PHYS 305 - INTRODUCTION TO MODERN PHYSICS Rotational kinematics Question Bank Set 4 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Rotational

kinematics

Question Bank - Set 4

Liberty University

Question 1

Question

A disc of radius 0.5 m is rotating with an angular velocity of 4 rad/s. A car

is initially at rest 5 m from the center of the disc. The car begins moving

towards the center of the disc with a constant acceleration of 2 m/s2. What is

the angular velocity of the car when it reaches the center of the disc?

Solution

Step 1: Find the initial angular velocity of the car. The initial angular velocity

of the car can be calculated using the formula ω=ω0+αt, where ω0is the initial

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

the car is initially at rest, ω0= 0. We also know that the acceleration is 2 m/s2

and the initial distance is 5 m. The linear acceleration can be related to the

angular acceleration using a=Rα, where ais the linear acceleration, Ris the

radius, and αis the angular acceleration. Solving for α, we get α=a

R=2

5= 0.4

rad/s2.

Step 2: Find the time taken for the car to reach the center. The time taken

for the car to reach the center of the disc can be calculated using the equation

s=ut +1

2at2, where sis the initial distance (5 m), uis the initial velocity (0

m/s), ais the acceleration (2 m/s2), and tis the time taken. Rearranging the

equation, we have t2+ 0.8t−5 = 0. Solving this quadratic equation, we get

t≈2.56 seconds.

Step 3: Find the final angular velocity of the car. Using the formula ω=

ω0+αt and plugging in the values ω0= 0, α= 0.4 rad/s2, and t= 2.56 s, we

get ω= 0 + 0.4×2.56 ≈1.024 rad/s.

Therefore, the angular velocity of the car when it reaches the center of the

disc is approximately 1.024 rad/s.

Question 2

Question

A disk with a radius of 0.5 meters is rotating with an angular velocity of 4

radians per second. At a certain instant, the angular acceleration of the disk is

calculated to be 2 radians per second squared. What is the linear acceleration

of a point on the rim of the disk at this instant?

Solution

Step 1: Calculate the linear velocity of a point on the rim of the disk using the

angular velocity. Step 2: Use the linear velocity and the angular acceleration to

find the linear acceleration.

Step 1: The linear velocity of a point on the rim of the disk can be calculated

using the formula:

v=rω

where ris the radius of the disk and ωis the angular velocity. Given that

r= 0.5 m and ω= 4 rad/s, we have:

v= 0.5×4 = 2 m/s

Step 2: The linear acceleration of a point on the rim of the disk can be

found using the formula:

a=rα

where ris the radius of the disk and αis the angular acceleration. Given that

r= 0.5 m and α= 2 rad/s

, we have:

a= 0.5×2 = 1 m/s2

Therefore, the linear acceleration of a point on the rim of the disk at this

instant is 1 m/s

Question 3

Question

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

5.0 rad/s2for 4.0 seconds. If the radius of the disk is 0.10 meters, what is the

linear acceleration of a point on the rim of the disk at t= 4.0 s?

Solution

Step 1: Find the angular velocity at t= 4.0 s using the equation ωf=ωi+αt,

where αis the angular acceleration. Given: Initial angular velocity, ωi= 0 rad/s

Angular acceleration, α= 5.0 rad/s2Time, t= 4.0 s

ωf=ωi+αt = 0 + 5.0×4.0 = 20.0 rad/s

Step 2: Calculate the linear velocity at the rim of the disk using v=rω.

Given: Radius, r= 0.10 m Angular velocity, ω= 20.0 rad/s

v=rω = 0.10 ×20.0=2.0 m/s

Step 3: Determine the linear acceleration of a point on the rim of the disk

using a=rα. Given: Radius, r= 0.10 m Angular acceleration, α= 5.0 rad/s2

a=rα = 0.10 ×5.0=0.50 m/s2

Therefore, the linear acceleration of a point on the rim of the disk at t= 4.0 s

is 0.50 m/s2.

Question 4

Question

A rigid body undergoing rotational motion starts from rest and accelerates

uniformly for 5.0 s reaching a final angular velocity of 8.0 rad/s. The body then

decelerates uniformly for the next 7.0 s until it comes to rest. Find: (a) the

angular acceleration during the acceleration phase, (b) the number of revolutions

the body makes during the deceleration phase.

Solution

(a) During the acceleration phase: Step 1: Determine the initial angular ve-

locity. The body starts from rest, so the initial angular velocity, ωi, is 0 rad/s.

Step 2: Calculate the angular acceleration, α. The final angular velocity,

ωf, is 8.0 rad/s. The time taken, t, is 5.0 s. The angular acceleration is given

by:

α=ωf−ωi

Substitute the given values:

α=8.0 rad/s −0 rad/s

5.0 s = 1.6 rad/s2

Therefore, the angular acceleration during the acceleration phase is 1.6 rad/s2.

(b) During the deceleration phase: Step 1: Determine the final angular

velocity. The body comes to rest, so the final angular velocity, ωf, is 0 rad/s.

Step 2: Calculate the angular acceleration, α′. The initial angular velocity,

ωi, is 8.0 rad/s. The time taken, t′, is 7.0 s. The angular acceleration during

deceleration is given by:

α′=ωf−ωi

t′

Substitute the given values:

α′=0 rad/s −8.0 rad/s

7.0 s =−1.1429 rad/s2

Step 3: Calculate the total angle rotated during deceleration. The number

of revolutions is given by:

Number of revolutions = 1

2π ω2

f−ω2

α′!

Substitute the given values:

Number of revolutions = 1

2π0−(8.0)2

−1.1429 ≈1

2π×56.0898 ≈8.93

Therefore, the body makes approximately 8.93 revolutions during the decelera-

tion phase.

Question 5

Question

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

the disc for 3 seconds, causing the disc to achieve an angular velocity of 6 rad/s.

What is the moment of inertia of the disc?

Solution

Step 1: Identify the given values: The net torque, τ= 4 N·m

The time, t= 3 seconds

The final angular velocity, ωf= 6 rad/s

The radius of the disc, r= 0.2 m

Step 2: Calculate the angular acceleration using the relation α=ωf−ωi

where ωiis the initial angular velocity (which is 0 in this case).

α=6−0

3= 2 rad/s2

Step 3: Use the formula τ=Iα to find the moment of inertia, I. Rearranging

the formula gives I=τ

α.

I=4

2= 2 kg ·m2

Step 4: Write the final answer: The moment of inertia of the disc is 2 kg ·m2.

Question 6

Question

A solid sphere of radius Rand mass Mis rolling 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, find an expression for its angular velocity ω

when it reaches the bottom.

Solution

Step 1: The forces acting on the solid sphere on the incline are gravity (Fgravity =

mg) acting downward, the normal force (Fnormal) acting perpendicular to the

incline, and the friction force (ffriction) acting parallel to the incline. By resolving

the gravitational force into components, we get:

mg sin θ=ffriction

Step 2: The friction force provides the torque needed to induce the sphere to

roll without slipping. The torque due to the friction force is given by τ=

ffriction ·R. The torque also creates an angular acceleration that relates to the

angular velocity and time through the equation τ=Iα, where Iis the moment

of inertia of the sphere and αis the angular acceleration. Step 3: For a solid

sphere rolling without slipping, the moment of inertia with respect to its center

of mass is I=2

5MR2. The angular acceleration α=a

R, where ais the linear

acceleration of the center of mass. Step 4: The linear acceleration aof the

center of mass of the sphere can be found using Newton’s second law, taking

into account the net force along the incline (Fnet, parallel =ma):

mg sin θ−ffriction =Ma

Step 5: Since the sphere is rolling without slipping, the linear acceleration ais

related to the angular acceleration αthrough a=Rα. Substituting this into

the previous equation yields:

mg sin θ−ffriction =MRα

Step 6: Combining the expressions for torque, angular acceleration, and net

force, we have:

ffriction ·R=2

5MR2ffriction

MR 

Solving for ffriction and then using it to find the angular velocity ωat the bottom

of the incline is the next step.

Question 7

Question

A space station is rotating about its axis, which is perpendicular to the ground

with an angular speed of 0.50 rad/s. A bolt at the outer edge of one of the

space station’s rotating arms comes loose and floats away perpendicular to the

axis of rotation. If the bolt is 10.0 m from the axis of rotation, how far from

the axis will it land on the ground?

Solution

Step 1: Identify the given values and convert angular speed to linear speed.

Given: Angular speed, ω= 0.50 rad/s Distance from the axis, r= 10.0 m We

know that linear speed, v=r·ω.

Step 2: Calculate the linear speed of the bolt. Substitute rand ωinto the

formula: v= 10.0 m ·0.50 rad/s = 5.0 m/s.

Step 3: Determine the time taken by the bolt to reach the ground. The bolt

moves horizontally with a speed of 5.0 m/s. Let’s assume the bolt takes time t

to reach the ground. The horizontal distance traversed by the bolt, x, is given

as x=v·t.

Step 4: Calculate the horizontal distance traveled by the bolt. The horizontal

distance travelled by the bolt is equal to the radius of rotation of the space

station. Therefore, x= 10.0 m.

Step 5: Find the time taken for the bolt to reach the ground. Substitute x

and vin the expression x=v·t: 10.0=5.0·t. Solving for t, we get t= 2.0

seconds.

Step 6: Calculate the distance from the axis where the bolt will land on the

ground. The distance covered horizontally by the bolt when it lands is equal to

the radius of rotation. Therefore, the bolt will land 10.0 m away from the axis.

Question 8

Question

A disc of radius Ris rotating at a constant angular velocity ωabout a fixed axis

passing through its center. A small bug starts at a distance rfrom the center

of the disc, at an angle of θ0from the radial line passing through the bug. The

bug then starts crawling along the disc at a constant speed vbug relative to the

disc. Determine the angular velocity of the bug with respect to the disc as a

function of the bug’s position.

Solution

Step 1: The velocity of the bug relative to the disc is tangent to the bug’s

circular path, forming an angle ϕwith the radial line. The radial component of

the bug’s velocity is zero, as the bug doesn’t move radially. The velocity of the

bug relative to the disc is given by:

vbug =qv2

r+v2

θ=vθ=rdθ

Step 2: Using the law of cosines, we can express the bug’s velocity as a

function of ωand θin terms of the bug’s angular velocity with respect to the

disc. The law of cosines gives:

bug =R2ω2+r2ω2

bug −2Rrωωbug cos(θ−ϕ)

Step 3: Substituting vbug =rωbug and vbug =rdθ

dt into the law of cosines

equation, we get:

r2dθ

dt 2

=R2ω2+r2ω2

bug −2Rrωωbug cosθ−arctanR

rtan θ

Step 4: Differentiating both sides with respect to time, we get:

2rdθ

d2θ

dt2=−2Rrωωbug sinθ−arctanR

rtan θdθ

Step 5: Simplifying, we find the angular acceleration of the bug with respect

to the disc: d2θ

dt2=−ωbug R

r−cos θ

sin θω

Question 9

Question

A disk of radius 0.1 m is rotating with an angular velocity of 10 rad/s. A small

pebble is stuck at the edge of the disk. How far will the pebble travel in 3

seconds?

Solution

Step 1: Calculate the linear velocity of the pebble on the edge of the disk. Since

the pebble is located at the edge of the disk, its linear velocity can be calculated

using the formula:

v=rω

where vis the linear velocity, ris the radius of the disk, and ωis the angular

velocity. Substituting r= 0.1 m and ω= 10 rad/s:

v= 0.1×10 = 1 m/s

Step 2: Calculate the distance the pebble will travel in 3 seconds. Since

distance traveled is equal to velocity multiplied by time:

Distance = v×t

Substitute v= 1 m/s and t= 3 s:

Distance = 1 ×3 = 3 meters

Therefore, the pebble will travel 3 meters in 3 seconds.

Question 10

Question

A solid sphere of radius Rstarts from rest and rolls without slipping down a 30

degree incline. Calculate the translational velocity of the center of mass of the

sphere when it has descended a height equal to h.

Solution

Step 1: Determine the final angular velocity of the sphere when it has descended

a height equal to h. Let vbe the linear velocity of the sphere’s center of mass,

ωbe the angular velocity of the sphere, and Ibe the moment of inertia of the

sphere.

The total kinetic energy of the sphere is given by the sum of its translational

and rotational kinetic energies:

KE =1

2Mv2+1

2Iω2

Since the sphere rolls without slipping, the linear velocity is related to the

angular velocity by v=Rω. Substituting this into the kinetic energy equation

gives:

KE =1

2M(Rω)2+1

2Iω2

Step 2: Equate the potential energy of the sphere initially at rest at height

hto its final kinetic energy.

The initial potential energy of the sphere at height his given by P E =Mgh.

Setting this equal to the kinetic energy, we have:

Mgh =1

2M(Rω)2+1

2Iω2

Step 3: Substitute the moment of inertia of a solid sphere and solve for the

final angular velocity.

For a solid sphere rotating about its center, the moment of inertia is I=

5MR2. Substituting this into the equation and rearranging yields:

Mgh =1

2M(Rω)2+1

22

5MR2ω2

Solving for ωgives:

ω=r5gh

The final linear velocity of the center of mass is then given by:

v=Rω =Rr5gh

7R=r5ghR

Question 11

Question

A solid sphere of radius Rand mass Mis rotating about a fixed axis with an

angular speed ωi. A small mass mof clay falls vertically onto the sphere and

sticks to it. If the final system rotates about the same axis with an angular

speed of ωf, determine the ratio ωf

ωiin terms of m,M, and R.

Solution

Step 1: Determine the moment of inertia of the initial system (sphere rotating

with angular speed ωi). The moment of inertia of a solid sphere rotating about

its axis is given by I=2

5MR2. The moment of inertia of the initial system is

Ii=2

5MR2.

Step 2: Determine the moment of inertia of the final system (system after

clay falls onto the sphere). Since the clay sticks to the sphere and rotates with

it, the moment of inertia of the final system is If=Ii+mR2, where mR2

represents the moment of inertia of the clay.

Step 3: Apply the principle of conservation of angular momentum. The

principle of conservation of angular momentum states that the initial angular

momentum is equal to the final angular momentum. We have:

Iiωi=Ifωf

Substitute the expressions for Iiand If:

5MR2ωi=2

5MR2+mR2ωf

Step 4: Solve for the ratio ωf

ωi. Divide both sides of the equation by 2

5MR2ωi:

ωf

ωi

5MR2

5MR2+mR2=2

2+5m

=2M

2M+ 5m

Therefore, the ratio ωf

ωiin terms of m,M, and Ris 2M

2M+5m.

Question 12

Question

A solid sphere of radius Rand moment of inertia Iabout its center of mass

rolls without slipping on a horizontal surface. Initially, the sphere is at rest. It

then accelerates uniformly for a distance dbefore coming to a stop. Find the

angular acceleration of the sphere during this acceleration phase.

Solution

Step 1: Since the sphere rolls without slipping, we can relate the linear and

angular quantities using the relationship v=Rω, where vis the linear velocity,

Ris the radius of the sphere, and ωis the angular velocity. Step 2: The

linear acceleration aof the sphere is equal to a=αR, where αis the angular

acceleration. Since the sphere starts from rest and then stops, the final velocity

vfis zero. Step 3: Using the kinematic equation v2

f=v2

i+ 2ad where viis the

initial velocity (zero) and dis the distance traveled, we can write 0 = 0 + 2ad.

Solving for a, we find a= 0. Step 4: Substituting a=αR, we have 0 = 2αRd.

Step 5: Solving for α, we find α= 0. Step 6: Therefore, the angular acceleration

of the sphere during the acceleration phase is 0 .

Question 13

Question

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

sitting on the edge of the disk.

1. What is the bug’s linear velocity?

2. If the bug starts moving towards the center of the disk at a rate of 0.1

m/s, what is the bug’s angular velocity when it is 0.1 m from the center?

Solution

1. To find the bug’s linear velocity, we can use the relationship between linear

velocity (v) and angular velocity (ω) in a rotating disk:

v=r·ω

where ris the radius of the disk and ωis the angular velocity.

Step 1: Substitute the given values into the formula:

v= 0.2 m ·5 rad/s = 1 m/s

So, the bug’s linear velocity is 1 m/s.

2. To find the bug’s angular velocity when it is 0.1 m from the center, we can

use the principle of conservation of angular momentum. Since the bug moves

towards the center of the disk, the moment of inertia of the bug-disk system

changes.

The conservation of angular momentum is given by:

I1·ω1=I2·ω2

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

Step 1: Initially, the moment of inertia I1of the bug-disk system is:

I1=mr2

Since the bug is at the edge of the disk, r1= 0.2 m.

Step 2: Finally, the moment of inertia I2when the bug is 0.1 m from the

center is:

I2=mr2

where r2= 0.1 m.

Step 3: Taking the conservation of angular momentum equation and sub-

stituting for I1,I2, and ω1calculated in part 1, we get:

m·0.22·5 = m·0.12·ω2

Step 4: Solve for ω2:

0.2·5 = 0.12·ω2

ω2=1

0.01 ·1·5 = 500 rad/s

So, the bug’s angular velocity when it is 0.1 m from the center is 500 rad/s.

Question 14

Question

A disk of radius Ris rolling without slipping on a horizontal surface. Start-

ing from rest, the disk accelerates uniformly for time t1and then decelerates

uniformly for time t2. If the total distance traveled by a point on the rim of

the disk during this time is d, determine the coefficient of kinetic friction µk

between the disk and the surface.

Solution

Let’s denote the angular acceleration of the disk as αand the angular velocity

at the end of the acceleration phase as ω1. The angular velocity at the end of

the deceleration phase is denoted as ω2. We also know that the angular velocity

ωand the linear velocity vare related by ω=v

Step 1: During the acceleration phase: Using the kinematic equation relat-

ing angular velocity, angular acceleration, and time:

ω1=αt1

Also, the linear velocity during the acceleration phase is v1=Rω1.

Step 2: During the deceleration phase: Similar to Step 1, we relate ω2and

α:

ω2=αt2

And the linear velocity during the deceleration phase is v2=Rω2.

Step 3: Total distance traveled: The total distance traveled by a point on

the rim of the disk can be calculated as:

d=v1t1+v2t2

Step 4: Expressing v1and v2: Substitute v1and v2into the distance equa-

tion:

d=Rω1t1+Rω2t2

Step 5: Substituting angular velocity expressions: Using the expressions

for ω1and ω2from Step 1 and Step 2:

d=Rαt2

1+Rαt2

Step 6: Determining αin terms of d: Since the disk is rolling without

slipping, we have the relationship α=µkg

d=µkgt2

1+µkgt2

Step 7: Solving for µk: Given dand the values of t1and t2, we can solve

for µk:

µk=d

g(t2

1+t2

Question 15

Question

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

down a 30◦incline. How far along the incline does the sphere slide before it

stops?

Solution

Step 1: We can use energy conservation to solve this problem. The potential

energy at the top of the incline is converted into translational and rotational

kinetic energy at the bottom.

Step 2: The potential energy at the top is given by P E =M gh, where

h=Rsin(30◦) is the height of the incline.

Step 3: At the bottom, the kinetic energy is given by KE =1

2Mv2

cm +1

2Iω2,

where vcm is the velocity of the center of mass and ωis the angular velocity.

Step 4: The moment of inertia of a solid sphere about its center is I=2

5MR2.

Step 5: Since the sphere is rolling without slipping, we have vcm =Rω.

Step 6: Equating the potential energy at the top to the kinetic energy at the

bottom, we have Mgh =1

2Mv2

cm +1

2Iω2.

Step 7: Substituting the moment of inertia and the relationship between vcm

and ω, we have Mgh =1

2M(v2

cm +2

5R2ω2).

Step 8: Substituting for vcm and rearranging, we get vcm =q5gh

Step 9: The distance the sphere slides along the incline can be found using

the equation of motion v2

cm =u2+ 2as, where u= 0 and a=gsin(30◦).

Step 10: Substituting the values, we have 5gh

7= 2ghs sin(30◦).

Step 11: Solving for s, we get s=5

14 R.

Therefore, the sphere slides 5

14 times its radius along the incline before it

stops.

Question 16

Question

A disk of radius Rstarts from rest and accelerates uniformly for a time t. If the

angular velocity of the disk after the acceleration is given by ω=αt2, where α

is a constant, determine the angular acceleration of the disk.

Solution

Step 1: We know that angular acceleration (α) is the rate of change of angular

velocity with respect to time. Mathematically, it is given by α=dω

dt .

Step 2: Given that the angular velocity of the disk is ω=αt2, we can find

the angular acceleration by differentiating this expression with respect to time.

Step 3: Differentiating ω=αt2with respect to t, we get:

α=dω

dt =d(αt2)

Step 4: Applying the power rule of differentiation, we have:

α= 2αt

Step 5: Thus, the angular acceleration of the disk is α= 2αt.

Question 17

Question

A disk initially at rest undergoes an angular acceleration of 2.0 rad/s2for 5.0 s.

If the moment of inertia of the disk is 0.10 kg m2, determine the final angular

velocity of the disk.

Solution

Step 1: We can use the equation of rotational motion to find the final angular

velocity of the disk. The equation is given by:

ωf=ωi+αt

where: - ωfis the final angular velocity, - ωiis the initial angular velocity (which

is 0 since the disk is initially at rest), - αis the angular acceleration, and - tis

the time for which the angular acceleration occurs.

Step 2: Substituting the given values into the equation, we get:

ωf= 0 + (2.0 rad/s2)(5.0 s)

Step 3: Now, calculate the final angular velocity:

ωf= 10.0 rad/s

So, the final angular velocity of the disk is ωf= 10.0 rad/s.

Question 18

Question

A disk of radius Rand mass Mis rotating about an axis perpendicular to

the disk through its center with an angular velocity ω. If a constant torque is

applied to the disk in the direction opposite to its rotation, causing it to come

to a complete stop after making ncomplete revolutions, find the magnitude of

the torque.

Solution

Step 1: Find the initial angular velocity ωiof the disk. The disk comes to a stop

after making ncomplete revolutions, so the final angular velocity is 0. Using

the equation of rotational kinematics:

ωf=ωi+αt

where αis the angular acceleration and tis the time taken to stop. Since the

final angular velocity is 0 and the angular acceleration is constant, ωi=−αt.

Step 2: Find the angular acceleration αand time t. The angular acceleration

is given by:

α=∆ω

∆t

Since the disk stops from an initial angular velocity of ωin ncomplete revolu-

tions, we have ∆ω=−ωand ∆t=2πn

ω. Therefore, α=−ω

2πn and t=πn

ω.

Step 3: Find the moment of inertia Iof the disk. The moment of inertia of

a disk rotating about its center is I=1

2MR2.

Step 4: Find the torque τapplied to the disk. The torque applied to the

disk is given by:

τ=Iα =1

2MR2·−ω

2πn =−M R2ω

4πn

Question 19

Question

A disk with radius Ris rotating with an angular velocity ω0. A small piece of

the disk with mass mand radius rbreaks off. Calculate the angular velocity of

the disk after the piece breaks off.

Solution

1. Calculate the moment of inertia of the original disk with radius R: Given

that the moment of inertia of a disk about its center is 1

2mR2, the moment of

inertia of the original disk is 1

2MR2, where Mis the total mass of the original

disk (which we assume is uniformly distributed).

2. Calculate the moment of inertia of the piece that broke off: Given that

the moment of inertia of a disk about its center is 1

2mr2, the moment of inertia

of the piece is 1

2mr2.

3. Calculate the moment of inertia of the system after the piece broke

off: The moment of inertia of the system after the piece broke off is Ifinal =

2MR2−1

2mr2.

4. Apply conservation of angular momentum: Before the piece breaks off, the

initial angular momentum is Linitial = (Iinitial)ω0, where Iinitial =1

2MR2. After

the piece breaks off, the final angular momentum is Lfinal = (Ifinal)ωfinal. Since

angular momentum is conserved, Linitial =Lfinal. This gives 1

2MR2ω0=

1

2MR2−1

2mr2ωfinal.

5. Solve for ωfinal: Solving for ωfinal, we get ωfinal =MR2ω0

MR2−mr2.

Question 20

Question

A disk of mass mand radius Ris initially at rest. A constant force Fis applied

tangent to the disk at a distance rfrom the center. The force causes the disk

to rotate without slipping. Find the angular acceleration of the disk in terms of

F,m,r, and R.

Solution

Step 1: The torque due to the applied force Fis given by τ=rF .

Step 2: The moment of inertia of a disk rotating about its center is I=

2mR2.

Step 3: The torque τis related to the angular acceleration αand moment

of inertia Iby τ=Iα.

Step 4: Using the values of torque and moment of inertia, we have rF =

2mR2α.

Step 5: Solving for the angular acceleration α, we get α=2rF

mR2.

Therefore, the angular acceleration of the disk in terms of F,m,r, and Ris

α=2rF

mR2.

Question 21

Question

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

acceleration of 2 rad/s2.

1. Find the angular speed of the disk after 3 seconds.

2. Determine the angular displacement of a point on the rim of the disk after

3 seconds.

Solution

1. First, let’s determine the angular speed of the disk after 3 seconds using the

equation for angular kinematics:

ωf=ωi+αt

where: - ωfis the final angular speed, - ωiis the initial angular speed (0 because

it starts from rest), - αis the angular acceleration (2 rad/s2in this case), - tis

the time elapsed (3 seconds).

Step 1: Plug in the known values into the equation.

ωf= 0 + 2 ×3

ωf= 6 rad/s

Therefore, the angular speed of the disk after 3 seconds is 6 rad/s.

2. To find the angular displacement of a point on the rim of the disk after

3 seconds, we use the equation:

θ=θi+ωit+1

2αt2

where: - θis the angular displacement, - θiis the initial angular displacement

(0 because we’re starting from rest), - ωiis the initial angular speed (0 as well),

-αis the angular acceleration (2 rad/s2), - tis the time elapsed (3 seconds).

Step 2: Substitute the known values into the equation.

θ=0+0+ 1

2×2×32

θ= 3 ×32

θ= 9 ×π

180 rad

θ≈0.157 rad

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

3 seconds is approximately 0.157 rad.

Question 22

Question

A wheel starts from rest and has a constant angular acceleration of 2.0 rad/s2.

How long will it take the wheel to reach an angular speed of 8.0 rad/s?

Solution

Step 1: Write down the given information. The angular acceleration, α=

2.0 rad/s2, and the final angular speed, ωf= 8.0 rad/s.

Step 2: Use the kinematic equation for rotational motion. The angular speed

as a function of time is given by:

ω=ω0+αt

Here, ω0is the initial angular speed (which is 0 since the wheel starts from rest).

Step 3: Plug in the values to find the time taken to reach ωf.

ωf=αt

t=ωf

t=8.0 rad/s

2.0 rad/s2= 4.0 s

So, it will take the wheel 4.0 seconds to reach an angular speed of 8.0 rad/s.

Question 23

Question

A wheel starts from rest and rotates with constant angular acceleration. After

3.0 seconds have passed, it has rotated through an angle of 45 radians. What

is the angular acceleration of the wheel?

Solution

Step 1: We are given the following information: Initial angular velocity, ω0= 0

(since the wheel starts from rest),

Time, t= 3.0 s,

Final angular position, θ= 45 rad.

Step 2: The angular displacement (∆θ) of an object undergoing constant

angular acceleration is given by the formula:

∆θ=ω0t+1

2αt2,

where αis the angular acceleration.

Step 3: Substituting the given values into the equation, we get:

45 rad = 0 + 1

2α(3.0 s)2.

Step 4: Simplifying the equation, we find:

45 = 9

2α.

Step 5: Solving for α, we have:

α=2×45

9= 10 rad/s2.

Step 6: Therefore, the angular acceleration of the wheel is 10 rad/s2.

Question 24

Question

A wheel initially at rest accelerates uniformly for 10 seconds and reaches an

angular velocity of 20 rad/s. After that, it decelerates uniformly for another

10 seconds and comes to a stop. If the total angular displacement during the

entire process is 600 radians, what is the magnitude of the angular acceleration

during the deceleration phase?

Solution

Step 1: Calculate the initial angular velocity during the acceleration phase using

the formula for uniformly accelerated angular motion:

ωinitial = 0

ωfinal = 20 rad/s

α=ωfinal −ωinitial

t=20 −0

10 = 2 rad/s2

Step 2: Calculate the angular displacement during the acceleration phase

using the formula:

θacceleration =1

2(ωinitial +ωfinal)×t

θacceleration =1

2(0 + 20) ×10 = 100 radians

Step 3: Calculate the angular displacement during the deceleration phase:

θdeceleration = total displacement −θacceleration = 600 −100 = 500 radians

Step 4: Calculate the final angular velocity during the deceleration phase

using the formula:

ω2=ω2

initial + 2αθ

ωinitial = 20 rad/s

θ= 500 radians

ω= 0

0 = 202+ 2α×500

400 = 1000α

α= 0.4 rad/s2

Therefore, the magnitude of the angular acceleration during the deceleration

phase is 0.4 rad/s

Question 25

Question

A wheel with radius 0.5 meters starts from rest and rotates with a constant

angular acceleration of 2 rad/s2. What is the angular velocity of the wheel after

3 seconds?

Solution

Step 1: We can use the equation for angular velocity in terms of angular accel-

eration and time:

ω=ω0+αt

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

0 since the wheel starts from rest), αis the angular acceleration (2 rad/s2), and

tis the time (3 seconds).

Step 2: Substitute the given values into the equation:

ω= 0 + 2 ×3 = 6 rad/s

Step 3: Therefore, the angular velocity of the wheel after 3 seconds is 6

rad/s.

Question 26

Question

A thin rod of length Land mass mis rotating about one end at a constant

angular speed ω. At a certain instant, a small piece of mass dm breaks off from

the rod at a distance xfrom the end of the rod. What is the kinetic energy of

the piece that breaks off?

Solution

1. To find the kinetic energy of the piece that breaks off, we first need to

determine the angular speed at which the piece was initially moving.

2. The initial angular momentum of the rod and the piece must be conserved.

Therefore, we have: Linitial =Lfinal

3. The initial angular momentum of the system is given by the rod’s angular

momentum: Linitial =Iω

where I=1

3mL2is the moment of inertia of the rod about the end it is

rotating about.

4. The final angular momentum of the system is the sum of the angular

momentum of the rod after the piece breaks off and the angular momentum of

the piece: Lfinal = (I−dm ·x2)(ω) + x(dm)(ωpiece)

5. Setting the initial angular momentum equal to the final angular momen-

tum, we have: Iω = (I−dm ·x2)(ω) + x(dm)(ωpiece)

6. Since the piece breaks off with the same angular velocity as the rod,

ωpiece =ω, and we can solve for ω.

7. Once we have ω, we can determine the kinetic energy of the piece that

breaks off: KE =1

2dm(ωx)2

Question 27

Question

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

1.5 rad/s2. How long will it take for the wheel to make 14 complete revolutions?

Solution

Step 1: Convert the number of revolutions to radians. Given that 1 revolution

is equivalent to 2πradians, 14 revolutions will be 14 ×2π= 28πradians.

Step 2: Use the kinematic equation for rotational motion. The kinematic

equation for rotational motion can be expressed as:

θ=θ0+ω0t+1

2αt2

where: θ= final angular position (radians), θ0= initial angular position (radi-

ans), ω0= initial angular velocity (rad/s), α= angular acceleration (rad/s2),

and t= time (s).

Step 3: Solve for the time required to make 28πradians. Given that the

wheel is initially at rest (ω0= 0 rad/s) and the initial angular position is 0, the

equation simplifies to:

θ=1

2αt2

Substitute the given values: θ= 28πradians and α= 1.5 rad/s2.

28π=1

2×1.5×t2

Step 4: Solve for t. Solve for t:

t2=28π

0.75 = 37.333

t=√37.333 = 6.11 s

Therefore, it will take approximately 6.11 seconds for the wheel to make 14

complete revolutions.

Question 28

Question

A solid cylinder of mass mand radius Ris initially at rest. A constant force F

is applied tangentially to the edge of the cylinder, causing it to rotate about its

central axis. If the force is applied for a time interval ∆t, determine the angular

velocity of the cylinder at the end of the time interval ∆t.

Solution

Step 1: Calculate the torque applied to the cylinder. The torque τapplied to

the cylinder is given by the equation:

τ=F R

where Fis the magnitude of the force applied tangentially to the edge of the

cylinder.

Step 2: Calculate the moment of inertia of the cylinder. The moment of

inertia Iof a solid cylinder rotating about its central axis is given by:

I=1

2mR2

Step 3: Apply Newton’s second law for rotational motion. The net torque

acting on the cylinder is equal to the moment of inertia times the angular

acceleration. We have:

τ=Iα

where αis the angular acceleration of the cylinder.

Step 4: Relate angular acceleration to angular velocity. Since the cylinder

starts from rest, the relationship between angular acceleration αand angular

velocity ωis given by:

α=∆ω

∆t

Step 5: Substitute the expressions for torque and moment of inertia into the

equation. Substitute τ=F R and I=1

2mR2into the equation τ=Iα:

F R =1

2mR2·∆ω

∆t

Step 6: Solve for the angular velocity ω. Solving for ω, we get:

ω=2F∆t

Therefore, the angular velocity of the cylinder at the end of the time interval

∆tis 2F∆t

Question 29

Question

A solid sphere of mass mand radius Ris rolling without slipping on a horizontal

surface. Initially, the sphere has an angular speed ω0. It then rolls up an incline

plane making an angle θwith the horizontal. What is the maximum height h

the center of the sphere reaches on the incline?

Solution

1. At the bottom of the incline, the total mechanical energy of the sphere is

given by the sum of its translational kinetic energy (1

2mv2) and its rotational

kinetic energy (1

2Iω2, where Iis the moment of inertia of a solid sphere about

its center):

Ebottom =1

2mv2

0+1

2Iω2

2. As the sphere reaches its maximum height, its kinetic energy is zero.

Therefore, the mechanical energy of the system at the maximum height is equal

to the gravitational potential energy at that height:

Emax height =mgh

3. Setting Ebottom =Emax height gives:

2mv2

0+1

2Iω2

0=mgh

4. The linear speed of the sphere when it reaches the incline is related to its

angular speed by v=Rω. Substituting v0=Rω0into the equation from step 3

yields: 1

2m(Rω0)2+1

2Iω2

0=mgh

5. The moment of inertia of a solid sphere about its center is 2

5mR2. Sub-

stituting I=2

5mR2into the equation from step 4 gives:

2mR2ω2

0+1

22

5mR2ω2

0=mgh

6. Simplifying the equation from step 5 yields:

10mR2ω2

0=mgh

7. Solving for the maximum height hgives:

h=7

10R2ω2

Therefore, the maximum height the center of the sphere reaches on the

incline is 7

10 R2ω2

Question 30

Question

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

acceleration of 1.5 rad/s2. Find the angular velocity of the disc after 3 seconds.

Solution

Step 1: We can use the kinematic equation for rotational motion to find the

angular velocity of the disc after 3 seconds. The equation relates the final

angular velocity (ωf), initial angular velocity (ωi), angular acceleration (α),

and time (t) as follows:

ωf=ωi+αt

Step 2: Since the disc starts from rest, the initial angular velocity ωiis 0.

Step 3: Substitute the given values into the equation to solve for the final

angular velocity:

ωf= 0 + (1.5 rad/s2)·3 s

Step 4: Calculate the final angular velocity:

ωf= 1.5 rad/s2·3 s = 4.5 rad/s

Answer: The angular velocity of the disc after 3 seconds is 4.5 rad/s.

Question 31

Question

A thin rod of length Lrotates about a pivot at one end with an angular speed ω.

A small object of mass mis attached to the other end. What is the maximum

angular speed of the rod when the object is directly above the pivot? Assume

the rod is massless and the object does not slip on the rod.

Solution

1. To find the maximum angular speed of the rod when the object is directly

above the pivot, we can consider conservation of angular momentum for the

system.

2. The initial angular momentum of the system is given by I1ω, where I1is

the moment of inertia of the rod with the object at a distance Lfrom the pivot.

3. The final angular momentum is I2ωmax, where I2is the moment of inertia

of the rod with the object at the pivot and ωmax is the maximum angular speed.

4. By conservation of angular momentum, we have:

I1ω=I2ωmax

5. The moment of inertia of a rod rotating about one end is 1

3mL2. Thus,

I1=1

3mL2.

6. When the object is at the pivot, by parallel axis theorem, the moment of

inertia of the rod-object system is I2=1

3mL2+mL2=4

3mL2.

7. Substituting I1,I2into the conservation of angular momentum equation,

we get: 1

3mL2ω=4

3mL2ωmax

8. Simplifying, we find:

ωmax =1

4ω

9. Therefore, the maximum angular speed of the rod when the object is

directly above the pivot is 1

4ω.

Question 32

Question

A disk with a radius of 0.2 m is initially at rest. It then accelerates with a

constant angular acceleration of 6 rad/s

for 4 seconds. What is the angular

velocity of the disk after 4 seconds?

Solution

Step 1: Calculate the angular displacement of the disk after 4 seconds using the

equation θ=1

2αt2, where αis the angular acceleration and tis the time.

Step 2: Substitute the values α= 6 rad/s

and t= 4 s into the equation to

find the angular displacement.

θ=1

2×6×(4)2= 48 rad

Step 3: Use the equation of angular velocity ω=ω0+αt to find the angular

velocity of the disk after 4 seconds. Since the disk starts from rest, the initial

angular velocity ω0is 0.

Step 4: Substitute the values α= 6 rad/s

and t= 4 s into the equation to

find the angular velocity.

ω= 0 + 6 ×4 = 24 rad/s

Therefore, the angular velocity of the disk after 4 seconds is 24 rad/s.

Question 33

Question

A solid cylinder of radius Rand mass Mrolls without slipping down a rough

incline that makes an angle θwith the horizontal. The cylinder starts from rest

at the top of the incline. What is the speed of the center of mass of the cylinder

when it reaches the bottom of the incline?

Solution

Step 1: First, we need to determine the acceleration of the cylinder down the

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

and the friction force. The net torque is given by τ=Iα, where Iis the

moment of inertia of the cylinder and αis its angular acceleration. This net

torque produces the angular acceleration α=τ

Step 2: The net torque causing the rotation is due to the gravitational

force and the friction force. The gravitational force acting on the cylinder is

decomposed into two components: one parallel to the incline (mg sin θ) and

the other perpendicular to the incline (mg cos θ), with mbeing the mass of the

cylinder and gthe acceleration due to gravity.

Step 3: The friction force acts up the incline to oppose the motion. The

maximum static friction is given by fmax =µsN, where µsis the coefficient of

static friction and Nis the normal force. The normal force and friction force

can be expressed in terms of the weight of the cylinder, mg.

Step 4: The rolling condition requires that the linear speed vof the center

of the cylinder is related to the angular speed ωby v=Rω. The acceleration

of the center of mass of the cylinder is given by a=Rα.

Step 5: Applying Newton’s second law for translational motion, we have

fnet =Ma, where fnet =ffriction −mg sin θis the net force acting on the

cylinder. Solving for agives a=ffriction

M−gsin θ.

Step 6: Combining the expressions for acceleration from both rotational and

translational dynamics, we can solve for the final velocity of the center of mass of

the cylinder when it reaches the bottom of the incline. This speed will depend

on the angle θ, the coefficients of friction, and the properties of the cylinder

(radius, mass, moment of inertia).

Question 34

Question

A disk with a radius of 0.2 m is spinning at an angular velocity of 5 rad/s.

The angular velocity decreases uniformly by 0.1 rad/s2. What is the angular

displacement of a point on the rim of the disk after 4 seconds?

Solution

Step 1: The final angular velocity after 4 seconds can be calculated using the

formula:

ωf=ωi+αt

where ωf= final angular velocity, ωi= initial angular velocity, α= angular

acceleration, and t= time.

Plugging in the given values:

ωf= 5 rad/s −0.1 rad/s2×4 s

ωf= 5 rad/s −0.4 rad/s

ωf= 4.6 rad/s

Step 2: The angular displacement can be calculated using the formula:

θ=ωit+1

2αt2

where θ= angular displacement, ωi= initial angular velocity, α= angular

acceleration, and t= time.

Plugging in the given values:

θ= 5 rad/s ×4 s + 1

2× −0.1 rad/s2×(4 s)2

θ= 20 rad −0.2 rad

θ= 19.8 rad

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

4 seconds is 19.8 radians.

Question 35

Question

A solid cylinder of radius Rand mass Mis initially at rest. A constant force F

is applied tangentially at the edge of the cylinder and acts for a time interval ∆t.

The cylinder rolls without slipping. Find the angular velocity of the cylinder

after time ∆t.

Solution

Step 1: Determine the torque applied by the force F.

The torque (τ) applied by a force (F) acting at a distance (r) from the axis of

rotation is given by:

τ=F r

Step 2: Find the acceleration of the cylinder.

The net torque applied to the cylinder causes angular acceleration (α). The

relationship between torque and angular acceleration is:

τ=Iα

where Iis the moment of inertia of the cylinder. For a solid cylinder rotating

about its axis, the moment of inertia is I=1

2MR2. Substituting this into the

equation, we get:

F r =1

2MR2α

Solving for α, we find:

α=2F

Step 3: Find the final angular velocity.

The relationship between angular acceleration (α), angular velocity (ω), and

time (∆t) is:

ω=ω0+α∆t

Since the cylinder is initially at rest (ω0= 0), the final angular velocity can be

found by:

ω=α∆t

Therefore, the angular velocity of the cylinder after time ∆tis:

ω=2F∆t

Question 2

Question

A disk with a radius of 0.5 meters is rotating with an angular velocity of 4

radians per second. At a certain instant, the angular acceleration of the disk is

calculated to be 2 radians per second squared. What is the linear acceleration

of a point on the rim of the disk at this instant?

Solution

Step 1: Calculate the linear velocity of a point on the rim of the disk using the

angular velocity. Step 2: Use the linear velocity and the angular acceleration to

find the linear acceleration.

Step 1: The linear velocity of a point on the rim of the disk can be calculated

using the formula:

v=rω

where ris the radius of the disk and ωis the angular velocity. Given that

r= 0.5 m and ω= 4 rad/s, we have:

v= 0.5×4 = 2 m/s

Step 2: The linear acceleration of a point on the rim of the disk can be

found using the formula:

a=rα

where ris the radius of the disk and αis the angular acceleration. Given that

r= 0.5 m and α= 2 rad/s

, we have:

a= 0.5×2 = 1 m/s2

Therefore, the linear acceleration of a point on the rim of the disk at this

instant is 1 m/s

Question 3

Question

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

5.0 rad/s2for 4.0 seconds. If the radius of the disk is 0.10 meters, what is the

linear acceleration of a point on the rim of the disk at t= 4.0 s?

Solution

Step 1: Find the angular velocity at t= 4.0 s using the equation ωf=ωi+αt,

where αis the angular acceleration. Given: Initial angular velocity, ωi= 0 rad/s

Angular acceleration, α= 5.0 rad/s2Time, t= 4.0 s

ωf=ωi+αt = 0 + 5.0×4.0 = 20.0 rad/s

Step 2: Calculate the linear velocity at the rim of the disk using v=rω.

Given: Radius, r= 0.10 m Angular velocity, ω= 20.0 rad/s

v=rω = 0.10 ×20.0=2.0 m/s

Step 3: Determine the linear acceleration of a point on the rim of the disk

using a=rα. Given: Radius, r= 0.10 m Angular acceleration, α= 5.0 rad/s2

a=rα = 0.10 ×5.0=0.50 m/s2

Therefore, the linear acceleration of a point on the rim of the disk at t= 4.0 s

is 0.50 m/s2.

Question 4

Question

A rigid body undergoing rotational motion starts from rest and accelerates

uniformly for 5.0 s reaching a final angular velocity of 8.0 rad/s. The body then

decelerates uniformly for the next 7.0 s until it comes to rest. Find: (a) the

angular acceleration during the acceleration phase, (b) the number of revolutions

the body makes during the deceleration phase.

Solution

(a) During the acceleration phase: Step 1: Determine the initial angular ve-

locity. The body starts from rest, so the initial angular velocity, ωi, is 0 rad/s.

Step 2: Calculate the angular acceleration, α. The final angular velocity,

ωf, is 8.0 rad/s. The time taken, t, is 5.0 s. The angular acceleration is given

by:

α=ωf−ωi

Substitute the given values:

α=8.0 rad/s −0 rad/s

5.0 s = 1.6 rad/s2

Therefore, the angular acceleration during the acceleration phase is 1.6 rad/s2.

(b) During the deceleration phase: Step 1: Determine the final angular

velocity. The body comes to rest, so the final angular velocity, ωf, is 0 rad/s.

Step 2: Calculate the angular acceleration, α′. The initial angular velocity,

ωi, is 8.0 rad/s. The time taken, t′, is 7.0 s. The angular acceleration during

deceleration is given by:

α′=ωf−ωi

t′

Substitute the given values:

α′=0 rad/s −8.0 rad/s

7.0 s =−1.1429 rad/s2

Step 3: Calculate the total angle rotated during deceleration. The number

of revolutions is given by:

Number of revolutions = 1

2π ω2

f−ω2

α′!

Substitute the given values:

Number of revolutions = 1

2π0−(8.0)2

−1.1429 ≈1

2π×56.0898 ≈8.93

Therefore, the body makes approximately 8.93 revolutions during the decelera-

tion phase.

Question 5

Question

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

the disc for 3 seconds, causing the disc to achieve an angular velocity of 6 rad/s.

What is the moment of inertia of the disc?

Solution

Step 1: Identify the given values: The net torque, τ= 4 N·m

The time, t= 3 seconds

The final angular velocity, ωf= 6 rad/s

The radius of the disc, r= 0.2 m

Step 2: Calculate the angular acceleration using the relation α=ωf−ωi

where ωiis the initial angular velocity (which is 0 in this case).

α=6−0

3= 2 rad/s2

Step 3: Use the formula τ=Iα to find the moment of inertia, I. Rearranging

the formula gives I=τ

α.

I=4

2= 2 kg ·m2

Step 4: Write the final answer: The moment of inertia of the disc is 2 kg ·m2.

Question 6

Question

A solid sphere of radius Rand mass Mis rolling 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, find an expression for its angular velocity ω

when it reaches the bottom.

Solution

Step 1: The forces acting on the solid sphere on the incline are gravity (Fgravity =

mg) acting downward, the normal force (Fnormal) acting perpendicular to the

incline, and the friction force (ffriction) acting parallel to the incline. By resolving

the gravitational force into components, we get:

mg sin θ=ffriction

Step 2: The friction force provides the torque needed to induce the sphere to

roll without slipping. The torque due to the friction force is given by τ=

ffriction ·R. The torque also creates an angular acceleration that relates to the

angular velocity and time through the equation τ=Iα, where Iis the moment

of inertia of the sphere and αis the angular acceleration. Step 3: For a solid

sphere rolling without slipping, the moment of inertia with respect to its center

of mass is I=2

5MR2. The angular acceleration α=a

R, where ais the linear

acceleration of the center of mass. Step 4: The linear acceleration aof the

center of mass of the sphere can be found using Newton’s second law, taking

into account the net force along the incline (Fnet, parallel =ma):

mg sin θ−ffriction =Ma

Step 5: Since the sphere is rolling without slipping, the linear acceleration ais

related to the angular acceleration αthrough a=Rα. Substituting this into

the previous equation yields:

mg sin θ−ffriction =MRα

Step 6: Combining the expressions for torque, angular acceleration, and net

force, we have:

ffriction ·R=2

5MR2ffriction

MR 

Solving for ffriction and then using it to find the angular velocity ωat the bottom

of the incline is the next step.

Question 7

Question

A space station is rotating about its axis, which is perpendicular to the ground

with an angular speed of 0.50 rad/s. A bolt at the outer edge of one of the

space station’s rotating arms comes loose and floats away perpendicular to the

axis of rotation. If the bolt is 10.0 m from the axis of rotation, how far from

the axis will it land on the ground?

Solution

Step 1: Identify the given values and convert angular speed to linear speed.

Given: Angular speed, ω= 0.50 rad/s Distance from the axis, r= 10.0 m We

know that linear speed, v=r·ω.

Step 2: Calculate the linear speed of the bolt. Substitute rand ωinto the

formula: v= 10.0 m ·0.50 rad/s = 5.0 m/s.

Step 3: Determine the time taken by the bolt to reach the ground. The bolt

moves horizontally with a speed of 5.0 m/s. Let’s assume the bolt takes time t

to reach the ground. The horizontal distance traversed by the bolt, x, is given

as x=v·t.

Step 4: Calculate the horizontal distance traveled by the bolt. The horizontal

distance travelled by the bolt is equal to the radius of rotation of the space

station. Therefore, x= 10.0 m.

Step 5: Find the time taken for the bolt to reach the ground. Substitute x

and vin the expression x=v·t: 10.0=5.0·t. Solving for t, we get t= 2.0

seconds.

Step 6: Calculate the distance from the axis where the bolt will land on the

ground. The distance covered horizontally by the bolt when it lands is equal to

the radius of rotation. Therefore, the bolt will land 10.0 m away from the axis.

Question 8

Question

A disc of radius Ris rotating at a constant angular velocity ωabout a fixed axis

passing through its center. A small bug starts at a distance rfrom the center

of the disc, at an angle of θ0from the radial line passing through the bug. The

bug then starts crawling along the disc at a constant speed vbug relative to the

disc. Determine the angular velocity of the bug with respect to the disc as a

function of the bug’s position.

Solution

Step 1: The velocity of the bug relative to the disc is tangent to the bug’s

circular path, forming an angle ϕwith the radial line. The radial component of

the bug’s velocity is zero, as the bug doesn’t move radially. The velocity of the

bug relative to the disc is given by:

vbug =qv2

r+v2

θ=vθ=rdθ

Step 2: Using the law of cosines, we can express the bug’s velocity as a

function of ωand θin terms of the bug’s angular velocity with respect to the

disc. The law of cosines gives:

bug =R2ω2+r2ω2

bug −2Rrωωbug cos(θ−ϕ)

Step 3: Substituting vbug =rωbug and vbug =rdθ

dt into the law of cosines

equation, we get:

r2dθ

dt 2

=R2ω2+r2ω2

bug −2Rrωωbug cosθ−arctanR

rtan θ

Step 4: Differentiating both sides with respect to time, we get:

2rdθ

d2θ

dt2=−2Rrωωbug sinθ−arctanR

rtan θdθ

Step 5: Simplifying, we find the angular acceleration of the bug with respect

to the disc: d2θ

dt2=−ωbug R

r−cos θ

sin θω

Question 9

Question

A disk of radius 0.1 m is rotating with an angular velocity of 10 rad/s. A small

pebble is stuck at the edge of the disk. How far will the pebble travel in 3

seconds?

Solution

Step 1: Calculate the linear velocity of the pebble on the edge of the disk. Since

the pebble is located at the edge of the disk, its linear velocity can be calculated

using the formula:

v=rω

where vis the linear velocity, ris the radius of the disk, and ωis the angular

velocity. Substituting r= 0.1 m and ω= 10 rad/s:

v= 0.1×10 = 1 m/s

Step 2: Calculate the distance the pebble will travel in 3 seconds. Since

distance traveled is equal to velocity multiplied by time:

Distance = v×t

Substitute v= 1 m/s and t= 3 s:

Distance = 1 ×3 = 3 meters

Therefore, the pebble will travel 3 meters in 3 seconds.

Question 10

Question

A solid sphere of radius Rstarts from rest and rolls without slipping down a 30

degree incline. Calculate the translational velocity of the center of mass of the

sphere when it has descended a height equal to h.

Solution

Step 1: Determine the final angular velocity of the sphere when it has descended

a height equal to h. Let vbe the linear velocity of the sphere’s center of mass,

ωbe the angular velocity of the sphere, and Ibe the moment of inertia of the

sphere.

The total kinetic energy of the sphere is given by the sum of its translational

and rotational kinetic energies:

KE =1

2Mv2+1

2Iω2

Since the sphere rolls without slipping, the linear velocity is related to the

angular velocity by v=Rω. Substituting this into the kinetic energy equation

gives:

KE =1

2M(Rω)2+1

2Iω2

Step 2: Equate the potential energy of the sphere initially at rest at height

hto its final kinetic energy.

The initial potential energy of the sphere at height his given by P E =Mgh.

Setting this equal to the kinetic energy, we have:

Mgh =1

2M(Rω)2+1

2Iω2

Step 3: Substitute the moment of inertia of a solid sphere and solve for the

final angular velocity.

For a solid sphere rotating about its center, the moment of inertia is I=

5MR2. Substituting this into the equation and rearranging yields:

Mgh =1

2M(Rω)2+1

22

5MR2ω2

Solving for ωgives:

ω=r5gh

The final linear velocity of the center of mass is then given by:

v=Rω =Rr5gh

7R=r5ghR

Question 11

Question

A solid sphere of radius Rand mass Mis rotating about a fixed axis with an

angular speed ωi. A small mass mof clay falls vertically onto the sphere and

sticks to it. If the final system rotates about the same axis with an angular

speed of ωf, determine the ratio ωf

ωiin terms of m,M, and R.

Solution

Step 1: Determine the moment of inertia of the initial system (sphere rotating

with angular speed ωi). The moment of inertia of a solid sphere rotating about

its axis is given by I=2

5MR2. The moment of inertia of the initial system is

Ii=2

5MR2.

Step 2: Determine the moment of inertia of the final system (system after

clay falls onto the sphere). Since the clay sticks to the sphere and rotates with

it, the moment of inertia of the final system is If=Ii+mR2, where mR2

represents the moment of inertia of the clay.

Step 3: Apply the principle of conservation of angular momentum. The

principle of conservation of angular momentum states that the initial angular

momentum is equal to the final angular momentum. We have:

Iiωi=Ifωf

Substitute the expressions for Iiand If:

5MR2ωi=2

5MR2+mR2ωf

Step 4: Solve for the ratio ωf

ωi. Divide both sides of the equation by 2

5MR2ωi:

ωf

ωi

5MR2

5MR2+mR2=2

2+5m

=2M

2M+ 5m

Therefore, the ratio ωf

ωiin terms of m,M, and Ris 2M

2M+5m.

Question 12

Question

A solid sphere of radius Rand moment of inertia Iabout its center of mass

rolls without slipping on a horizontal surface. Initially, the sphere is at rest. It

then accelerates uniformly for a distance dbefore coming to a stop. Find the

angular acceleration of the sphere during this acceleration phase.

Solution

Step 1: Since the sphere rolls without slipping, we can relate the linear and

angular quantities using the relationship v=Rω, where vis the linear velocity,

Ris the radius of the sphere, and ωis the angular velocity. Step 2: The

linear acceleration aof the sphere is equal to a=αR, where αis the angular

acceleration. Since the sphere starts from rest and then stops, the final velocity

vfis zero. Step 3: Using the kinematic equation v2

f=v2

i+ 2ad where viis the

initial velocity (zero) and dis the distance traveled, we can write 0 = 0 + 2ad.

Solving for a, we find a= 0. Step 4: Substituting a=αR, we have 0 = 2αRd.

Step 5: Solving for α, we find α= 0. Step 6: Therefore, the angular acceleration

of the sphere during the acceleration phase is 0 .

Question 13

Question

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

sitting on the edge of the disk.

1. What is the bug’s linear velocity?

2. If the bug starts moving towards the center of the disk at a rate of 0.1

m/s, what is the bug’s angular velocity when it is 0.1 m from the center?

Solution

1. To find the bug’s linear velocity, we can use the relationship between linear

velocity (v) and angular velocity (ω) in a rotating disk:

v=r·ω

where ris the radius of the disk and ωis the angular velocity.

Step 1: Substitute the given values into the formula:

v= 0.2 m ·5 rad/s = 1 m/s

So, the bug’s linear velocity is 1 m/s.

2. To find the bug’s angular velocity when it is 0.1 m from the center, we can

use the principle of conservation of angular momentum. Since the bug moves

towards the center of the disk, the moment of inertia of the bug-disk system

changes.

The conservation of angular momentum is given by:

I1·ω1=I2·ω2

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

Step 1: Initially, the moment of inertia I1of the bug-disk system is:

I1=mr2

Since the bug is at the edge of the disk, r1= 0.2 m.

Step 2: Finally, the moment of inertia I2when the bug is 0.1 m from the

center is:

I2=mr2

where r2= 0.1 m.

Step 3: Taking the conservation of angular momentum equation and sub-

stituting for I1,I2, and ω1calculated in part 1, we get:

m·0.22·5 = m·0.12·ω2

Step 4: Solve for ω2:

0.2·5 = 0.12·ω2

ω2=1

0.01 ·1·5 = 500 rad/s

So, the bug’s angular velocity when it is 0.1 m from the center is 500 rad/s.

Question 14

Question

A disk of radius Ris rolling without slipping on a horizontal surface. Start-

ing from rest, the disk accelerates uniformly for time t1and then decelerates

uniformly for time t2. If the total distance traveled by a point on the rim of

the disk during this time is d, determine the coefficient of kinetic friction µk

between the disk and the surface.

Solution

Let’s denote the angular acceleration of the disk as αand the angular velocity

at the end of the acceleration phase as ω1. The angular velocity at the end of

the deceleration phase is denoted as ω2. We also know that the angular velocity

ωand the linear velocity vare related by ω=v

Step 1: During the acceleration phase: Using the kinematic equation relat-

ing angular velocity, angular acceleration, and time:

ω1=αt1

Also, the linear velocity during the acceleration phase is v1=Rω1.

Step 2: During the deceleration phase: Similar to Step 1, we relate ω2and

α:

ω2=αt2

And the linear velocity during the deceleration phase is v2=Rω2.

Step 3: Total distance traveled: The total distance traveled by a point on

the rim of the disk can be calculated as:

d=v1t1+v2t2

Step 4: Expressing v1and v2: Substitute v1and v2into the distance equa-

tion:

d=Rω1t1+Rω2t2

Step 5: Substituting angular velocity expressions: Using the expressions

for ω1and ω2from Step 1 and Step 2:

d=Rαt2

1+Rαt2

Step 6: Determining αin terms of d: Since the disk is rolling without

slipping, we have the relationship α=µkg

d=µkgt2

1+µkgt2

Step 7: Solving for µk: Given dand the values of t1and t2, we can solve

for µk:

µk=d

g(t2

1+t2

Question 15

Question

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

down a 30◦incline. How far along the incline does the sphere slide before it

stops?

Solution

Step 1: We can use energy conservation to solve this problem. The potential

energy at the top of the incline is converted into translational and rotational

kinetic energy at the bottom.

Step 2: The potential energy at the top is given by P E =M gh, where

h=Rsin(30◦) is the height of the incline.

Step 3: At the bottom, the kinetic energy is given by KE =1

2Mv2

cm +1

2Iω2,

where vcm is the velocity of the center of mass and ωis the angular velocity.

Step 4: The moment of inertia of a solid sphere about its center is I=2

5MR2.

Step 5: Since the sphere is rolling without slipping, we have vcm =Rω.

Step 6: Equating the potential energy at the top to the kinetic energy at the

bottom, we have Mgh =1

2Mv2

cm +1

2Iω2.

Step 7: Substituting the moment of inertia and the relationship between vcm

and ω, we have Mgh =1

2M(v2

cm +2

5R2ω2).

Step 8: Substituting for vcm and rearranging, we get vcm =q5gh

Step 9: The distance the sphere slides along the incline can be found using

the equation of motion v2

cm =u2+ 2as, where u= 0 and a=gsin(30◦).

Step 10: Substituting the values, we have 5gh

7= 2ghs sin(30◦).

Step 11: Solving for s, we get s=5

14 R.

Therefore, the sphere slides 5

14 times its radius along the incline before it

stops.

Question 16

Question

A disk of radius Rstarts from rest and accelerates uniformly for a time t. If the

angular velocity of the disk after the acceleration is given by ω=αt2, where α

is a constant, determine the angular acceleration of the disk.

Solution

Step 1: We know that angular acceleration (α) is the rate of change of angular

velocity with respect to time. Mathematically, it is given by α=dω

dt .

Step 2: Given that the angular velocity of the disk is ω=αt2, we can find

the angular acceleration by differentiating this expression with respect to time.

Step 3: Differentiating ω=αt2with respect to t, we get:

α=dω

dt =d(αt2)

Step 4: Applying the power rule of differentiation, we have:

α= 2αt

Step 5: Thus, the angular acceleration of the disk is α= 2αt.

Question 17

Question

A disk initially at rest undergoes an angular acceleration of 2.0 rad/s2for 5.0 s.

If the moment of inertia of the disk is 0.10 kg m2, determine the final angular

velocity of the disk.

Solution

Step 1: We can use the equation of rotational motion to find the final angular

velocity of the disk. The equation is given by:

ωf=ωi+αt

where: - ωfis the final angular velocity, - ωiis the initial angular velocity (which

is 0 since the disk is initially at rest), - αis the angular acceleration, and - tis

the time for which the angular acceleration occurs.

Step 2: Substituting the given values into the equation, we get:

ωf= 0 + (2.0 rad/s2)(5.0 s)

Step 3: Now, calculate the final angular velocity:

ωf= 10.0 rad/s

So, the final angular velocity of the disk is ωf= 10.0 rad/s.

Question 18

Question

A disk of radius Rand mass Mis rotating about an axis perpendicular to

the disk through its center with an angular velocity ω. If a constant torque is

applied to the disk in the direction opposite to its rotation, causing it to come

to a complete stop after making ncomplete revolutions, find the magnitude of

the torque.

Solution

Step 1: Find the initial angular velocity ωiof the disk. The disk comes to a stop

after making ncomplete revolutions, so the final angular velocity is 0. Using

the equation of rotational kinematics:

ωf=ωi+αt

where αis the angular acceleration and tis the time taken to stop. Since the

final angular velocity is 0 and the angular acceleration is constant, ωi=−αt.

Step 2: Find the angular acceleration αand time t. The angular acceleration

is given by:

α=∆ω

∆t

Since the disk stops from an initial angular velocity of ωin ncomplete revolu-

tions, we have ∆ω=−ωand ∆t=2πn

ω. Therefore, α=−ω

2πn and t=πn

ω.

Step 3: Find the moment of inertia Iof the disk. The moment of inertia of

a disk rotating about its center is I=1

2MR2.

Step 4: Find the torque τapplied to the disk. The torque applied to the

disk is given by:

τ=Iα =1

2MR2·−ω

2πn =−M R2ω

4πn

Question 19

Question

A disk with radius Ris rotating with an angular velocity ω0. A small piece of

the disk with mass mand radius rbreaks off. Calculate the angular velocity of

the disk after the piece breaks off.

Solution

1. Calculate the moment of inertia of the original disk with radius R: Given

that the moment of inertia of a disk about its center is 1

2mR2, the moment of

inertia of the original disk is 1

2MR2, where Mis the total mass of the original

disk (which we assume is uniformly distributed).

2. Calculate the moment of inertia of the piece that broke off: Given that

the moment of inertia of a disk about its center is 1

2mr2, the moment of inertia

of the piece is 1

2mr2.

3. Calculate the moment of inertia of the system after the piece broke

off: The moment of inertia of the system after the piece broke off is Ifinal =

2MR2−1

2mr2.

4. Apply conservation of angular momentum: Before the piece breaks off, the

initial angular momentum is Linitial = (Iinitial)ω0, where Iinitial =1

2MR2. After

the piece breaks off, the final angular momentum is Lfinal = (Ifinal)ωfinal. Since

angular momentum is conserved, Linitial =Lfinal. This gives 1

2MR2ω0=

1

2MR2−1

2mr2ωfinal.

5. Solve for ωfinal: Solving for ωfinal, we get ωfinal =MR2ω0

MR2−mr2.

Question 20

Question

A disk of mass mand radius Ris initially at rest. A constant force Fis applied

tangent to the disk at a distance rfrom the center. The force causes the disk

to rotate without slipping. Find the angular acceleration of the disk in terms of

F,m,r, and R.

Solution

Step 1: The torque due to the applied force Fis given by τ=rF .

Step 2: The moment of inertia of a disk rotating about its center is I=

2mR2.

Step 3: The torque τis related to the angular acceleration αand moment

of inertia Iby τ=Iα.

Step 4: Using the values of torque and moment of inertia, we have rF =

2mR2α.

Step 5: Solving for the angular acceleration α, we get α=2rF

mR2.

Therefore, the angular acceleration of the disk in terms of F,m,r, and Ris

α=2rF

mR2.

Question 21

Question

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

acceleration of 2 rad/s2.

1. Find the angular speed of the disk after 3 seconds.

2. Determine the angular displacement of a point on the rim of the disk after

3 seconds.

Solution

1. First, let’s determine the angular speed of the disk after 3 seconds using the

equation for angular kinematics:

ωf=ωi+αt

where: - ωfis the final angular speed, - ωiis the initial angular speed (0 because

it starts from rest), - αis the angular acceleration (2 rad/s2in this case), - tis

the time elapsed (3 seconds).

Step 1: Plug in the known values into the equation.

ωf= 0 + 2 ×3

ωf= 6 rad/s

Therefore, the angular speed of the disk after 3 seconds is 6 rad/s.

2. To find the angular displacement of a point on the rim of the disk after

3 seconds, we use the equation:

θ=θi+ωit+1

2αt2

where: - θis the angular displacement, - θiis the initial angular displacement

(0 because we’re starting from rest), - ωiis the initial angular speed (0 as well),

-αis the angular acceleration (2 rad/s2), - tis the time elapsed (3 seconds).

Step 2: Substitute the known values into the equation.

θ=0+0+ 1

2×2×32

θ= 3 ×32

θ= 9 ×π

180 rad

θ≈0.157 rad

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

3 seconds is approximately 0.157 rad.

Question 22

Question

A wheel starts from rest and has a constant angular acceleration of 2.0 rad/s2.

How long will it take the wheel to reach an angular speed of 8.0 rad/s?

Solution

Step 1: Write down the given information. The angular acceleration, α=

2.0 rad/s2, and the final angular speed, ωf= 8.0 rad/s.

Step 2: Use the kinematic equation for rotational motion. The angular speed

as a function of time is given by:

ω=ω0+αt

Here, ω0is the initial angular speed (which is 0 since the wheel starts from rest).

Step 3: Plug in the values to find the time taken to reach ωf.

ωf=αt

t=ωf

t=8.0 rad/s

2.0 rad/s2= 4.0 s

So, it will take the wheel 4.0 seconds to reach an angular speed of 8.0 rad/s.

Question 23

Question

A wheel starts from rest and rotates with constant angular acceleration. After

3.0 seconds have passed, it has rotated through an angle of 45 radians. What

is the angular acceleration of the wheel?

Solution

Step 1: We are given the following information: Initial angular velocity, ω0= 0

(since the wheel starts from rest),

Time, t= 3.0 s,

Final angular position, θ= 45 rad.

Step 2: The angular displacement (∆θ) of an object undergoing constant

angular acceleration is given by the formula:

∆θ=ω0t+1

2αt2,

where αis the angular acceleration.

Step 3: Substituting the given values into the equation, we get:

45 rad = 0 + 1

2α(3.0 s)2.

Step 4: Simplifying the equation, we find:

45 = 9

2α.

Step 5: Solving for α, we have:

α=2×45

9= 10 rad/s2.

Step 6: Therefore, the angular acceleration of the wheel is 10 rad/s2.

Question 24

Question

A wheel initially at rest accelerates uniformly for 10 seconds and reaches an

angular velocity of 20 rad/s. After that, it decelerates uniformly for another

10 seconds and comes to a stop. If the total angular displacement during the

entire process is 600 radians, what is the magnitude of the angular acceleration

during the deceleration phase?

Solution

Step 1: Calculate the initial angular velocity during the acceleration phase using

the formula for uniformly accelerated angular motion:

ωinitial = 0

ωfinal = 20 rad/s

α=ωfinal −ωinitial

t=20 −0

10 = 2 rad/s2

Step 2: Calculate the angular displacement during the acceleration phase

using the formula:

θacceleration =1

2(ωinitial +ωfinal)×t

θacceleration =1

2(0 + 20) ×10 = 100 radians

Step 3: Calculate the angular displacement during the deceleration phase:

θdeceleration = total displacement −θacceleration = 600 −100 = 500 radians

Step 4: Calculate the final angular velocity during the deceleration phase

using the formula:

ω2=ω2

initial + 2αθ

ωinitial = 20 rad/s

θ= 500 radians

ω= 0

0 = 202+ 2α×500

400 = 1000α

α= 0.4 rad/s2

Therefore, the magnitude of the angular acceleration during the deceleration

phase is 0.4 rad/s

Question 25

Question

A wheel with radius 0.5 meters starts from rest and rotates with a constant

angular acceleration of 2 rad/s2. What is the angular velocity of the wheel after

3 seconds?

Solution

Step 1: We can use the equation for angular velocity in terms of angular accel-

eration and time:

ω=ω0+αt

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

0 since the wheel starts from rest), αis the angular acceleration (2 rad/s2), and

tis the time (3 seconds).

Step 2: Substitute the given values into the equation:

ω= 0 + 2 ×3 = 6 rad/s

Step 3: Therefore, the angular velocity of the wheel after 3 seconds is 6

rad/s.

Question 26

Question

A thin rod of length Land mass mis rotating about one end at a constant

angular speed ω. At a certain instant, a small piece of mass dm breaks off from

the rod at a distance xfrom the end of the rod. What is the kinetic energy of

the piece that breaks off?

Solution

1. To find the kinetic energy of the piece that breaks off, we first need to

determine the angular speed at which the piece was initially moving.

2. The initial angular momentum of the rod and the piece must be conserved.

Therefore, we have: Linitial =Lfinal

3. The initial angular momentum of the system is given by the rod’s angular

momentum: Linitial =Iω

where I=1

3mL2is the moment of inertia of the rod about the end it is

rotating about.

4. The final angular momentum of the system is the sum of the angular

momentum of the rod after the piece breaks off and the angular momentum of

the piece: Lfinal = (I−dm ·x2)(ω) + x(dm)(ωpiece)

5. Setting the initial angular momentum equal to the final angular momen-

tum, we have: Iω = (I−dm ·x2)(ω) + x(dm)(ωpiece)

6. Since the piece breaks off with the same angular velocity as the rod,

ωpiece =ω, and we can solve for ω.

7. Once we have ω, we can determine the kinetic energy of the piece that

breaks off: KE =1

2dm(ωx)2

Question 27

Question

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

1.5 rad/s2. How long will it take for the wheel to make 14 complete revolutions?

Solution

Step 1: Convert the number of revolutions to radians. Given that 1 revolution

is equivalent to 2πradians, 14 revolutions will be 14 ×2π= 28πradians.

Step 2: Use the kinematic equation for rotational motion. The kinematic

equation for rotational motion can be expressed as:

θ=θ0+ω0t+1

2αt2

where: θ= final angular position (radians), θ0= initial angular position (radi-

ans), ω0= initial angular velocity (rad/s), α= angular acceleration (rad/s2),

and t= time (s).

Step 3: Solve for the time required to make 28πradians. Given that the

wheel is initially at rest (ω0= 0 rad/s) and the initial angular position is 0, the

equation simplifies to:

θ=1

2αt2

Substitute the given values: θ= 28πradians and α= 1.5 rad/s2.

28π=1

2×1.5×t2

Step 4: Solve for t. Solve for t:

t2=28π

0.75 = 37.333

t=√37.333 = 6.11 s

Therefore, it will take approximately 6.11 seconds for the wheel to make 14

complete revolutions.

Question 28

Question

A solid cylinder of mass mand radius Ris initially at rest. A constant force F

is applied tangentially to the edge of the cylinder, causing it to rotate about its

central axis. If the force is applied for a time interval ∆t, determine the angular

velocity of the cylinder at the end of the time interval ∆t.

Solution

Step 1: Calculate the torque applied to the cylinder. The torque τapplied to

the cylinder is given by the equation:

τ=F R

where Fis the magnitude of the force applied tangentially to the edge of the

cylinder.

Step 2: Calculate the moment of inertia of the cylinder. The moment of

inertia Iof a solid cylinder rotating about its central axis is given by:

I=1

2mR2

Step 3: Apply Newton’s second law for rotational motion. The net torque

acting on the cylinder is equal to the moment of inertia times the angular

acceleration. We have:

τ=Iα

where αis the angular acceleration of the cylinder.

Step 4: Relate angular acceleration to angular velocity. Since the cylinder

starts from rest, the relationship between angular acceleration αand angular

velocity ωis given by:

α=∆ω

∆t

Step 5: Substitute the expressions for torque and moment of inertia into the

equation. Substitute τ=F R and I=1

2mR2into the equation τ=Iα:

F R =1

2mR2·∆ω

∆t

Step 6: Solve for the angular velocity ω. Solving for ω, we get:

ω=2F∆t

Therefore, the angular velocity of the cylinder at the end of the time interval

∆tis 2F∆t

Question 29

Question

A solid sphere of mass mand radius Ris rolling without slipping on a horizontal

surface. Initially, the sphere has an angular speed ω0. It then rolls up an incline

plane making an angle θwith the horizontal. What is the maximum height h

the center of the sphere reaches on the incline?

Solution

1. At the bottom of the incline, the total mechanical energy of the sphere is

given by the sum of its translational kinetic energy (1

2mv2) and its rotational

kinetic energy (1

2Iω2, where Iis the moment of inertia of a solid sphere about

its center):

Ebottom =1

2mv2

0+1

2Iω2

2. As the sphere reaches its maximum height, its kinetic energy is zero.

Therefore, the mechanical energy of the system at the maximum height is equal

to the gravitational potential energy at that height:

Emax height =mgh

3. Setting Ebottom =Emax height gives:

2mv2

0+1

2Iω2

0=mgh

4. The linear speed of the sphere when it reaches the incline is related to its

angular speed by v=Rω. Substituting v0=Rω0into the equation from step 3

yields: 1

2m(Rω0)2+1

2Iω2

0=mgh

5. The moment of inertia of a solid sphere about its center is 2

5mR2. Sub-

stituting I=2

5mR2into the equation from step 4 gives:

2mR2ω2

0+1

22

5mR2ω2

0=mgh

6. Simplifying the equation from step 5 yields:

10mR2ω2

0=mgh

7. Solving for the maximum height hgives:

h=7

10R2ω2

Therefore, the maximum height the center of the sphere reaches on the

incline is 7

10 R2ω2

Question 30

Question

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

acceleration of 1.5 rad/s2. Find the angular velocity of the disc after 3 seconds.

Solution

Step 1: We can use the kinematic equation for rotational motion to find the

angular velocity of the disc after 3 seconds. The equation relates the final

angular velocity (ωf), initial angular velocity (ωi), angular acceleration (α),

and time (t) as follows:

ωf=ωi+αt

Step 2: Since the disc starts from rest, the initial angular velocity ωiis 0.

Step 3: Substitute the given values into the equation to solve for the final

angular velocity:

ωf= 0 + (1.5 rad/s2)·3 s

Step 4: Calculate the final angular velocity:

ωf= 1.5 rad/s2·3 s = 4.5 rad/s

Answer: The angular velocity of the disc after 3 seconds is 4.5 rad/s.

Question 31

Question

A thin rod of length Lrotates about a pivot at one end with an angular speed ω.

A small object of mass mis attached to the other end. What is the maximum

angular speed of the rod when the object is directly above the pivot? Assume

the rod is massless and the object does not slip on the rod.

Solution

1. To find the maximum angular speed of the rod when the object is directly

above the pivot, we can consider conservation of angular momentum for the

system.

2. The initial angular momentum of the system is given by I1ω, where I1is

the moment of inertia of the rod with the object at a distance Lfrom the pivot.

3. The final angular momentum is I2ωmax, where I2is the moment of inertia

of the rod with the object at the pivot and ωmax is the maximum angular speed.

4. By conservation of angular momentum, we have:

I1ω=I2ωmax

5. The moment of inertia of a rod rotating about one end is 1

3mL2. Thus,

I1=1

3mL2.

6. When the object is at the pivot, by parallel axis theorem, the moment of

inertia of the rod-object system is I2=1

3mL2+mL2=4

3mL2.

7. Substituting I1,I2into the conservation of angular momentum equation,

we get: 1

3mL2ω=4

3mL2ωmax

8. Simplifying, we find:

ωmax =1

4ω

9. Therefore, the maximum angular speed of the rod when the object is

directly above the pivot is 1

4ω.

Question 32

Question

A disk with a radius of 0.2 m is initially at rest. It then accelerates with a

constant angular acceleration of 6 rad/s

for 4 seconds. What is the angular

velocity of the disk after 4 seconds?

Solution

Step 1: Calculate the angular displacement of the disk after 4 seconds using the

equation θ=1

2αt2, where αis the angular acceleration and tis the time.

Step 2: Substitute the values α= 6 rad/s

and t= 4 s into the equation to

find the angular displacement.

θ=1

2×6×(4)2= 48 rad

Step 3: Use the equation of angular velocity ω=ω0+αt to find the angular

velocity of the disk after 4 seconds. Since the disk starts from rest, the initial

angular velocity ω0is 0.

Step 4: Substitute the values α= 6 rad/s

and t= 4 s into the equation to

find the angular velocity.

ω= 0 + 6 ×4 = 24 rad/s

Therefore, the angular velocity of the disk after 4 seconds is 24 rad/s.

Question 33

Question

A solid cylinder of radius Rand mass Mrolls without slipping down a rough

incline that makes an angle θwith the horizontal. The cylinder starts from rest

at the top of the incline. What is the speed of the center of mass of the cylinder

when it reaches the bottom of the incline?

Solution

Step 1: First, we need to determine the acceleration of the cylinder down the

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

and the friction force. The net torque is given by τ=Iα, where Iis the

moment of inertia of the cylinder and αis its angular acceleration. This net

torque produces the angular acceleration α=τ

Step 2: The net torque causing the rotation is due to the gravitational

force and the friction force. The gravitational force acting on the cylinder is

decomposed into two components: one parallel to the incline (mg sin θ) and

the other perpendicular to the incline (mg cos θ), with mbeing the mass of the

cylinder and gthe acceleration due to gravity.

Step 3: The friction force acts up the incline to oppose the motion. The

maximum static friction is given by fmax =µsN, where µsis the coefficient of

static friction and Nis the normal force. The normal force and friction force

can be expressed in terms of the weight of the cylinder, mg.

Step 4: The rolling condition requires that the linear speed vof the center

of the cylinder is related to the angular speed ωby v=Rω. The acceleration

of the center of mass of the cylinder is given by a=Rα.

Step 5: Applying Newton’s second law for translational motion, we have

fnet =Ma, where fnet =ffriction −mg sin θis the net force acting on the

cylinder. Solving for agives a=ffriction

M−gsin θ.

Step 6: Combining the expressions for acceleration from both rotational and

translational dynamics, we can solve for the final velocity of the center of mass of

the cylinder when it reaches the bottom of the incline. This speed will depend

on the angle θ, the coefficients of friction, and the properties of the cylinder

(radius, mass, moment of inertia).

Question 34

Question

A disk with a radius of 0.2 m is spinning at an angular velocity of 5 rad/s.

The angular velocity decreases uniformly by 0.1 rad/s2. What is the angular

displacement of a point on the rim of the disk after 4 seconds?

Solution

Step 1: The final angular velocity after 4 seconds can be calculated using the

formula:

ωf=ωi+αt

where ωf= final angular velocity, ωi= initial angular velocity, α= angular

acceleration, and t= time.

Plugging in the given values:

ωf= 5 rad/s −0.1 rad/s2×4 s

ωf= 5 rad/s −0.4 rad/s

ωf= 4.6 rad/s

Step 2: The angular displacement can be calculated using the formula:

θ=ωit+1

2αt2

where θ= angular displacement, ωi= initial angular velocity, α= angular

acceleration, and t= time.

Plugging in the given values:

θ= 5 rad/s ×4 s + 1

2× −0.1 rad/s2×(4 s)2

θ= 20 rad −0.2 rad

θ= 19.8 rad

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

4 seconds is 19.8 radians.

Question 35

Question

A solid cylinder of radius Rand mass Mis initially at rest. A constant force F

is applied tangentially at the edge of the cylinder and acts for a time interval ∆t.

The cylinder rolls without slipping. Find the angular velocity of the cylinder

after time ∆t.

Solution

Step 1: Determine the torque applied by the force F.

The torque (τ) applied by a force (F) acting at a distance (r) from the axis of

rotation is given by:

τ=F r

Step 2: Find the acceleration of the cylinder.

The net torque applied to the cylinder causes angular acceleration (α). The

relationship between torque and angular acceleration is:

τ=Iα

where Iis the moment of inertia of the cylinder. For a solid cylinder rotating

about its axis, the moment of inertia is I=1

2MR2. Substituting this into the

equation, we get:

F r =1

2MR2α

Solving for α, we find:

α=2F

Step 3: Find the final angular velocity.

The relationship between angular acceleration (α), angular velocity (ω), and

time (∆t) is:

ω=ω0+α∆t

Since the cylinder is initially at rest (ω0= 0), the final angular velocity can be

found by:

ω=α∆t

Therefore, the angular velocity of the cylinder after time ∆tis:

ω=2F∆t

PHYS 305 - INTRODUCTION TO MODERN PHYSICS Rotational kinematics Question Bank Set 4

Students also viewed