1(0)2+C1= 5

C1= 5

Step 3: Substitute the value of C1back into the equation to find the angular

velocity at t= 4 s.

Substitute C1= 5 into the equation −0.1t2+C1=ω:

−0.1(4)2+ 5 = ω

−0.1(16) + 5 = ω

−1.6 + 5 = ω

ω= 3.4 rad/s

Therefore, the angular velocity of the disc after 4 seconds is 3.4 rad/s.

Question 9

Question

A solid cylinder of mass Mand radius Ris initially at rest on a horizontal

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

to roll without slipping. Calculate the time it takes for the cylinder to reach a

linear speed of v.

Solution

Step 1: First, we need to find the acceleration of the cylinder. The friction force

on the cylinder will provide the torque needed for the rotation and the force

needed for the linear acceleration. Since the cylinder rolls without slipping, we

have the relationship a=αR, where ais the linear acceleration and αis the

angular acceleration.

Step 2: The net torque on the cylinder is equal to F R −ffr, where ffis the

friction force. This torque is also equal to Iα, where Iis the moment of inertia

of the cylinder. For a solid cylinder rotating about its axis, I=1

2MR2.

Step 3: The friction force is equal to µN, where µis the coefficient of friction

and Nis the normal force. For a cylinder, N=Mg. Substituting these values

into our torque equation and equating it to Iα, we get F R −µM gR =1

2MR2α.

Step 4: Since the cylinder rolls without slipping, the linear acceleration a

is equal to the angular acceleration αR. Therefore, we can rewrite the above

equation in terms of a:F R −µMgR =1

2MRa.

Step 5: Given that F−ff=M a for the linear motion, we can substitute

the expression for ffto get F−µMg = (M+1

2M)a.

Step 6: Solving for the linear acceleration a, we find a=F−µMg

2M. Now that

we have the acceleration, we can use the kinematic equation v=u+at to find

the time tit takes for the cylinder to reach a linear speed of v. Since the cylinder

starts from rest, u= 0, so v=at implies t=v

a. Substituting the expression for

agives us the final answer.

Question 10

Question

A wheel initially at rest has an angular acceleration of 3.0 rad/s2. If the wheel

turns through 45 rad, what is its final angular velocity?

Solution

Step 1: Given parameters: The initial angular velocity, ωi, is 0 rad/s. The

angular acceleration, α, is 3.0 rad/s2. The final angle rotated, θ, is 45 rad.

Step 2: Using the equation for angular motion: We can use the relationship

between angular displacement, initial angular velocity, angular acceleration, and

final angular velocity:

ω2

f=ω2

i+ 2αθ

Step 3: Substituting the known values: Plugging in the values we have:

ω2

f= (0)2+ 2(3.0)(45)

ω2

f= 270

Step 4: Finding the final angular velocity: Taking the square root of both

sides:

ωf=√270 ≈16.43 rad/s

Step 5: Therefore, the final angular velocity of the wheel is approximately

16.43 rad/s.

Question 11

Question

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

2.0 rad/s2. If the wheel makes 10 complete revolutions, calculate the time it

takes for the wheel to make these revolutions.

Solution

Step 1: Identify the given values and the unknowns. Let abe the angular

acceleration of the wheel, which is 2.0 rad/s2, and let Nbe the number of

complete revolutions made by the wheel, which is 10. We are asked to find the

time tit takes for the wheel to make these revolutions.

Step 2: Convert the number of revolutions into radians. One complete revo-

lution is equivalent to 2πradians. Therefore, 10 complete revolutions correspond

to 10 ×2π= 20πradians.

Step 3: Use the equation for angular displacement with constant angular

acceleration. The equation to relate angular displacement, angular acceleration,

initial angular velocity, and time is:

θ=ωit+1

2at2

where θis the angular displacement, ωiis the initial angular velocity (which is

0 since the wheel starts at rest), ais the angular acceleration, and tis the time.

We can simplify this equation to:

θ=1

2at2

Substitute in the known values:

20π=1

2×2.0×t2

Step 4: Solve for the time t. Solve for tin the equation:

20π=t2

t=√20π≈6.32 s

Therefore, it takes approximately 6.32 seconds for the wheel to make 10

complete revolutions.

Question 12

Question

A thin hoop with radius Rand mass Mis rolling without slipping on a horizontal

surface. Initially, the hoop is at rest. A particle of mass mis moving horizontally

at speed vtoward the hoop. The particle collides with the hoop and sticks to

it. Find the angular velocity of the system just after the collision.

Solution

Let the initial speed of the particle be vand the angular velocity of the system

just after the collision be ω.

Step 1: Let’s first determine the initial angular momentum of the system.

Since the hoop is initially at rest, the initial angular momentum of the system

is solely due to the particle and is given by

Linitial =m·(R+ 0) ·v=mRv

Step 2: After the collision, the mass of the system is M+mand the new

angular velocity of the hoop-particle system is ω. The final angular momentum

of the system is given by

Lfinal = (M+m)R2·ω

Step 3: Since angular momentum is conserved in the absence of external

torques,

Linitial =Lfinal

mRv = (M+m)R2·ω

Step 4: Solving for ωgives

ω=mRv

(M+m)R2=mv

M+m

Question 13

Question

A solid sphere with radius Rand mass Mrolls without slipping down an inclined

plane making an angle θwith the horizontal. The sphere starts from rest at the

top of the incline. What is its linear acceleration after it has rolled a distance

hdown the incline?

Solution

Step 1: Finding the angular acceleration

The net torque on the sphere about its center of mass can be calculated as:

τ=Iα

where Iis the moment of inertia of the sphere about its center of mass, αis the

angular acceleration, and τis the net torque acting on the sphere. Since the

sphere is rolling without slipping, we have:

α=a

where ais the linear acceleration of the sphere and Ris its radius.

The net torque acting on the sphere is due to the force of gravity and the

normal force. The component of the gravitational force parallel to the incline

causes a torque:

τgravity =MgR sin θ

The torque due to the normal force is perpendicular to the radius of the sphere

and therefore does not contribute to the torque about the center of mass.

Setting the net torque equal to Iα, we have:

MgR sin θ=2

5MR2a

Which simplifies to:

a=5

2gsin θ

Step 2: Finding the linear acceleration

The linear acceleration of the sphere down the incline is the same as the

acceleration due to gravity along the incline since there is no slipping. Using

trigonometry, we can find the acceleration down the incline:

adown incline =gsin θ

Therefore, the linear acceleration of the sphere after it has rolled a distance

hdown the incline is a=5

2gsin θ.

Question 14

Question

A disk of radius 0.2 m starts from rest and undergoes an angular acceleration

of 4.0 rad/s2. What is the angular velocity of the disk after 3 seconds?

Solution

Step 1: We can use the equation for angular motion to find the angular ve-

locity. The equation relating angular velocity, initial angular velocity, angular

acceleration, and time is:

ωf=ωi+αt

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

angular acceleration, and tis the time.

Step 2: Given that the disk starts from rest, the initial angular velocity is

0. Thus, we have:

ωf= 0 + (4.0 rad/s2)(3 s)

Step 3: Calculating the final angular velocity:

ωf= 12 rad/s

Step 4: Therefore, the angular velocity of the disk after 3 seconds is 12 rad/s .

Question 15

Question

A disk with a radius of 0.5 meters starts from rest and accelerates with a constant

angular acceleration of 2 rad/s2. Find the angular velocity of the disk after it

has rotated through an angle of π/3 radians.

Solution

Step 1: Determine the angular velocity using the kinematic equation for rota-

tional motion:

ω2

f=ω2

i+ 2αθ

where - ωiis the initial angular velocity (0 rad/s, since the disk starts from

rest), - αis the angular acceleration (2 rad/s2), and - θis the angle rotated

through (π/3 radians). Thus,

ω2

f= 0 + 2(2)(π/3)

ω2

f= 8π/3

ωf=p8π/3

ωf=2√6π

3rad/s

Therefore, the angular velocity of the disk after rotating through π/3 radians

is 2√6π

3rad/s.

Question 16

Question

A uniform disk of radius Rand mass Mstarts from rest and rolls down a

frictionless incline that makes an angle θwith the horizontal. What is the

linear speed of the center of the disk when it reaches the bottom of the incline?

Solution

Step 1: Start by calculating the moment of inertia of the disk rolling down

the incline. The moment of inertia of a disk about its center is I=1

2MR2.

Using the parallel axis theorem, the moment of inertia about the center of

mass is Icm =I+Md2, where dis the perpendicular distance between the

center of mass and the rotation axis. For a disk, this distance is d=R

2. So,

Icm =1

2MR2+M(R

2)2=3

4MR2.

Step 2: Find the acceleration of the disk down the incline. The gravita-

tional force component down the incline is mg sin θ, which provides a torque,

causing angular acceleration. The torque is τ=Icmα, where αis the angular

acceleration. The angular acceleration is related to the linear acceleration aby

a=αR. So, τ=Icm a

R. The torque is also equal to Rmg sin θ, so we have

Rmg sin θ=3

4MR2a

R. Solving for a, we get a=4

3gsin θ.

Step 3: Calculate the linear speed at the bottom of the incline. At the

bottom, the linear speed vis related to the angular speed ωby v=Rω. Using

the kinematic equation v2=u2+ 2as, where uis the initial velocity (which is 0

in this case), ais the acceleration, and sis the linear distance traveled, we get

v=√2as. Substituting a=4

3gsin θand s=R(height of the incline), we find

v=q2·4

3gsin θ·R=q8

3gR sin θ.

Question 17

Question

A thin rod of length Land mass Mis rotating about an axis perpendicular to the

rod and passing through one end with an angular velocity ω. How much work

is required to stop the rod from rotating? Assume the rod’s mass is uniformly

distributed along its length.

Solution

1. The rotational kinetic energy of the rotating rod is given by

K=1

2Iω2

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

thin rod rotating about an axis perpendicular to the rod and passing through

one end, the moment of inertia is I=1

3ML2.

2. Substituting the moment of inertia into the equation for kinetic energy,

we have

K=1

21

3ML2ω2=1

6Mω2L2

3. To stop the rotating rod, we need to remove all of the kinetic energy.

Hence, the work required is equal to the initial kinetic energy of the system.

Therefore, the work done to stop the rod is

W=1

6Mω2L2

Question 18

Question

A disc of radius 0.2 m rotates with a constant angular acceleration of 2 rad/s2.

At t= 0, the angular velocity of the disc is 1 rad/s and the disc is rotating

counterclockwise. Find the angular velocity of the disc after 4 seconds.

Solution

Step 1: Determine the angular displacement of the disc after 4 seconds. Given

that the disc is rotating with a constant angular acceleration, the angular dis-

placement θcan be calculated using the equation:

θ=ωit+1

2αt2,

where ωiis the initial angular velocity, αis the angular acceleration, and tis

the time. Plugging in the values: ωi= 1 rad/s, α= 2 rad/s2, and t= 4 s, we

get:

θ= 1(4) + 1

2·2·42= 4 + 16 = 20 rad.

Step 2: Find the final angular velocity of the disc. The final angular velocity

ωfcan be calculated using the equation:

ωf=ωi+αt,

where ωiis the initial angular velocity, αis the angular acceleration, and tis

the time. Plugging in the values: ωi= 1 rad/s, α= 2 rad/s2, and t= 4 s, we

get:

ωf= 1 + 2 ·4 = 1 + 8 = 9 rad/s.

Therefore, the angular velocity of the disc after 4 seconds is 9 rad/s.

Question 19

Question

A wheel with an initial angular velocity of 10 rad/s is acted upon by a constant

angular acceleration of −2 rad/s2. How long will it take for the wheel to come

to a complete stop?

Solution

Step 1: Find the angular velocity at which the wheel stops.

Given: Initial angular velocity, ω0= 10 rad/s Angular acceleration, α=

−2 rad/s2

The final angular velocity, ωf, at which the wheel stops can be found using

the equation of rotational motion:

ωf=ω0+αt

Substitute the known values:

ωf= 10 rad/s −2t

Step 2: Solve for the time tat which the wheel stops.

When the wheel comes to a complete stop, ωf= 0. Therefore, we have:

0 = 10 rad/s −2t

Solving for t:

2t= 10 rad/s

t=10

2= 5 s

Therefore, it will take 5 seconds for the wheel to come to a complete stop.

Question 20

Question

A solid sphere of radius Rand mass Mstarts from rest and rolls without

slipping down an inclined plane making an angle θwith the horizontal. Find

the acceleration of the center of mass of the sphere.

Solution

Step 1: The forces acting on the sphere are the gravitational force Mg sin(θ)

down the plane, the normal force Nperpendicular to the plane (which provides

the necessary centripetal force for rolling without slipping), and the friction

force fup the plane.

Step 2: The net force acting on the sphere is:

f−Mg sin(θ) = Macm

Step 3: The torque about the center of mass causing the rotational acceler-

ation is due to the friction force. The torque equation is:

fR =Iα

Step 4: For a solid sphere of radius Rand mass M, the moment of inertia

I=2

5MR2. The rotational acceleration α=acm

Step 5: Substituting the moment of inertia Iand rotational acceleration α

into the torque equation gives:

fR =2

5MR2·acm

Step 6: Simplifying the equation above, we have:

f=2

5Macm

Step 7: Now we can substitute the expression for finto the net force equa-

tion: 2

5Macm −Mg sin(θ) = Macm

Step 8: Solving for the acceleration of the center of mass acm, we find:

acm =5

7gsin(θ)

Question 21

Question

A solid sphere of mass Mand radius Ris initially at rest on a horizontal

surface. It starts rolling without slipping due to an applied constant force 

acting horizontally on it from a certain distance above the ground. The sphere

reaches the ground with an angular velocity ω. What is the speed of the sphere’s

center of mass just before it hits the ground, in terms of M,R,g(acceleration

due to gravity), ω, and any other necessary constants?

Solution

Step 1: Find the acceleration of the center of mass of the sphere

The net torque acting on the sphere is due to the applied force 

F, which

only acts horizontally. The moment of inertia of a solid sphere about its center

is 2

5MR2. Thus, we can use the equation for rotational motion to find the

resultant torque:

τ=Iα

where

τ=F R =2

5MR2·α

Since the sphere is rolling without slipping, we have the relationship α=a

where ais the linear acceleration of the center of mass of the sphere. Therefore,

F R =2

5MR2·a

F=2

5Ma

The net force acting on the sphere is responsible for its acceleration. Therefore,

the net force is given by

F−mg =Ma

Substitute the expression for Fin terms of ainto the above equation:

5Ma −mg =Ma

5a−g=a

a=5

Step 2: Find the speed of the center of mass just before hitting the ground

The final speed vof the center of mass just before it hits the ground is given

by the equation of motion:

v2=u2+ 2as

where uis the initial speed of the center of mass (which is 0 in this case), sis

the distance traveled by the center of mass, and ais the acceleration we found

in Step 1. Rearranging the equation, we get:

v=√2as =r2·5

7g·R

Therefore, the speed of the sphere’s center of mass just before hitting the

ground is r10

7gR .

Question 22

Question

A solid sphere of radius Rrolls without slipping down an incline of angle θ. If

the sphere starts from rest, determine the linear acceleration of the center of

mass of the sphere after it has rolled a distance dalong the incline.

Solution

Step 1: The moment of inertia for a solid sphere rotating about an axis through

its center is given by I=2

5MR2, where Mis the mass of the sphere. Step 2:

The torque on the sphere is due to the gravitational force mg sin(θ) acting at

a distance Rfrom the center. The torque equation is τ=Iα =Rmg sin(θ).

Step 3: The linear acceleration aof the center of mass can be related to the

angular acceleration αusing the equation a=Rα since the sphere is rolling

without slipping. Step 4: We can relate aand αusing the kinematic equation

of rotational motion α=a

R. Step 5: Substitute τ=Iα into the torque equation

Rmg sin(θ) = 2

5MR2·a

R. Step 6: Simplifying the equation Rmg sin(θ) = 2

5Ma,

we can solve for ato find a=5

2gsin(θ), which is the linear acceleration of the

center of mass of the sphere after it has rolled a distance dalong the incline.

Question 23

Question

A wheel starts from rest and rotates with a constant angular acceleration of 2.5

rad/s2. If the diameter of the wheel is 60 cm, determine the time it takes for

the wheel to reach an angular speed of 10 rad/s.

Solution

Step 1: First, we need to find the angular acceleration in terms of the radius of

the wheel. Given that the diameter is 60 cm, the radius rcan be calculated as

r=60

2= 30 cm = 0.3 m.

Step 2: We know that the angular acceleration αis equal to at

r, where atis

the tangential acceleration and ris the radius of the wheel. So, we can rewrite

the given angular acceleration as: α= 2.5 rad/s2=at

0.3.

Step 3: Next, we need to find the tangential acceleration at. Since at=r·α,

we have at= 0.3×2.5 = 0.75 m/s2.

Step 4: To find the time it takes for the wheel to reach an angular speed

of 10 rad/s, we can use the equation ω=ω0+αt, where ωis the final angular

speed, ω0is the initial angular speed, αis the angular acceleration, and tis the

time.

Step 5: Setting ω= 10 rad/s, ω0= 0 rad/s, and α= 2.5 rad/s2, we get:

10 = 0 + 2.5t. Solving for t, we find that t=10

2.5= 4 seconds.

Therefore, it takes 4 seconds for the wheel to reach an angular speed of 10

rad/s.

Question 24

Question

A wheel of radius 2.5 m starts from rest and accelerates with a constant angular

acceleration of 0.5 rad/s2.

1. What is the angular speed of the wheel after 3 seconds?

2. How many revolutions has the wheel completed after 6 seconds?

Solution

1. Let’s first find the angular speed of the wheel after 3 seconds using the

equation:

ω=ω0+αt

where ωis the final angular speed, ω0is the initial angular speed (which is 0 in

this case), αis the angular acceleration, and tis the time.

ω= 0 + (0.5 rad/s2)(3 s)

ω= 1.5 rad/s

So, after 3 seconds, the angular speed of the wheel is 1.5 rad/s.

2. Next, we need to find the number of revolutions completed by the wheel

after 6 seconds. We can calculate this by first finding the total angle rotated by

the wheel in 6 seconds using the equation:

θ=ω0t+1

2αt2

where θis the total angle rotated, ω0is the initial angular speed, αis the angular

acceleration, and tis the time.

θ= 0 ×6 + 1

2×0.5×(6)2

θ= 9 rad

Now, we can find the number of revolutions completed by dividing the total

angle by 2π:

Number of revolutions = 9 rad

2π

Number of revolutions ≈1.43 rev

Therefore, after 6 seconds, the wheel has completed approximately 1.43 revolu-

tions.

Question 25

Question

A thin hoop of radius Rand mass Mrolls without slipping down an inclined

plane that makes an angle θwith the horizontal. The hoop starts from rest at

the top of the incline. What is the linear acceleration of the center of the hoop

after it has rolled a distance ddown the incline?

Solution

Step 1: First, we’ll find the angular acceleration of the hoop as it rolls down the

incline. Given that the hoop is rolling without slipping, the linear acceleration

aof the center of mass is related to the angular acceleration αby a=Rα.

Step 2: The torque about the center of mass of the hoop is caused by the

force of gravity. The torque τis given by τ=R·sin(θ)·mg =Iα, where Iis

the moment of inertia for a hoop of mass Mand radius Rabout its center of

mass. For a hoop, I=MR2.

Step 3: Solving for αin the torque equation, we have R·sin(θ)·mg =MR2α,

which simplifies to α=R·sin(θ)·g

R= sin(θ)·g.

Step 4: Now, we can find the linear acceleration of the center of the hoop

using a=Rα. Substituting α= sin(θ)·g, we get a=R·sin(θ)·g.

Step 5: Finally, the linear acceleration of the center of the hoop after it has

rolled a distance ddown the incline is the component of aparallel to the incline.

This acceleration is aparallel =a·sin(θ) = R·sin2(θ)·g.

Question 26

Question

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

bug starts at the edge of the disk and crawls at a constant speed of 0.1 m/s

towards the center. How far has the bug traveled when it reaches the center of

the disk?