PHYS 305 - INTRODUCTION TO MODERN PHYSICS Rotational kinematics Question Bank Set 3 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Rotational

kinematics

Question Bank - Set 3

Liberty University

Question 1

Question

A disc of radius Ris rotating with an angular velocity ω0about an axis perpen-

dicular to its surface. The angular velocity decreases at a constant rate αuntil

it comes to rest. Find an expression for the time tit takes for the disc to stop

rotating.

Solution

Step 1: Let ω(t) be the angular velocity of the disc at time t. The angular

acceleration αis negative since the angular velocity is decreasing. We have the

equation for angular acceleration:

α=dω

Step 2: Integrate both sides of the equation with respect to tto find the

angular velocity as a function of time:

Zαdt =Zdω

dt dt

−αt +C1=ω(t)

Step 3: At t= 0, the initial angular velocity is ω0, so we have the initial

condition:

ω0=ω(0) = C1

Step 4: Substitute back into the equation from Step 2:

ω(t) = ω0−αt

Step 5: The disc stops when ω(t) = 0, so we solve for t:

0 = ω0−αt

t=ω0

Therefore, the time it takes for the disc to stop rotating is t=ω0

α.

Question 2

Question

A solid sphere of radius Rand mass Mis rolling without slipping on a horizontal

surface. At time t= 0, it is given an initial angular velocity ω0about its

symmetry axis. What is the angular velocity of the sphere after it has rolled a

distance dalong the surface?

Solution

Step 1: The kinetic energy of the rolling sphere at any time can be given by the

sum of its translational and rotational kinetic energies:

KE =1

2Mv2+1

2Iω2,

where vis the speed of the center of mass and ωis the angular velocity, and I

is the moment of inertia of the sphere.

Step 2: Since the sphere is rolling without slipping, the speed of the center

of mass vis related to the angular velocity by v=Rω.

Step 3: The moment of inertia of a solid sphere about its symmetry axis is

I=2

5MR2.

Step 4: Substituting v=Rω and I=2

5MR2into the kinetic energy equa-

tion, we get

KE =1

2M(Rω)2+1

22

5MR2ω2.

Step 5: Simplifying the kinetic energy equation, we have

KE =1

2MR2ω21 + 2

5.

Step 6: After the sphere has rolled a distance d, the work done by the

frictional force must be equal to the change in kinetic energy. This leads to the

equation

ffrictiond= ∆KE.

Step 7: The frictional force ffriction can be expressed as

ffriction =µkineticN,

where µkinetic is the coefficient of kinetic friction and Nis the normal force.

Step 8: The normal force Nis equal in magnitude but opposite in direction

to the gravitational force Mg.

Step 9: Another relationship between the frictional force and the normal

force can be written as ffriction =µkineticM g.

Step 10: Substituting the expressions for ffriction and ∆KE into the work-

energy equation, we get

µkineticM gd =KEfinal −KEinitial.

Step 11: Solving for the final angular velocity ω, we find

ω=ω07µkineticgd

5Rω2

0+ 7µkineticgd.

Therefore, the angular velocity of the sphere after it has rolled a distance d

along the surface is ω=ω07µkinetic gd

5Rω2

0+7µkinetic gd .

Question 3

Question

A solid disk with radius Rrolls without slipping along a horizontal surface.

Initially, the center of mass of the disk is at point Aand the disk has an angular

speed of ω0. As the disk rolls, it moves to point Band comes to a stop. What

is the angular displacement of the disk during this motion from Ato B?

Solution

Step 1: We know that when a disk rolls without slipping, the linear speed of

the center of mass is given by v=Rω, where vis the linear speed of the center

of mass, Ris the radius of the disk, and ωis the angular speed of the disk.

Step 2: The disk comes to a stop at point B, so its final linear speed becomes

0. Let ∆θbe the angular displacement of the disk from point Ato B.

Step 3: The linear speed of the disk at point Ais v0=Rω0, and at point B

it is 0. Thus, the change in linear speed is ∆v= 0 −Rω0=−Rω0.

Step 4: The linear acceleration of the disk can be calculated using the equa-

tion a=∆v

∆t, where ais the linear acceleration and ∆tis the time taken for the

disk to come to a stop.

Step 5: Since the disk rolls without slipping, the linear acceleration is also

equal to Rα, where αis the angular acceleration of the disk.

Step 6: Combining the two equations, we have Rα =−Rω0

∆t, which simplifies

to α=−ω0

∆t.

Step 7: The angular displacement can be calculated using the equation ∆θ=

ω0∆t+1

2α∆t2.

Step 8: Substituting the values we have, ∆θ=ω0∆t+1

2−ω0

∆t∆t2.

Step 9: Simplifying further, ∆θ=ω0∆t−1

2ω0∆t=1

2ω0∆t.

Step 10: To find ∆t, we can use the fact that the linear acceleration is

constant, so ∆t=vf−vi

a, where vfis the final linear speed (0 in this case), viis

the initial linear speed (Rω0), and ais the linear acceleration.

Step 11: Substituting the values, ∆t=0−Rω0

−Rω0

∆t

= ∆t2.

Step 12: Solving for ∆t, we get ∆t= 1.

Step 13: Finally, substituting ∆t= 1 into the equation for ∆θ, we find

∆θ=1

2ω0×1 = 1

2ω0.

Therefore, the angular displacement of the disk during its motion from point

Ato Bis 1

2ω0.

Question 4

Question

A disc of radius Ris rotating about its axis with an angular velocity ω. A small

piece of the disc of mass mdetaches from the edge and starts sliding outward

while conserving its angular velocity. What is the angular velocity of the small

piece when its distance from the axis is 2R?

Solution

Step 1: Let’s denote the angular velocity of the small piece at a distance r

from the axis as ω′. The moment of inertia of the small piece about the axis is

I=mr2.

Step 2: We can use the conservation of angular momentum for the small

piece. Initially, the angular momentum of the small piece with respect to the

axis is Linitial =Iω.

Step 3: As the small piece moves outward to a distance of 2R, its moment of

inertia becomes I′=m(2R)2= 4mR2. The conservation of angular momentum

gives us Linitial =Lfinal.

Step 4: Therefore, Iω =I′ω′. Substituting the expressions for Iand I′, we

have mr2ω= 4mR2ω′.

Step 5: Plugging in the values r= 2Rinto the equation above, we get

m(2R)2ω= 4mR2ω′.

Step 6: Simplifying further, we find 4R2ω= 4R2ω′. This simplifies to

ω=ω′.

Step 7: Thus, the angular velocity of the small piece when its distance from

the axis is 2Ris equal to the initial angular velocity ω.

Question 5

Question

A thin-walled hollow sphere of radius 0.5 m and mass 2 kg is rotating about

a vertical axis through its center at 10 rad/s. A small ball of mass 0.1 kg is

dropped vertically onto the sphere and sticks to it. Find the new angular speed

of the system.

Solution

Step 1: Let’s first find the initial angular momentum of the system. The ini-

tial angular momentum (Li) of the system is given by the sum of the angular

momentum of the sphere and the angular momentum of the falling ball.

The angular momentum of the sphere:

Lsphere =Isphereωsphere

where Isphere is the moment of inertia of the sphere and ωsphere is the initial

angular velocity of the sphere.

The moment of inertia of a thin-walled hollow sphere is I=2

3mr2, where m

is the mass of the sphere and ris the radius.

Plugging in the values:

Isphere =2

3×2×(0.5)2=1

3kg m2

Given that ωsphere = 10 rad/s, we can find the angular momentum of the

sphere.

Step 2: The angular momentum of the falling ball just before it hits and

sticks to the sphere is mgh, where mis the mass of the ball, gis the acceleration

due to gravity, and his the height from which the ball is dropped.

Given that the height from which the ball is dropped is negligible compared

to the radius of the sphere, the initial angular momentum of the ball is mgh = 0.

Therefore, the initial angular momentum of the system (Li) is equal to the

angular momentum of the sphere.

Step 3: Let’s now find the final moment of inertia (If) of the system after

the ball sticks to the rotating sphere.

The moment of inertia of the sphere and the ball as a system is:

Isystem =Isphere +mr2

where mis the mass of the ball.

Plugging in the values:

Isystem =1

3+ 0.1×(0.5)2=1

3+ 0.025 = 1

3+1

40 =9

40 kg m2

Step 4: Using the principle of conservation of angular momentum, the final

angular speed of the system (ωf) can be calculated as:

Li=Lf

Isphereωsphere =Isystemωf

Solving for ωf:

ωf=Isphereωsphere

Isystem

Plugging in the values:

ωf=

3×10

=40

3×1

3=40

9= 4.4 rad/s

Therefore, the new angular speed of the system after the ball sticks to the

sphere is 4.4 rad/s.

Question 6

Question

A wheel rotating at 100 rad/s comes to rest after rotating through 100 revolu-

tions. Find the magnitude of the angular acceleration of the wheel.

Solution

Step 1: First, we calculate the initial angular velocity of the wheel in terms of

radians per second. Since 1 revolution is equivalent to 2πradians, the initial

angular velocity is given by:

ωi= 100 rad/s ×2πrad

1 rev = 200πrad/s

Step 2: Next, we calculate the final angular velocity of the wheel, which is

0 rad/s since it comes to rest.

Step 3: We use the equation of angular motion to relate the initial and final

angular velocities, angular acceleration, and angular displacement:

ω2

f=ω2

i+ 2αθ

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

velocity (200πrad/s), - αis the angular acceleration (to be determined), - θis

the angular displacement in radians (100 rev ×2πrad/rev).

Step 4: Substituting the known values into the equation gives:

0 = (200π)2+ 2α(100 ×2π)

Step 5: Solving for the angular acceleration, we find:

α=−(200π)2

200 ×2π=−200πrad/s2

Therefore, the magnitude of the angular acceleration of the wheel is 200π

rad/s2.

Question 7

Question

A thin cylindrical rod of mass mand radius Ris hinged at one end and set

into motion by applying a force perpendicular to the rod at a distance rfrom

the hinge. Initially, the rod is at rest in a horizontal position. Find the angular

acceleration of the rod as a function of time t.

Solution

1. The torque on the rod about the hinge can be calculated as τ=rF , where

Fis the applied force. The moment of inertia of a thin cylindrical rod rotating

about its end is I=1

3mR2, so the angular acceleration αcan be expressed as

τ=Iα.

2. Substituting the expressions for torque and moment of inertia into the

equation above, we have rF =1

3mR2α.

3. Rearranging the equation gives us the expression for angular acceleration:

α=3rF

mR2.

4. As the force Fis a constant, the angular acceleration αis also constant

and does not depend on time t.

Question 8

Question

A disc of radius 0.2 m is initially at rest. A constant torque of 4 N·m is applied

to the disc for 6 seconds, causing it to reach an angular velocity of 8 rad/s.

Calculate the moment of inertia of the disc.

Solution

Step 1: Recall the relationship between torque τ, moment of inertia I, angular

acceleration α, and radius r:

τ=Iα

Step 2: We are given that the torque applied is 4 N·m and the angular

acceleration can be calculated using the formula:

α=∆ω

∆t

Step 3: Substituting the given values into the formula:

α=8 rad/s −0

6 s =8 rad/s

6 s = 1.33 rad/s2

Step 4: Substitute the torque and angular acceleration into the torque for-

mula to solve for the moment of inertia I:

I=τ

α=4 N ·m

1.33 rad/s2= 3.01 kg ·m2

Step 5: The moment of inertia of the disc is 3.01 kg·m2.

Question 9

Question

A rotating platform initially at rest, experiences a constant angular acceleration

of 0.05 rad/s2. At time t= 6 s, the platform has rotated through an angle of 1

rad. What is the angular velocity of the platform at t= 6 s?

Solution

Step 1: First, let’s find the angular velocity of the platform at t= 6 s using the

equation of rotational kinematics:

ω=ω0+αt

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

0 rad/s since the platform starts at rest), αis the angular acceleration, and tis

the time.

Step 2: Substitute the given values into the equation:

ω= 0 + (0.05 rad/s2)×6 s

ω= 0.3 rad/s

Therefore, the angular velocity of the platform at t= 6 s is 0.3 rad/s.

Question 10

Question

A thin rod of mass Mand length Lis pivoted about one end and released from

rest in a vertical position. What is the angular speed ωof the rod when it

swings by 30◦below the horizontal?

Solution

Step 1: To find the angular speed ωof the rod when it swings by 30◦below

the horizontal, we will use conservation of energy. Initially, the rod is at rest

in a vertical position, and when it swings below the horizontal, all its potential

energy will convert into kinetic energy.

Step 2: The potential energy of the rod initially is Ui=M gh, where his

the height of the center of mass and h=L

Step 3: The initial potential energy Uiis equal to the final kinetic energy

Kf=1

2Iω2, where Iis the moment of inertia of the rod about the pivot point.

Step 4: The moment of inertia of a rod rotating about one end is I=1

3ML2.

Step 5: Setting the initial potential energy equal to the final kinetic energy,

we have Mgh =1

21

3ML2ω2.

Step 6: Substituting in the values of h,I, and Mgives MgL =1

6ML2ω2.

Step 7: Solving for ω, we find ω=q12g

Step 8: Therefore, the angular speed of the rod when it swings by 30◦below

the horizontal is ω=q12g

Question 11

Question

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

velocity ω. If a small piece of mass mis attached to the free end of the rod,

determine the angular velocity of the system.

Solution

Let’s denote the moment of inertia of the rod about its end as Irod, the moment

of inertia of the mass mabout the end of the rod as Imass, and the total moment

of inertia of the system as Itotal.

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

thin rod rotating about one end is given by Irod =1

3ML2.

Step 2: Find the moment of inertia of the mass. Since the mass mis

attached to the free end of the rod, Imass =m·L2.

Step 3: Find the total moment of inertia of the system. The total moment

of inertia of the system is the sum of the moments of inertia of the rod and the

mass, Itotal =Irod +Imass.

Step 4: Apply the conservation of angular momentum. The initial angular

momentum of the system is equal to the final angular momentum, which means

Irodω=Itotalωfinal.

Step 5: Solve for the final angular velocity ωfinal. Substitute the expressions

for Irod,Imass, and Itotal into the conservation of angular momentum equation

and solve for ωfinal.

3ML2ω=1

3ML2+mL2ωfinal

ωfinal =

3M+m

3M+mω

ωfinal =ω

Therefore, the angular velocity of the system with the added mass remains

the same as the initial angular velocity ω.

Question 12

Question

A disk of radius Ris initially at rest. A constant tangential force Ftis applied

at the edge of the disk for a given time interval, causing the disk to rotate. Find

an expression for the angular velocity ωof the disk in terms of Ft,R, and the

moment of inertia Iof the disk during this time interval.

Solution

Step 1: We start by using Newton’s second law for rotation, which states that

the net torque τacting on an object is equal to the moment of inertia Itimes

the angular acceleration α. The torque is given by τ=R·Ft, where Ris the

radius of the disk.

Step 2: We substitute this expression for torque back into Newton’s second

law for rotation to get R·Ft=I·α. We also know that α=dω

dt , where ωis the

angular velocity of the disk.

Step 3: From the definition of moment of inertia for a disk (rotating around

its symmetry axis), we have I=1

2MR2, where Mis the mass of the disk.

Step 4: Substituting the expression for Iinto the equation R·Ft=I·α, we

have R·Ft=1

2MR2·dω

dt .

Step 5: We can simplify this expression to obtain Ft

2=dω

dt . Integrating both

sides with respect to tgives RFt

2dt =Rdω.

Step 6: This simplifies to Ft

2t=ω+C, where Cis a constant of integration.

Since the disk starts from rest, ω= 0 at t= 0, so the constant Cis 0.

Step 7: Therefore, the expression for the angular velocity ωof the disk in

terms of Ft,R, and Iduring the time interval twhen the force is applied is

given by ω=Ft

2t.

Thus, the angular velocity ωof the disk is directly proportional to the tan-

gential force Ftand the time tfor which the force is applied, and inversely

proportional to the moment of inertia Iof the disk.

Question 13

Question

A disc of radius rand mass mis rotating about a fixed axis passing through its

center. If the angular velocity is given by ω(t) = αt +βt2−γt3, where α,β,

and γare constants, find the expression for the angular acceleration of the disc

as a function of time.

Solution

Step 1: To find the angular acceleration α(t), we need to differentiate the angular

velocity function ω(t) with respect to time.

Step 1: Given the angular velocity function ω(t) = αt +βt2−γt3,

Differentiate with respect to time: α(t) = dω

dt =α+ 2βt −3γt2.

Step 2: Therefore, the expression for the angular acceleration α(t) as a

function of time is α(t) = α+ 2βt −3γt2.

Question 14

Question

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

acceleration of 2 rad/s2for 5 seconds. Calculate the angular velocity of the disc

after this time period.

Solution

Step 1: Find the angular displacement of the disc. Given that the disc starts

from rest, the angular displacement θcan be calculated using the kinematic

equation:

θ=1

2αt2

θ=1

2×2×(52)

θ= 25 rad

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

of the disc can be found using the kinematic equation:

ωf=ωi+αt

ωf= 0 + 2 ×5

ωf= 10 rad/s

Therefore, the angular velocity of the disc after 5 seconds is 10 rad/s.

Question 15

Question

A wheel initially at rest accelerates uniformly for 10 seconds, reaching an angular

velocity of 30 rad/s. It then decelerates uniformly for 5 seconds until it comes

to a stop. If the total angle rotated during this whole process is 450 radians,

what is the average angular acceleration of the wheel during the entire motion?

Solution

Step 1: Find the angular acceleration during the first part of the motion. Given

that the initial angular velocity (ωi= 0 rad/s), the final angular velocity (ωf=

30 rad/s), and the time duration (t= 10 s), we can use the equation for angular

acceleration:

α=∆ω

∆t

α=30 rad/s −0 rad/s

10 s = 3 rad/s2

Step 2: Find the angle rotated during the first part of the motion. We can

use the equation for angular displacement:

θ=ωit+1

2αt2

Substitute the known values to find θ:

θ= 0 ×10 + 1

2×3×(10)2= 150 radians

Step 3: Find the angular acceleration during the second part of the motion.

Since the wheel decelerates uniformly, the final angular velocity (ωf= 0 rad/s),

the initial angular velocity (ωi= 30 rad/s), and the time duration (t= 5 s) can

be used to find the angular acceleration:

α=∆ω

∆t

α=0 rad/s −30 rad/s

5 s =−6 rad/s2(negative sign indicates deceleration)

Step 4: Find the angle rotated during the second part of the motion. Using

the same angular displacement equation:

θ=ωit+1

2αt2

Substitute the known values to find θ:

θ= 30 ×5 + 1

2×(−6) ×(5)2= 75 radians

Step 5: Determine the average angular acceleration. The total angle rotated

is 450 radians, and the total time duration is 15 seconds. The average angular

acceleration can be found using the total change in angular velocity and the

total time:

Average α=ωtotal

ttotal

Average α=30 rad/s −0 rad/s

15 s = 2 rad/s2

Therefore, the average angular acceleration of the wheel during the entire

motion is 2 rad/s2.

Question 16

Question

A disc of mass Mand radius Ris rotating with an angular velocity ω. A small

object of mass mand radius rfalls gently onto the disc and sticks to it. The

moment of inertia of the disc alone is I. Determine the final angular velocity of

the system.

Solution

Step 1: Initially, the disc is rotating with angular velocity ωand the small

object falls gently onto it and comes to rest relative to the disc. Conservation of

angular momentum can be used to determine the final angular velocity of the

system.

Step 2: The initial angular momentum of the disc-object system is Iω.

Step 3: The final angular momentum of the system is (I+m(r+R)2)Ω,

where Ω is the final angular velocity of the system.

Step 4: Since angular momentum is conserved, we have Iω = (I+m(r+

R)2)Ω.

Step 5: Solving for Ω, we get Ω = Iω

I+m(r+R)2.

Step 6: Substituting the given values of the moment of inertia Ias well as

the radii of the disc and the object, we find the final angular velocity of the

system as Ω = Iω

I+m(r+R)2.

Question 17

Question

A disk of radius Rrolls without slipping along a horizontal surface. Starting

from rest, the disk accelerates uniformly for a time tand reaches a final angular

speed of ω. Find the distance dit has rolled along the surface in terms of R,t,

and ω.

Solution

Step 1: Find the angular acceleration αof the disk. The final angular speed ω

can be related to the initial angular speed ω0= 0, the time t, and the angular

acceleration αthrough the equation:

ω=ω0+αt

Since the initial angular speed is 0, we have:

ω=αt

Therefore, the angular acceleration is:

α=ω

Step 2: Find the distance drolled by the disk. The distance rolled by the disk

can be determined by considering the relationship between linear and angular

quantities. The linear distance rolled dis related to the angle turned by the

disk by the equation:

d=Rθ

where θis the angle turned by the disk. The angle turned by the disk can be

expressed in terms of angular speed and time using the equation:

θ=1

2αt2

Substitute the expression for αfound in Step 1:

θ=1

2ω

tt2=ω

Therefore, the distance rolled by the disk is:

d=R·ω

2t=1

2Rωt

Thus, the distance rolled by the disk in terms of R,t, and ωis 1

2Rωt.

Question 18

Question

A thin circular ring of radius Ris rotating with an angular velocity ωabout an

axis passing through its center and perpendicular to the plane of the ring. A

small object of mass mis placed gently on the surface of the ring. What is the

minimum coefficient of static friction between the object and the ring that will

prevent the object from slipping?

Solution

1. The forces acting on the object on the rotating ring include gravitational

force, normal force, static frictional force, and a pseudo force due to the non-

inertial frame of reference. The object undergoes circular motion with radius

R, so the sum of forces in the radial direction should provide the necessary

centripetal force:

mg −N=mv2

2. Since the object is at rest relative to the rotating frame, the centripetal

acceleration must be zero. Therefore, the net force in the radial direction is also

zero:

N=mg

3. The maximum frictional force is given by fmax =µN, where µis the coef-

ficient of static friction. The object will begin to slip when the frictional force

approaches this maximum value:

µminN=mv2

4. Substituting the expression for Nfrom step 2:

µminmg =mv2

5. Cancelling mass mfrom both sides:

µming=v2

6. The tangential velocity of the object is related to the angular velocity of the

ring:

v=Rω

7. Substituting this into the equation from step 5:

µming=(Rω)2

R=Rω2

8. Therefore, the minimum coefficient of static friction needed to prevent slip-

ping is µmin =ω2/g.

Question 19

Question

A solid cylinder of radius R, mass M, and moment of inertia I=1

2MR2is

initially at rest. A constant force Fis applied tangentially to the edge of the

cylinder for a distance d, causing the cylinder to move. What is the angular

velocity of the cylinder after the force is applied and the distance it travels? Use

that the kinetic energy of a rotating object is given by K=1

2Iω2.

Solution

Step 1: Calculate the torque applied to the cylinder. The torque, τ, applied to

the cylinder is given by τ=F r, where ris the radius of the cylinder. Given

that τ=Iα, where αis the angular acceleration, we can write F r =Iα.

Step 2: Find the angular acceleration of the cylinder. Since the cylinder

starts from rest, the initial angular velocity, ωi, is 0. The final angular velocity,

ωf, can be found using the relationship ω2

f=ω2

i+ 2αθ, where θis the angular

displacement. Here, θ=d

R, so we have ω2

f= 2αd

R.

Step 3: Solve for the angular acceleration. Substitute F r =Iα into the

equation F r =Iα to solve for α:F·R=I·α, giving F·R=1

2MR2α. Thus,

α=2F

Step 4: Calculate the final angular velocity. Substitute α=2F

Minto the

equation ω2

f= 2αd

R:ω2

f= 2 2F

Md

R, so ωf=q4F d

Step 5: Find the distance traveled by the cylinder. We know that v=ωR,

so the linear velocity at the edge of the cylinder is v=ωfR. The distance

traveled, d, is given by d=vt, so d=ωfR·t=q4F d

M·R·d

Therefore, the angular velocity of the cylinder after the application of force

is q4F d

Mrad/s, and the distance traveled by the cylinder is 2F d

Mmeters.

Question 20

Question

A disk of radius 0.5 m is rotating about an axis through its center with an

angular acceleration of 3 rad/s2. At t= 2 s, the angular velocity of the disk is

4 rad/s in the counterclockwise direction. If the angular acceleration is constant,

determine the angular velocity of the disk when t= 4 s.

Solution

Step 1: Find the initial angular velocity of the disk using the angular acceleration

and time provided. Given: r= 0.5 m, α= 3 rad/s2,t= 2 s, and ω0= 0 rad/s.

The angular velocity at t= 2 s can be found using the equation ω=ω0+αt:

ω= 0 + 3 ×2 = 6 rad/s

Step 2: Find the angular displacement of the disk from t= 2 s to t= 4 s.

The angular displacement θcan be found using the equation θ=ω0t+1

2αt2:

θ= 6 ×2 + 1

2×3×22= 12 + 6 = 18 rad

Step 3: Use the angular displacement to find the angular velocity of the

disk at t= 4 s. The final angular velocity can be found using the equation

ω2=ω2

0+ 2αθ:

ω2= 62+ 2 ×3×18 = 36 + 36 = 72

ω=√72 = 6√2≈8.49 rad/s

Therefore, the angular velocity of the disk when t= 4 s is approximately

8.49 rad/s in the counterclockwise direction.

Question 21

Question

A solid sphere of radius Ris rolling without slipping along a horizontal surface.

The sphere is given an initial angular speed ω0and an initial linear speed v0at

the same time. If the sphere rolls without slipping for a distance d, determine

the final values of its linear speed and angular speed.

Solution

Let’s denote the final linear speed of the sphere as vfand the final angular speed

as ωf.

Step 1: Relate the linear speed and angular speed: Since the sphere is rolling

without slipping, we have the following relationship between linear speed and

angular speed:

vf=ωfR

Step 2: Apply conservation of energy: The initial kinetic energy of the

sphere consists of both translational and rotational kinetic energy. The final

kinetic energy of the sphere will also consist of translational and rotational

kinetic energy. Therefore, we can write:

2Iω2

0+1

2mv2

0=1

2Iω2

f+1

2mv2

Step 3: Substitute vf=ωfRinto the conservation of energy equation:

2Iω2

0+1

2m(v0)2=1

2Ivf

R2+1

2m(vf)2

Step 4: Substitute I=2

5mR2into the equation:

5mR2ω2

0+1

2mv2

0=1

5mv2

f+1

2mv2

Step 5: Simplify and solve for vf:

10mv2

0=3

5mv2

f=7

6v2

vf=r7

6v0

Step 6: Find ωf:

vf=ωfR

ωf=vf

R=p7/6v0

Therefore, the final linear speed of the sphere is q7

6v0and the final angular

speed is √7/6v0

Question 22

Question

A disk with radius Rstarts from rest and accelerates with a constant angular

acceleration αfor a time t. Find an expression for the angle through which the

disk turns during this time.

Solution

Step 1: We can start by finding the angular velocity of the disk at time t.

We know that angular acceleration is defined as the rate of change of angular

velocity. Therefore, we have

α=ω

Solving for ω, we get

ω=αt

Step 2: Next, we can find the angle turned by the disk in time t. The

relationship between angular displacement θ, angular velocity ω, and time tis

given by

θ=ωt

Substitute in the expression for ωwe found earlier:

θ= (αt)t=αt2

Step 3: Therefore, the expression for the angle through which the disk turns

during the time tis θ=αt2.

Question 23

Question

A wheel initially at rest accelerates with a constant angular acceleration of

2.0 rad/s2for 10 seconds. If the wheel makes 5 complete revolutions during this

time, what is the wheel’s angular velocity at the end of the 10 seconds?

Solution

Step 1: Calculate the final angular displacement of the wheel during the 10-

second interval. We know that angular displacement θis related to the number

of revolutions made by the wheel:

θ= 2π×number of revolutions

Given that the wheel makes 5 complete revolutions, θ= 2π×5 = 10πrad.

Step 2: Use the formula for angular displacement with constant angular

acceleration to find the final angular velocity. The formula relating angular

displacement, initial angular velocity, angular acceleration, and time is:

θ=ωit+1

2αt2

Where: - θ= 10πrad - ωi= 0 rad/s (initial angular velocity) - α= 2.0 rad/s2

(angular acceleration) - t= 10 s Plugging in the values:

10π= 0 ×10 + 1

2×2.0×102

10π= 100

From this, we find that the final angular velocity is 10 rad/s.

Question 24

Question

A solid sphere of mass 2 kg and radius 0.4 m starts from rest and rolls without

slipping down a 30-degree incline. How fast is the sphere rolling after it has

traveled 3 meters along the incline? (Neglect air resistance and friction)

Solution

Step 1: We can start by calculating the acceleration of the sphere along the

incline. The net acceleration of the sphere can be found by considering the

forces acting on it. The component of the gravitational force parallel to the

incline will cause linear acceleration (acm) and angular acceleration (α).

Step 2: The component of the gravitational force parallel to the incline can

be calculated as:

F|| =mg sin(30◦)

Step 3: The linear acceleration of the center of mass (acm) can be found

using Newton’s second law:

macm =mg sin(30◦)

acm =gsin(30◦)

Step 4: The angular acceleration (α) can be found using the relation between

linear acceleration and angular acceleration for a rolling object:

acm =Rα

gsin(30◦) = Rα

Step 5: The equation for rolling without slipping relates linear velocity (v)

and angular velocity (ω) as:

v=Rω

Step 6: The centripetal acceleration of a rolling object is given by:

acp =Rα

Therefore, v2=Rα

Step 7: Now that we have the linear acceleration, we can find the velocity

of the sphere after traveling 3 meters. We can use the kinematic equation:

v2=u2+ 2acms

where uis the initial velocity, and sis the distance traveled.

Step 8: Substituting the given values into the equation, we can solve for the

final velocity (v).

Question 25

Question

A disk with a radius of 0.1 m is spinning counterclockwise at an angular velocity

of 10 rad/s. A point on the edge of the disk releases a bug, which lands 0.04 m

from the center of the disk. What is the bug’s speed immediately after landing?

Solution

1. First, calculate the linear velocity of the point on the edge of the disk using

the formula v=ωr, where vis the linear velocity, ωis the angular velocity, and

ris the radius of the disk.

2. Substitute the values: v= 10 rad/s ×0.1 m = 1 m/s.

3. Next, find the linear velocity component tangential to the circular path

at the point where the bug lands. This can be found using the formula vt=rω,

where vtis the tangential velocity, ωis the angular velocity, and ris the distance

from the center where the bug lands.

4. Substitute the values: vt= 0.04 m ×10 rad/s = 0.4 m/s.

5. The bug’s total speed is the vector sum of the linear velocity and the tan-

gential velocity at the landing point. This can be found using the Pythagorean

theorem: vtotal =pv2+v2

6. Substitute the values: vtotal =√12+ 0.42≈√1.16 ≈1.08 m/s.

7. Therefore, the bug’s speed immediately after landing is approximately

1.08 m/s.

Question 26

Question

A thin rod of length Land mass Mis free to rotate in a vertical plane about a

pivot at one end. The rod is released from rest at an angle θ0with the vertical.

What is the angular velocity of the rod when it is in the horizontal position?

Solution

Step 1: First, let’s determine the initial potential and kinetic energy of the

system. The initial potential energy of the rod is given by:

Ui=−MgL cos θ0

The initial kinetic energy of the rod is zero since it is released from rest.

Step 2: Next, let’s determine the final potential and kinetic energy of the

system when the rod is in the horizontal position. The final potential energy of

the rod is given by:

Uf=−MgL cos(90◦)=0

The final kinetic energy of the rod consists of both translational and rotational

kinetic energy. Since the rod is horizontal, all the initial potential energy is

converted into kinetic energy.

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

energy, so we have:

MgL cos θ0=1

2Iω2

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

Step 4: The moment of inertia of a rod rotating about one end is I=1

3ML2.

Substituting this into the equation, we get:

MgL cos θ0=1

21

3ML2ω2

Step 5: Solving for ω, we find:

ω=r3gcos θ0

Therefore, the angular velocity of the rod when it is in the horizontal position

is q3gcos θ0

2L.

Question 27

Question

A disk of radius 0.15 m starts from rest and has an angular acceleration of

1.5 rad/s2.

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

2. How many revolutions has the disk made after 6 seconds?

Solution

1. We can use the equation for angular motion:

ωf=ωi+αt

where ωfis the final angular speed, ωiis the initial angular speed (which is 0

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

Step 1: Substitute the known values into the equation.

ωf= 0 + (1.5 rad/s2)(3 s)

ωf= 4.5 rad/s

So, the angular speed of the disk after 3 seconds is 4.5 rad/s.

2. To find the number of revolutions, we need to first calculate the total

angular displacement of the disk after 6 seconds:

Step 2: Use the equation for angular displacement:

θ=1

2αt2

where θis the angular displacement.

Substitute the known values:

θ=1

2(1.5 rad/s2)(6 s)2

θ= 27πrad

To determine the number of revolutions, we need to convert this into revo-

lutions:

Number of revolutions = Total angular displacement

2π

Number of revolutions = 27π

2π= 13.5 rev

Therefore, after 6 seconds, the disk has made 13.5 revolutions.

Question 28

Question

A uniform disk of radius Rand mass Mis rotating about a vertical axis passing

through its center with an angular velocity ω0. A small object of mass mslides

without friction along a radial slot in the disk from the rim to the center. What

is the angular velocity of the disk-object system after the object reaches the

center?

Solution

1. Conservation of angular momentum: Initially, the total angular mo-

mentum of the system is given by the angular momentum of the disk and the

angular momentum of the object:

Linitial =I0ω0+mR2ω0=I0ω0+mR2v

2. Conservation of mechanical energy: The initial mechanical energy

of the system is equal to the final mechanical energy:

Einitial =1

2I0ω2

0+1

2mv2=Efinal =1

2Ifω2

3. From conservation of angular momentum:

I0ω0+mR2ω0=Ifωf

I0ω0+mR2v

R=Ifωf

4. The moment of inertia of the disk-object system after the object reaches

the center is:

If=I0+mR2

5. Substituting Ifinto the equation from step 3:

I0ω0+mR2ω0= (I0+mR2)ωf

(I0+mR2)ωf=I0ω0+mR2ω0

6. Solving for ωf:

ωf=I0ω0+mR2ω0

I0+mR2

Question 29

Question

An object starts from rest and rotates with a constant angular acceleration of

2.5 rad/s2. If the object reaches an angular velocity of 7.0 rad/s after rotating

through an angle of 5.0 rad, determine the angular velocity of the object when

it has rotated through an angle of 12 rad.

Solution

Step 1: Determine the final angular velocity using the given information.

Given: α= 2.5 rad/s2, ∆θ= 5.0 rad, vi= 0 rad/s, and vf= 7.0 rad/s. We can

use the equation connecting angular velocity, initial angular velocity, angular

acceleration, and angular displacement:

f=v2

i+ 2α∆θ

Substitute the given values:

(7.0)2= 02+ 2(2.5)(5.0)

49 = 25

Since the equation is invalid, there seems to be a mistake in the given values.

Let’s reevaluate.

Step 2: Reevaluate the given values.

It seems there was a mistake in the calculation in Step 1. Let’s recalculate the

final angular velocity using the correct equation:

vf=qv2

i+ 2α∆θ

vf=p0 + 2(2.5)(5.0)

vf=√25 = 5.0 rad/s

Step 3: Determine the angular velocity when the object has rotated through

an angle of 12 rad.

Using the equation relating final angular velocity, initial angular velocity, angu-

lar acceleration, and angular displacement:

f=v2

i+ 2α∆θ

f= 0 + 2(2.5)(12)

vf=√60 ≈7.75 rad/s

Therefore, the angular velocity of the object when it has rotated through an

angle of 12 rad is approximately 7.75 rad/s.

Question 30

Question

A circular disk of radius Rrests on a frictionless horizontal surface. A constant

force Fis applied to the rim of the disk perpendicular to the radius as shown

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

where Iis the moment of inertia of the disk about its center.

Solution

Step 1: The torque acting on the disk is given by τ=F r, where ris the radius

of the disk. By Newton’s second law for rotation, we have:

τ=Iα

where αis the angular acceleration of the disk and Iis the moment of inertia.

Step 2: Substituting τ=F r and rearranging the equation, we get:

F r =Iα

Step 3: The moment of inertia of a disk about its center is I=1

2MR2, where

Mis the mass of the disk. Substituting this into the equation, we have:

F r =1

2MR2α

Step 4: We can relate the linear acceleration aat the rim of the disk to the

angular acceleration αusing a=Rα. Substituting this into the equation, we

get:

F=1

2Ma

Step 5: Since the force Fis causing the disk to rotate, we can relate ato

Fby ma =F, where mis the mass of the disk. Substituting this into the

equation, we get:

F=1

2MF

m

Step 6: Solving for α, we find:

α=2F

Therefore, the angular acceleration of the disk is 2F

mR .

Question 31

Question

A solid sphere of radius Rstarts from rest and rolls without slipping down a

ramp that makes an angle θwith the horizontal. Determine the linear acceler-

ation of the center of the sphere as a function of its position ymeasured down

the ramp.

Solution

Step 1: Draw a free body diagram of the sphere. At any position ydown the

ramp, the forces acting on the sphere are its weight mg, the normal force N,

and the frictional force fs.

Step 2: Break the weight force into components. The weight mg can be

broken into two components parallel and perpendicular to the ramp. The com-

ponent parallel to the ramp is mg sin(θ) and the component perpendicular to

the ramp is mg cos(θ).

Step 3: Write down the equations of motion for the sphere. Using Newton’s

second law, we have two equations: a) In the direction parallel to the ramp

(x-direction):

max=fs

b) In the direction perpendicular to the ramp (y-direction):

may=N−mg cos(θ)

Step 4: Express the acceleration components in terms of angular accelera-

tion. Since the sphere is rolling without slipping, the linear acceleration aof

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

Step 5: Relate linear and angular acceleration. The linear acceleration ais

related to the angular acceleration αby a=Rα.

Step 6: Solve for linear acceleration as a function of position y. Using the

equations from step 3 and the relation between linear and angular acceleration,

we can solve for the linear acceleration as a function of y.ax=Rα =Rax

R=ax

ay=N−mg cos(θ) = mayTherefore, the linear acceleration aof the center

of the sphere as a function of its position yis a=mg sin(θ)

1 + 2

at any position

down the ramp.

Question 32

Question

A solid disk of radius 0.5 m is rolling along a horizontal surface without slipping.

The disk is initially at rest but then accelerates at a constant rate of 4 rad/s

for 3 seconds. Determine the angular velocity of the disk after 3 seconds and

the distance it travels during this time.

Solution

Step 1: Find the angular velocity of the disk after 3 seconds using the equation

of motion for rotational motion.

ωf=ωi+αt

where: ωf= final angular velocity, ωi= initial angular velocity (0 since the disk

starts from rest), α= angular acceleration (4 rad/s

), t= time (3 seconds).

Substitute the given values into the equation:

ωf= 0 + 4 ×3

ωf= 12 rad/s

So, the angular velocity of the disk after 3 seconds is 12 rad/s.

Step 2: Find the distance the disk travels during this time using the equation

for the distance traveled by a point on a rotating object.

s=rθ

where: s= distance traveled, r= radius of the disk (0.5 m), θ= angular

displacement.

From kinematics, the angular displacement (θ) can be found using the equa-

tion:

θ=ωit+1

2αt2

Substitute the given values into the equation:

θ= 0 ×3 + 1

2×4×(3)2

θ= 18 rad

Now, find the distance traveled by the disk:

s= 0.5×18

s= 9 m

Therefore, the angular velocity of the disk after 3 seconds is 12 rad/s and

the distance it travels during this time is 9 meters.

Question 33

Question

An object initially at rest starts rotating about a fixed axis with a constant

angular acceleration of 2.0 rad/s2. After 3.0 seconds, it has rotated through an

angle of 4.0 rad. Determine the angular velocity of the object at this time.

Solution

Step 1: Write down the known values. Given: - Angular acceleration, α=

2.0 rad/s2- Initial angular velocity, ω0= 0 (object starts at rest) - Time taken,

t= 3.0 s - Angle rotated, ∆θ= 4.0 rad

Step 2: Calculate the final angular velocity using the equation for angular

displacement in terms of initial angular velocity, angular acceleration, and time:

∆θ=ω0t+1

2αt2

Plugging in the known values:

4.0 rad = 0 + 1

2×2.0 rad/s2×(3.0 s)2

4.0 rad = 3.0 rad

Step 3: Calculate the final angular velocity, ωusing the equation:

ω=ω0+αt

Plugging in the known values:

ω= 0 + 2.0 rad/s2×3.0 s

ω= 6.0 rad/s

Therefore, the angular velocity of the object at this time is ω= 6.0 rad/s.

Question 34

Question

A thin rod of length Land mass Mis pivoted about one end. A force Fis

applied perpendicular to the rod at a distance dfrom the pivot point. If the

rod is initially at rest, find the angular acceleration of the rod.

Solution

Step 1: Begin by analyzing the torque acting on the rod. The torque (τ) acting

on the rod is given by:

τ=F·d·sin θ

where θis the angle between the force and the lever arm, d. Since the rod is at

rest initially, the net torque must be equal to the moment of inertia (I) times

the angular acceleration (α). Therefore, we have:

τ=I·α

Step 2: Express the moment of inertia and the angular acceleration. For a

thin rod rotating about one end, the moment of inertia is given by:

I=1

3ML2

Substitute this into the torque equation:

F·d·sin θ=1

3ML2α

Step 3: Determine the acceleration. The force component perpendicular to

the rod can be expressed as:

F⊥=Fsin θ

Since the rod is at rest, the sum of the torques is zero, leading to:

F·d·sin θ= 0.5ML2α

Step 4: Solve for the angular acceleration. Now, we can solve for the angular

acceleration:

α=2F·d·sin θ

ML2

Therefore, the angular acceleration of the rod is 2F·d·sin θ

ML2.

Question 35

Question

A solid sphere is initially at rest and then rolls without slipping down a 30-

degree incline. If the sphere’s mass is 2 kg and its radius is 0.1 m, what is its

linear acceleration down the incline?

Solution

Step 1: The net torque about the center of mass of the sphere can be calculated

using the formula τ=Iα, where τis the net torque, Iis the moment of inertia

of a solid sphere (2

5mR2), and αis the angular acceleration.

Step 2: To find the angular acceleration, we can use the kinematic equation

τnet =Iα =2

5mR2α=mR2a, where ais the linear acceleration.

Step 3: The net torque about the center of mass is given by τnet =mgsin(θ)R,

where mis the mass of the sphere, gis the acceleration due to gravity, and θis

the incline angle.

Step 4: Setting τnet equal to mR2a, we have mgsin(θ)R=mR2a.

Step 5: Solving for a, we get a=gsin(θ)R.

Step 6: Substituting g= 9.8 m/s2,θ= 30◦, and R= 0.1 m into the equation,

we find a= 9.8 m/s2×sin(30◦)×0.1 m.

Step 7: Calculating the linear acceleration, we get a= 9.8×1

2×0.1 =

0.49 m/s2.

Therefore, the linear acceleration of the sphere down the incline is 0.49 m/s2.

Question 5

Question

A thin-walled hollow sphere of radius 0.5 m and mass 2 kg is rotating about

a vertical axis through its center at 10 rad/s. A small ball of mass 0.1 kg is

dropped vertically onto the sphere and sticks to it. Find the new angular speed

of the system.

Solution

Step 1: Let’s first find the initial angular momentum of the system. The ini-

tial angular momentum (Li) of the system is given by the sum of the angular

momentum of the sphere and the angular momentum of the falling ball.

The angular momentum of the sphere:

Lsphere =Isphereωsphere

where Isphere is the moment of inertia of the sphere and ωsphere is the initial

angular velocity of the sphere.

The moment of inertia of a thin-walled hollow sphere is I=2

3mr2, where m

is the mass of the sphere and ris the radius.

Plugging in the values:

Isphere =2

3×2×(0.5)2=1

3kg m2

Given that ωsphere = 10 rad/s, we can find the angular momentum of the

sphere.

Step 2: The angular momentum of the falling ball just before it hits and

sticks to the sphere is mgh, where mis the mass of the ball, gis the acceleration

due to gravity, and his the height from which the ball is dropped.

Given that the height from which the ball is dropped is negligible compared

to the radius of the sphere, the initial angular momentum of the ball is mgh = 0.

Therefore, the initial angular momentum of the system (Li) is equal to the

angular momentum of the sphere.

Step 3: Let’s now find the final moment of inertia (If) of the system after

the ball sticks to the rotating sphere.

The moment of inertia of the sphere and the ball as a system is:

Isystem =Isphere +mr2

where mis the mass of the ball.

Plugging in the values:

Isystem =1

3+ 0.1×(0.5)2=1

3+ 0.025 = 1

3+1

40 =9

40 kg m2

Step 4: Using the principle of conservation of angular momentum, the final

angular speed of the system (ωf) can be calculated as:

Li=Lf

Isphereωsphere =Isystemωf

Solving for ωf:

ωf=Isphereωsphere

Isystem

Plugging in the values:

ωf=

3×10

=40

3×1

3=40

9= 4.4 rad/s

Therefore, the new angular speed of the system after the ball sticks to the

sphere is 4.4 rad/s.

Question 6

Question

A wheel rotating at 100 rad/s comes to rest after rotating through 100 revolu-

tions. Find the magnitude of the angular acceleration of the wheel.

Solution

Step 1: First, we calculate the initial angular velocity of the wheel in terms of

radians per second. Since 1 revolution is equivalent to 2πradians, the initial

angular velocity is given by:

ωi= 100 rad/s ×2πrad

1 rev = 200πrad/s

Step 2: Next, we calculate the final angular velocity of the wheel, which is

0 rad/s since it comes to rest.

Step 3: We use the equation of angular motion to relate the initial and final

angular velocities, angular acceleration, and angular displacement:

ω2

f=ω2

i+ 2αθ

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

velocity (200πrad/s), - αis the angular acceleration (to be determined), - θis

the angular displacement in radians (100 rev ×2πrad/rev).

Step 4: Substituting the known values into the equation gives:

0 = (200π)2+ 2α(100 ×2π)

Step 5: Solving for the angular acceleration, we find:

α=−(200π)2

200 ×2π=−200πrad/s2

Therefore, the magnitude of the angular acceleration of the wheel is 200π

rad/s2.

Question 7

Question

A thin cylindrical rod of mass mand radius Ris hinged at one end and set

into motion by applying a force perpendicular to the rod at a distance rfrom

the hinge. Initially, the rod is at rest in a horizontal position. Find the angular

acceleration of the rod as a function of time t.

Solution

1. The torque on the rod about the hinge can be calculated as τ=rF , where

Fis the applied force. The moment of inertia of a thin cylindrical rod rotating

about its end is I=1

3mR2, so the angular acceleration αcan be expressed as

τ=Iα.

2. Substituting the expressions for torque and moment of inertia into the

equation above, we have rF =1

3mR2α.

3. Rearranging the equation gives us the expression for angular acceleration:

α=3rF

mR2.

4. As the force Fis a constant, the angular acceleration αis also constant

and does not depend on time t.

Question 8

Question

A disc of radius 0.2 m is initially at rest. A constant torque of 4 N·m is applied

to the disc for 6 seconds, causing it to reach an angular velocity of 8 rad/s.

Calculate the moment of inertia of the disc.

Solution

Step 1: Recall the relationship between torque τ, moment of inertia I, angular

acceleration α, and radius r:

τ=Iα

Step 2: We are given that the torque applied is 4 N·m and the angular

acceleration can be calculated using the formula:

α=∆ω

∆t

Step 3: Substituting the given values into the formula:

α=8 rad/s −0

6 s =8 rad/s

6 s = 1.33 rad/s2

Step 4: Substitute the torque and angular acceleration into the torque for-

mula to solve for the moment of inertia I:

I=τ

α=4 N ·m

1.33 rad/s2= 3.01 kg ·m2

Step 5: The moment of inertia of the disc is 3.01 kg·m2.

Question 9

Question

A rotating platform initially at rest, experiences a constant angular acceleration

of 0.05 rad/s2. At time t= 6 s, the platform has rotated through an angle of 1

rad. What is the angular velocity of the platform at t= 6 s?

Solution

Step 1: First, let’s find the angular velocity of the platform at t= 6 s using the

equation of rotational kinematics:

ω=ω0+αt

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

0 rad/s since the platform starts at rest), αis the angular acceleration, and tis

the time.

Step 2: Substitute the given values into the equation:

ω= 0 + (0.05 rad/s2)×6 s

ω= 0.3 rad/s

Therefore, the angular velocity of the platform at t= 6 s is 0.3 rad/s.

Question 10

Question

A thin rod of mass Mand length Lis pivoted about one end and released from

rest in a vertical position. What is the angular speed ωof the rod when it

swings by 30◦below the horizontal?

Solution

Step 1: To find the angular speed ωof the rod when it swings by 30◦below

the horizontal, we will use conservation of energy. Initially, the rod is at rest

in a vertical position, and when it swings below the horizontal, all its potential

energy will convert into kinetic energy.

Step 2: The potential energy of the rod initially is Ui=M gh, where his

the height of the center of mass and h=L

Step 3: The initial potential energy Uiis equal to the final kinetic energy

Kf=1

2Iω2, where Iis the moment of inertia of the rod about the pivot point.

Step 4: The moment of inertia of a rod rotating about one end is I=1

3ML2.

Step 5: Setting the initial potential energy equal to the final kinetic energy,

we have Mgh =1

21

3ML2ω2.

Step 6: Substituting in the values of h,I, and Mgives MgL =1

6ML2ω2.

Step 7: Solving for ω, we find ω=q12g

Step 8: Therefore, the angular speed of the rod when it swings by 30◦below

the horizontal is ω=q12g

Question 11

Question

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

velocity ω. If a small piece of mass mis attached to the free end of the rod,

determine the angular velocity of the system.

Solution

Let’s denote the moment of inertia of the rod about its end as Irod, the moment

of inertia of the mass mabout the end of the rod as Imass, and the total moment

of inertia of the system as Itotal.

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

thin rod rotating about one end is given by Irod =1

3ML2.

Step 2: Find the moment of inertia of the mass. Since the mass mis

attached to the free end of the rod, Imass =m·L2.

Step 3: Find the total moment of inertia of the system. The total moment

of inertia of the system is the sum of the moments of inertia of the rod and the

mass, Itotal =Irod +Imass.

Step 4: Apply the conservation of angular momentum. The initial angular

momentum of the system is equal to the final angular momentum, which means

Irodω=Itotalωfinal.

Step 5: Solve for the final angular velocity ωfinal. Substitute the expressions

for Irod,Imass, and Itotal into the conservation of angular momentum equation

and solve for ωfinal.

3ML2ω=1

3ML2+mL2ωfinal

ωfinal =

3M+m

3M+mω

ωfinal =ω

Therefore, the angular velocity of the system with the added mass remains

the same as the initial angular velocity ω.

Question 12

Question

A disk of radius Ris initially at rest. A constant tangential force Ftis applied

at the edge of the disk for a given time interval, causing the disk to rotate. Find

an expression for the angular velocity ωof the disk in terms of Ft,R, and the

moment of inertia Iof the disk during this time interval.

Solution

Step 1: We start by using Newton’s second law for rotation, which states that

the net torque τacting on an object is equal to the moment of inertia Itimes

the angular acceleration α. The torque is given by τ=R·Ft, where Ris the

radius of the disk.

Step 2: We substitute this expression for torque back into Newton’s second

law for rotation to get R·Ft=I·α. We also know that α=dω

dt , where ωis the

angular velocity of the disk.

Step 3: From the definition of moment of inertia for a disk (rotating around

its symmetry axis), we have I=1

2MR2, where Mis the mass of the disk.

Step 4: Substituting the expression for Iinto the equation R·Ft=I·α, we

have R·Ft=1

2MR2·dω

dt .

Step 5: We can simplify this expression to obtain Ft

2=dω

dt . Integrating both

sides with respect to tgives RFt

2dt =Rdω.

Step 6: This simplifies to Ft

2t=ω+C, where Cis a constant of integration.

Since the disk starts from rest, ω= 0 at t= 0, so the constant Cis 0.

Step 7: Therefore, the expression for the angular velocity ωof the disk in

terms of Ft,R, and Iduring the time interval twhen the force is applied is

given by ω=Ft

2t.

Thus, the angular velocity ωof the disk is directly proportional to the tan-

gential force Ftand the time tfor which the force is applied, and inversely

proportional to the moment of inertia Iof the disk.

Question 13

Question

A disc of radius rand mass mis rotating about a fixed axis passing through its

center. If the angular velocity is given by ω(t) = αt +βt2−γt3, where α,β,

and γare constants, find the expression for the angular acceleration of the disc

as a function of time.

Solution

Step 1: To find the angular acceleration α(t), we need to differentiate the angular

velocity function ω(t) with respect to time.

Step 1: Given the angular velocity function ω(t) = αt +βt2−γt3,

Differentiate with respect to time: α(t) = dω

dt =α+ 2βt −3γt2.

Step 2: Therefore, the expression for the angular acceleration α(t) as a

function of time is α(t) = α+ 2βt −3γt2.

Question 14

Question

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

acceleration of 2 rad/s2for 5 seconds. Calculate the angular velocity of the disc

after this time period.

Solution

Step 1: Find the angular displacement of the disc. Given that the disc starts

from rest, the angular displacement θcan be calculated using the kinematic

equation:

θ=1

2αt2

θ=1

2×2×(52)

θ= 25 rad

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

of the disc can be found using the kinematic equation:

ωf=ωi+αt

ωf= 0 + 2 ×5

ωf= 10 rad/s

Therefore, the angular velocity of the disc after 5 seconds is 10 rad/s.

Question 15

Question

A wheel initially at rest accelerates uniformly for 10 seconds, reaching an angular

velocity of 30 rad/s. It then decelerates uniformly for 5 seconds until it comes

to a stop. If the total angle rotated during this whole process is 450 radians,

what is the average angular acceleration of the wheel during the entire motion?

Solution

Step 1: Find the angular acceleration during the first part of the motion. Given

that the initial angular velocity (ωi= 0 rad/s), the final angular velocity (ωf=

30 rad/s), and the time duration (t= 10 s), we can use the equation for angular

acceleration:

α=∆ω

∆t

α=30 rad/s −0 rad/s

10 s = 3 rad/s2

Step 2: Find the angle rotated during the first part of the motion. We can

use the equation for angular displacement:

θ=ωit+1

2αt2

Substitute the known values to find θ:

θ= 0 ×10 + 1

2×3×(10)2= 150 radians

Step 3: Find the angular acceleration during the second part of the motion.

Since the wheel decelerates uniformly, the final angular velocity (ωf= 0 rad/s),

the initial angular velocity (ωi= 30 rad/s), and the time duration (t= 5 s) can

be used to find the angular acceleration:

α=∆ω

∆t

α=0 rad/s −30 rad/s

5 s =−6 rad/s2(negative sign indicates deceleration)

Step 4: Find the angle rotated during the second part of the motion. Using

the same angular displacement equation:

θ=ωit+1

2αt2

Substitute the known values to find θ:

θ= 30 ×5 + 1

2×(−6) ×(5)2= 75 radians

Step 5: Determine the average angular acceleration. The total angle rotated

is 450 radians, and the total time duration is 15 seconds. The average angular

acceleration can be found using the total change in angular velocity and the

total time:

Average α=ωtotal

ttotal

Average α=30 rad/s −0 rad/s

15 s = 2 rad/s2

Therefore, the average angular acceleration of the wheel during the entire

motion is 2 rad/s2.

Question 16

Question

A disc of mass Mand radius Ris rotating with an angular velocity ω. A small

object of mass mand radius rfalls gently onto the disc and sticks to it. The

moment of inertia of the disc alone is I. Determine the final angular velocity of

the system.

Solution

Step 1: Initially, the disc is rotating with angular velocity ωand the small

object falls gently onto it and comes to rest relative to the disc. Conservation of

angular momentum can be used to determine the final angular velocity of the

system.

Step 2: The initial angular momentum of the disc-object system is Iω.

Step 3: The final angular momentum of the system is (I+m(r+R)2)Ω,

where Ω is the final angular velocity of the system.

Step 4: Since angular momentum is conserved, we have Iω = (I+m(r+

R)2)Ω.

Step 5: Solving for Ω, we get Ω = Iω

I+m(r+R)2.

Step 6: Substituting the given values of the moment of inertia Ias well as

the radii of the disc and the object, we find the final angular velocity of the

system as Ω = Iω

I+m(r+R)2.

Question 17

Question

A disk of radius Rrolls without slipping along a horizontal surface. Starting

from rest, the disk accelerates uniformly for a time tand reaches a final angular

speed of ω. Find the distance dit has rolled along the surface in terms of R,t,

and ω.

Solution

Step 1: Find the angular acceleration αof the disk. The final angular speed ω

can be related to the initial angular speed ω0= 0, the time t, and the angular

acceleration αthrough the equation:

ω=ω0+αt

Since the initial angular speed is 0, we have:

ω=αt

Therefore, the angular acceleration is:

α=ω

Step 2: Find the distance drolled by the disk. The distance rolled by the disk

can be determined by considering the relationship between linear and angular

quantities. The linear distance rolled dis related to the angle turned by the

disk by the equation:

d=Rθ

where θis the angle turned by the disk. The angle turned by the disk can be

expressed in terms of angular speed and time using the equation:

θ=1

2αt2

Substitute the expression for αfound in Step 1:

θ=1

2ω

tt2=ω

Therefore, the distance rolled by the disk is:

d=R·ω

2t=1

2Rωt

Thus, the distance rolled by the disk in terms of R,t, and ωis 1

2Rωt.

Question 18

Question

A thin circular ring of radius Ris rotating with an angular velocity ωabout an

axis passing through its center and perpendicular to the plane of the ring. A

small object of mass mis placed gently on the surface of the ring. What is the

minimum coefficient of static friction between the object and the ring that will

prevent the object from slipping?

Solution

1. The forces acting on the object on the rotating ring include gravitational

force, normal force, static frictional force, and a pseudo force due to the non-

inertial frame of reference. The object undergoes circular motion with radius

R, so the sum of forces in the radial direction should provide the necessary

centripetal force:

mg −N=mv2

2. Since the object is at rest relative to the rotating frame, the centripetal

acceleration must be zero. Therefore, the net force in the radial direction is also

zero:

N=mg

3. The maximum frictional force is given by fmax =µN, where µis the coef-

ficient of static friction. The object will begin to slip when the frictional force

approaches this maximum value:

µminN=mv2

4. Substituting the expression for Nfrom step 2:

µminmg =mv2

5. Cancelling mass mfrom both sides:

µming=v2

6. The tangential velocity of the object is related to the angular velocity of the

ring:

v=Rω

7. Substituting this into the equation from step 5:

µming=(Rω)2

R=Rω2

8. Therefore, the minimum coefficient of static friction needed to prevent slip-

ping is µmin =ω2/g.

Question 19

Question

A solid cylinder of radius R, mass M, and moment of inertia I=1

2MR2is

initially at rest. A constant force Fis applied tangentially to the edge of the

cylinder for a distance d, causing the cylinder to move. What is the angular

velocity of the cylinder after the force is applied and the distance it travels? Use

that the kinetic energy of a rotating object is given by K=1

2Iω2.

Solution

Step 1: Calculate the torque applied to the cylinder. The torque, τ, applied to

the cylinder is given by τ=F r, where ris the radius of the cylinder. Given

that τ=Iα, where αis the angular acceleration, we can write F r =Iα.

Step 2: Find the angular acceleration of the cylinder. Since the cylinder

starts from rest, the initial angular velocity, ωi, is 0. The final angular velocity,

ωf, can be found using the relationship ω2

f=ω2

i+ 2αθ, where θis the angular

displacement. Here, θ=d

R, so we have ω2

f= 2αd

R.

Step 3: Solve for the angular acceleration. Substitute F r =Iα into the

equation F r =Iα to solve for α:F·R=I·α, giving F·R=1

2MR2α. Thus,

α=2F

Step 4: Calculate the final angular velocity. Substitute α=2F

Minto the

equation ω2

f= 2αd

R:ω2

f= 2 2F

Md

R, so ωf=q4F d

Step 5: Find the distance traveled by the cylinder. We know that v=ωR,

so the linear velocity at the edge of the cylinder is v=ωfR. The distance

traveled, d, is given by d=vt, so d=ωfR·t=q4F d

M·R·d

Therefore, the angular velocity of the cylinder after the application of force

is q4F d

Mrad/s, and the distance traveled by the cylinder is 2F d

Mmeters.

Question 20

Question

A disk of radius 0.5 m is rotating about an axis through its center with an

angular acceleration of 3 rad/s2. At t= 2 s, the angular velocity of the disk is

4 rad/s in the counterclockwise direction. If the angular acceleration is constant,

determine the angular velocity of the disk when t= 4 s.

Solution

Step 1: Find the initial angular velocity of the disk using the angular acceleration

and time provided. Given: r= 0.5 m, α= 3 rad/s2,t= 2 s, and ω0= 0 rad/s.

The angular velocity at t= 2 s can be found using the equation ω=ω0+αt:

ω= 0 + 3 ×2 = 6 rad/s

Step 2: Find the angular displacement of the disk from t= 2 s to t= 4 s.

The angular displacement θcan be found using the equation θ=ω0t+1

2αt2:

θ= 6 ×2 + 1

2×3×22= 12 + 6 = 18 rad

Step 3: Use the angular displacement to find the angular velocity of the

disk at t= 4 s. The final angular velocity can be found using the equation

ω2=ω2

0+ 2αθ:

ω2= 62+ 2 ×3×18 = 36 + 36 = 72

ω=√72 = 6√2≈8.49 rad/s

Therefore, the angular velocity of the disk when t= 4 s is approximately

8.49 rad/s in the counterclockwise direction.

Question 21

Question

A solid sphere of radius Ris rolling without slipping along a horizontal surface.

The sphere is given an initial angular speed ω0and an initial linear speed v0at

the same time. If the sphere rolls without slipping for a distance d, determine

the final values of its linear speed and angular speed.

Solution

Let’s denote the final linear speed of the sphere as vfand the final angular speed

as ωf.

Step 1: Relate the linear speed and angular speed: Since the sphere is rolling

without slipping, we have the following relationship between linear speed and

angular speed:

vf=ωfR

Step 2: Apply conservation of energy: The initial kinetic energy of the

sphere consists of both translational and rotational kinetic energy. The final

kinetic energy of the sphere will also consist of translational and rotational

kinetic energy. Therefore, we can write:

2Iω2

0+1

2mv2

0=1

2Iω2

f+1

2mv2

Step 3: Substitute vf=ωfRinto the conservation of energy equation:

2Iω2

0+1

2m(v0)2=1

2Ivf

R2+1

2m(vf)2

Step 4: Substitute I=2

5mR2into the equation:

5mR2ω2

0+1

2mv2

0=1

5mv2

f+1

2mv2

Step 5: Simplify and solve for vf:

10mv2

0=3

5mv2

f=7

6v2

vf=r7

6v0

Step 6: Find ωf:

vf=ωfR

ωf=vf

R=p7/6v0

Therefore, the final linear speed of the sphere is q7

6v0and the final angular

speed is √7/6v0

Question 22

Question

A disk with radius Rstarts from rest and accelerates with a constant angular

acceleration αfor a time t. Find an expression for the angle through which the

disk turns during this time.

Solution

Step 1: We can start by finding the angular velocity of the disk at time t.

We know that angular acceleration is defined as the rate of change of angular

velocity. Therefore, we have

α=ω

Solving for ω, we get

ω=αt

Step 2: Next, we can find the angle turned by the disk in time t. The

relationship between angular displacement θ, angular velocity ω, and time tis

given by

θ=ωt

Substitute in the expression for ωwe found earlier:

θ= (αt)t=αt2

Step 3: Therefore, the expression for the angle through which the disk turns

during the time tis θ=αt2.

Question 23

Question

A wheel initially at rest accelerates with a constant angular acceleration of

2.0 rad/s2for 10 seconds. If the wheel makes 5 complete revolutions during this

time, what is the wheel’s angular velocity at the end of the 10 seconds?

Solution

Step 1: Calculate the final angular displacement of the wheel during the 10-

second interval. We know that angular displacement θis related to the number

of revolutions made by the wheel:

θ= 2π×number of revolutions

Given that the wheel makes 5 complete revolutions, θ= 2π×5 = 10πrad.

Step 2: Use the formula for angular displacement with constant angular

acceleration to find the final angular velocity. The formula relating angular

displacement, initial angular velocity, angular acceleration, and time is:

θ=ωit+1

2αt2

Where: - θ= 10πrad - ωi= 0 rad/s (initial angular velocity) - α= 2.0 rad/s2

(angular acceleration) - t= 10 s Plugging in the values:

10π= 0 ×10 + 1

2×2.0×102

10π= 100

From this, we find that the final angular velocity is 10 rad/s.

Question 24

Question

A solid sphere of mass 2 kg and radius 0.4 m starts from rest and rolls without

slipping down a 30-degree incline. How fast is the sphere rolling after it has

traveled 3 meters along the incline? (Neglect air resistance and friction)

Solution

Step 1: We can start by calculating the acceleration of the sphere along the

incline. The net acceleration of the sphere can be found by considering the

forces acting on it. The component of the gravitational force parallel to the

incline will cause linear acceleration (acm) and angular acceleration (α).

Step 2: The component of the gravitational force parallel to the incline can

be calculated as:

F|| =mg sin(30◦)

Step 3: The linear acceleration of the center of mass (acm) can be found

using Newton’s second law:

macm =mg sin(30◦)

acm =gsin(30◦)

Step 4: The angular acceleration (α) can be found using the relation between

linear acceleration and angular acceleration for a rolling object:

acm =Rα

gsin(30◦) = Rα

Step 5: The equation for rolling without slipping relates linear velocity (v)

and angular velocity (ω) as:

v=Rω

Step 6: The centripetal acceleration of a rolling object is given by:

acp =Rα

Therefore, v2=Rα

Step 7: Now that we have the linear acceleration, we can find the velocity

of the sphere after traveling 3 meters. We can use the kinematic equation:

v2=u2+ 2acms

where uis the initial velocity, and sis the distance traveled.

Step 8: Substituting the given values into the equation, we can solve for the

final velocity (v).

Question 25

Question

A disk with a radius of 0.1 m is spinning counterclockwise at an angular velocity

of 10 rad/s. A point on the edge of the disk releases a bug, which lands 0.04 m

from the center of the disk. What is the bug’s speed immediately after landing?

Solution

1. First, calculate the linear velocity of the point on the edge of the disk using

the formula v=ωr, where vis the linear velocity, ωis the angular velocity, and

ris the radius of the disk.

2. Substitute the values: v= 10 rad/s ×0.1 m = 1 m/s.

3. Next, find the linear velocity component tangential to the circular path

at the point where the bug lands. This can be found using the formula vt=rω,

where vtis the tangential velocity, ωis the angular velocity, and ris the distance

from the center where the bug lands.

4. Substitute the values: vt= 0.04 m ×10 rad/s = 0.4 m/s.

5. The bug’s total speed is the vector sum of the linear velocity and the tan-

gential velocity at the landing point. This can be found using the Pythagorean

theorem: vtotal =pv2+v2

6. Substitute the values: vtotal =√12+ 0.42≈√1.16 ≈1.08 m/s.

7. Therefore, the bug’s speed immediately after landing is approximately

1.08 m/s.

Question 26

Question

A thin rod of length Land mass Mis free to rotate in a vertical plane about a

pivot at one end. The rod is released from rest at an angle θ0with the vertical.

What is the angular velocity of the rod when it is in the horizontal position?

Solution

Step 1: First, let’s determine the initial potential and kinetic energy of the

system. The initial potential energy of the rod is given by:

Ui=−MgL cos θ0

The initial kinetic energy of the rod is zero since it is released from rest.

Step 2: Next, let’s determine the final potential and kinetic energy of the

system when the rod is in the horizontal position. The final potential energy of

the rod is given by:

Uf=−MgL cos(90◦)=0

The final kinetic energy of the rod consists of both translational and rotational

kinetic energy. Since the rod is horizontal, all the initial potential energy is

converted into kinetic energy.

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

energy, so we have:

MgL cos θ0=1

2Iω2

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

Step 4: The moment of inertia of a rod rotating about one end is I=1

3ML2.

Substituting this into the equation, we get:

MgL cos θ0=1

21

3ML2ω2

Step 5: Solving for ω, we find:

ω=r3gcos θ0

Therefore, the angular velocity of the rod when it is in the horizontal position

is q3gcos θ0

2L.

Question 27

Question

A disk of radius 0.15 m starts from rest and has an angular acceleration of

1.5 rad/s2.

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

2. How many revolutions has the disk made after 6 seconds?

Solution

1. We can use the equation for angular motion:

ωf=ωi+αt

where ωfis the final angular speed, ωiis the initial angular speed (which is 0

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

Step 1: Substitute the known values into the equation.

ωf= 0 + (1.5 rad/s2)(3 s)

ωf= 4.5 rad/s

So, the angular speed of the disk after 3 seconds is 4.5 rad/s.

2. To find the number of revolutions, we need to first calculate the total

angular displacement of the disk after 6 seconds:

Step 2: Use the equation for angular displacement:

θ=1

2αt2

where θis the angular displacement.

Substitute the known values:

θ=1

2(1.5 rad/s2)(6 s)2

θ= 27πrad

To determine the number of revolutions, we need to convert this into revo-

lutions:

Number of revolutions = Total angular displacement

2π

Number of revolutions = 27π

2π= 13.5 rev

Therefore, after 6 seconds, the disk has made 13.5 revolutions.

Question 28

Question

A uniform disk of radius Rand mass Mis rotating about a vertical axis passing

through its center with an angular velocity ω0. A small object of mass mslides

without friction along a radial slot in the disk from the rim to the center. What

is the angular velocity of the disk-object system after the object reaches the

center?

Solution

1. Conservation of angular momentum: Initially, the total angular mo-

mentum of the system is given by the angular momentum of the disk and the

angular momentum of the object:

Linitial =I0ω0+mR2ω0=I0ω0+mR2v

2. Conservation of mechanical energy: The initial mechanical energy

of the system is equal to the final mechanical energy:

Einitial =1

2I0ω2

0+1

2mv2=Efinal =1

2Ifω2

3. From conservation of angular momentum:

I0ω0+mR2ω0=Ifωf

I0ω0+mR2v

R=Ifωf

4. The moment of inertia of the disk-object system after the object reaches

the center is:

If=I0+mR2

5. Substituting Ifinto the equation from step 3:

I0ω0+mR2ω0= (I0+mR2)ωf

(I0+mR2)ωf=I0ω0+mR2ω0

6. Solving for ωf:

ωf=I0ω0+mR2ω0

I0+mR2

Question 29

Question

An object starts from rest and rotates with a constant angular acceleration of

2.5 rad/s2. If the object reaches an angular velocity of 7.0 rad/s after rotating

through an angle of 5.0 rad, determine the angular velocity of the object when

it has rotated through an angle of 12 rad.

Solution

Step 1: Determine the final angular velocity using the given information.

Given: α= 2.5 rad/s2, ∆θ= 5.0 rad, vi= 0 rad/s, and vf= 7.0 rad/s. We can

use the equation connecting angular velocity, initial angular velocity, angular

acceleration, and angular displacement:

f=v2

i+ 2α∆θ

Substitute the given values:

(7.0)2= 02+ 2(2.5)(5.0)

49 = 25

Since the equation is invalid, there seems to be a mistake in the given values.

Let’s reevaluate.

Step 2: Reevaluate the given values.

It seems there was a mistake in the calculation in Step 1. Let’s recalculate the

final angular velocity using the correct equation:

vf=qv2

i+ 2α∆θ

vf=p0 + 2(2.5)(5.0)

vf=√25 = 5.0 rad/s

Step 3: Determine the angular velocity when the object has rotated through

an angle of 12 rad.

Using the equation relating final angular velocity, initial angular velocity, angu-

lar acceleration, and angular displacement:

f=v2

i+ 2α∆θ

f= 0 + 2(2.5)(12)

vf=√60 ≈7.75 rad/s

Therefore, the angular velocity of the object when it has rotated through an

angle of 12 rad is approximately 7.75 rad/s.

Question 30

Question

A circular disk of radius Rrests on a frictionless horizontal surface. A constant

force Fis applied to the rim of the disk perpendicular to the radius as shown

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

where Iis the moment of inertia of the disk about its center.

Solution

Step 1: The torque acting on the disk is given by τ=F r, where ris the radius

of the disk. By Newton’s second law for rotation, we have:

τ=Iα

where αis the angular acceleration of the disk and Iis the moment of inertia.

Step 2: Substituting τ=F r and rearranging the equation, we get:

F r =Iα

Step 3: The moment of inertia of a disk about its center is I=1

2MR2, where

Mis the mass of the disk. Substituting this into the equation, we have:

F r =1

2MR2α

Step 4: We can relate the linear acceleration aat the rim of the disk to the

angular acceleration αusing a=Rα. Substituting this into the equation, we

get:

F=1

2Ma

Step 5: Since the force Fis causing the disk to rotate, we can relate ato

Fby ma =F, where mis the mass of the disk. Substituting this into the

equation, we get:

F=1

2MF

m

Step 6: Solving for α, we find:

α=2F

Therefore, the angular acceleration of the disk is 2F

mR .

Question 31

Question

A solid sphere of radius Rstarts from rest and rolls without slipping down a

ramp that makes an angle θwith the horizontal. Determine the linear acceler-

ation of the center of the sphere as a function of its position ymeasured down

the ramp.

Solution

Step 1: Draw a free body diagram of the sphere. At any position ydown the

ramp, the forces acting on the sphere are its weight mg, the normal force N,

and the frictional force fs.

Step 2: Break the weight force into components. The weight mg can be

broken into two components parallel and perpendicular to the ramp. The com-

ponent parallel to the ramp is mg sin(θ) and the component perpendicular to

the ramp is mg cos(θ).

Step 3: Write down the equations of motion for the sphere. Using Newton’s

second law, we have two equations: a) In the direction parallel to the ramp

(x-direction):

max=fs

b) In the direction perpendicular to the ramp (y-direction):

may=N−mg cos(θ)

Step 4: Express the acceleration components in terms of angular accelera-

tion. Since the sphere is rolling without slipping, the linear acceleration aof

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

Step 5: Relate linear and angular acceleration. The linear acceleration ais

related to the angular acceleration αby a=Rα.

Step 6: Solve for linear acceleration as a function of position y. Using the

equations from step 3 and the relation between linear and angular acceleration,

we can solve for the linear acceleration as a function of y.ax=Rα =Rax

R=ax

ay=N−mg cos(θ) = mayTherefore, the linear acceleration aof the center

of the sphere as a function of its position yis a=mg sin(θ)

1 + 2

at any position

down the ramp.

Question 32

Question

A solid disk of radius 0.5 m is rolling along a horizontal surface without slipping.

The disk is initially at rest but then accelerates at a constant rate of 4 rad/s

for 3 seconds. Determine the angular velocity of the disk after 3 seconds and

the distance it travels during this time.

Solution

Step 1: Find the angular velocity of the disk after 3 seconds using the equation

of motion for rotational motion.

ωf=ωi+αt

where: ωf= final angular velocity, ωi= initial angular velocity (0 since the disk

starts from rest), α= angular acceleration (4 rad/s

), t= time (3 seconds).

Substitute the given values into the equation:

ωf= 0 + 4 ×3

ωf= 12 rad/s

So, the angular velocity of the disk after 3 seconds is 12 rad/s.

Step 2: Find the distance the disk travels during this time using the equation

for the distance traveled by a point on a rotating object.

s=rθ

where: s= distance traveled, r= radius of the disk (0.5 m), θ= angular

displacement.

From kinematics, the angular displacement (θ) can be found using the equa-

tion:

θ=ωit+1

2αt2

Substitute the given values into the equation:

θ= 0 ×3 + 1

2×4×(3)2

θ= 18 rad

Now, find the distance traveled by the disk:

s= 0.5×18

s= 9 m

Therefore, the angular velocity of the disk after 3 seconds is 12 rad/s and

the distance it travels during this time is 9 meters.

Question 33

Question

An object initially at rest starts rotating about a fixed axis with a constant

angular acceleration of 2.0 rad/s2. After 3.0 seconds, it has rotated through an

angle of 4.0 rad. Determine the angular velocity of the object at this time.

Solution

Step 1: Write down the known values. Given: - Angular acceleration, α=

2.0 rad/s2- Initial angular velocity, ω0= 0 (object starts at rest) - Time taken,

t= 3.0 s - Angle rotated, ∆θ= 4.0 rad

Step 2: Calculate the final angular velocity using the equation for angular

displacement in terms of initial angular velocity, angular acceleration, and time:

∆θ=ω0t+1

2αt2

Plugging in the known values:

4.0 rad = 0 + 1

2×2.0 rad/s2×(3.0 s)2

4.0 rad = 3.0 rad

Step 3: Calculate the final angular velocity, ωusing the equation:

ω=ω0+αt

Plugging in the known values:

ω= 0 + 2.0 rad/s2×3.0 s

ω= 6.0 rad/s

Therefore, the angular velocity of the object at this time is ω= 6.0 rad/s.

Question 34

Question

A thin rod of length Land mass Mis pivoted about one end. A force Fis

applied perpendicular to the rod at a distance dfrom the pivot point. If the

rod is initially at rest, find the angular acceleration of the rod.

Solution

Step 1: Begin by analyzing the torque acting on the rod. The torque (τ) acting

on the rod is given by:

τ=F·d·sin θ

where θis the angle between the force and the lever arm, d. Since the rod is at

rest initially, the net torque must be equal to the moment of inertia (I) times

the angular acceleration (α). Therefore, we have:

τ=I·α

Step 2: Express the moment of inertia and the angular acceleration. For a

thin rod rotating about one end, the moment of inertia is given by:

I=1

3ML2

Substitute this into the torque equation:

F·d·sin θ=1

3ML2α

Step 3: Determine the acceleration. The force component perpendicular to

the rod can be expressed as:

F⊥=Fsin θ

Since the rod is at rest, the sum of the torques is zero, leading to:

F·d·sin θ= 0.5ML2α

Step 4: Solve for the angular acceleration. Now, we can solve for the angular

acceleration:

α=2F·d·sin θ

ML2

Therefore, the angular acceleration of the rod is 2F·d·sin θ

ML2.

Question 35

Question

A solid sphere is initially at rest and then rolls without slipping down a 30-

degree incline. If the sphere’s mass is 2 kg and its radius is 0.1 m, what is its

linear acceleration down the incline?

Solution

Step 1: The net torque about the center of mass of the sphere can be calculated

using the formula τ=Iα, where τis the net torque, Iis the moment of inertia

of a solid sphere (2

5mR2), and αis the angular acceleration.

Step 2: To find the angular acceleration, we can use the kinematic equation

τnet =Iα =2

5mR2α=mR2a, where ais the linear acceleration.

Step 3: The net torque about the center of mass is given by τnet =mgsin(θ)R,

where mis the mass of the sphere, gis the acceleration due to gravity, and θis

the incline angle.

Step 4: Setting τnet equal to mR2a, we have mgsin(θ)R=mR2a.

Step 5: Solving for a, we get a=gsin(θ)R.

Step 6: Substituting g= 9.8 m/s2,θ= 30◦, and R= 0.1 m into the equation,

we find a= 9.8 m/s2×sin(30◦)×0.1 m.

Step 7: Calculating the linear acceleration, we get a= 9.8×1

2×0.1 =

0.49 m/s2.

Therefore, the linear acceleration of the sphere down the incline is 0.49 m/s2.

PHYS 305 - INTRODUCTION TO MODERN PHYSICS Rotational kinematics Question Bank Set 3

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