PHYS 305 - INTRODUCTION TO MODERN PHYSICS Rotational kinematics Question Bank Set 7 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Rotational

kinematics

Question Bank - Set 7

Liberty University

Question 1

Question

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

angular speed ω. What is the angular momentum of the rod about its center of

mass?

Solution

Let’s first find the angular momentum of a small element of mass dm at a

distance xfrom the rotating end.

Step 1: The angular momentum dL of this small element dm is given by

dL =r×dp, where ris the distance of the element from the axis of rotation,

and dp is the linear momentum of the element.

Step 2: The distance rof this small element dm from the center of mass is

L/2−x.

Step 3: The linear momentum dp of this small element dm is v×dm, where

vis the tangential velocity of the element.

Step 4: The tangential velocity vof this small element dm is given by

v=ωx.

Step 5: Therefore, the angular momentum dL of this small element dm is

dL = (L/2−x)×(ωx dm).

Step 6: Integrating over the entire rod, we get the total angular momentum

Labout the center of mass:

L=ZL

(L/2−x)×(ωx dm) = ωZL

(Lx/2−x2)dm

Step 7: Using the definition of center of mass, we can replace dm with

Ldx:

L=Mω

LZL

(Lx/2−x2)dx

Step 8: Solving the integral, we get:

L=Mω

LL2

4−L3

3=1

4MωL

Therefore, the angular momentum of the rod about its center of mass is

4MωL.

Question 2

Question

A wheel initially at rest accelerates uniformly for 5.0 s and has an angular

velocity of 15 rad/s. If the angular acceleration is 3.0 rad/s2, what is the

angular displacement of the wheel during this time period?

Solution

Step 1: First, we find the angular velocity of the wheel after 5.0 s using the

formula:

ω=ω0+αt

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

angular acceleration, and tis the time duration. Substitute the given values:

ω0= 0 rad/s, α= 3.0 rad/s2, and t= 5.0 s into the formula to get:

ω= 0 + (3.0 rad/s2)×5.0 s = 15 rad/s

Step 2: Next, we calculate the angular displacement using the formula:

θ=ω0t+1

2αt2

Substitute the given values: ω0= 0 rad/s, α= 3.0 rad/s2, and t= 5.0 s into

the formula to get:

θ= 0 ×5.0 + 1

2×3.0×(5.0)2= 37.5 rad

Therefore, the angular displacement of the wheel during this time period is

37.5 radians.

Question 3

Question

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

about a horizontal axis through its end. The rod is released from rest in a

horizontal position. What is the angular speed of the rod when it makes an

angle θwith the vertical?

Solution

Step 1: Draw a free body diagram of the rod.

Step 2: Write the torque equation about the pivot point. The only force

acting on the rod is the gravitational force, which exerts a torque about the

pivot point. The torque produced by the gravitational force is given by τ=

−mgr sin θ, where mis the mass of a small element of length dr of the rod

located at distance rfrom the pivot point.

Step 3: Find the moment of inertia of the rod. The moment of inertia of the

rod about the pivot point is I=1

3ML2.

Step 4: Apply Newton’s second law for rotation: τ=Iα. Substitute the

torque and moment of inertia into the equation: −mgr sin θ=1

3ML2α

Step 5: Relate angular acceleration to angular speed. The angular acceler-

ation α=dω

dt , where ωis the angular speed of the rod.

Step 6: Integrate to relate angle and angular speed. Integrate both sides

with respect to time: R−mgr sin θ dt =R1

3ML2dω

Step 7: Use the initial conditions to solve the integral. When the rod is at

an angle θ, its center of mass is at a height h=L/2 cos θ. The speed of the

center of mass is v=√2gh, and the velocity of the end is v=rω.

Step 8: Solve for the angular speed. Solve for ωusing the equations found

in Step 7.

Step 9: Substitute the known values to find the final expression for the

angular speed. This depends on θand constants like g.

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

vertical is given by the final expression derived.

Question 4

Question

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

at the edge of the disk for a time t. After the force is removed, the disk comes

to rest again after moving a distance 3

4R. Find the coefficient of kinetic friction

between the disk and the surface.

Solution

Step 1: Find the angular acceleration. Let Ibe the moment of inertia of the

disk. The torque due to the force Fis given by τ=F R. Using Iα =τ, where

αis the angular acceleration, we have:

Iα =F R

α=F R

Step 2: Find the angular velocity. The final angular velocity ωfcan be found

with the formula ω2

f=ω2

0+2αθ, where ω0is the initial angular velocity and θis

the angle through which the disk turns. Since the disk starts from rest, ω0= 0.

The angle through which the disk turns is θ=3

4R.

ω2

f= 2 ·F R

I·3

ωf=r6F

IR2

Step 3: Find the coefficient of kinetic friction. At the point where the disk

stops, the friction force fprovides the torque to stop the disk. The torque due

to friction is τf=fR. Since the disk stops, the total torque must be zero.

Therefore:

fR =Iα

fR =I·ωf

Given that the disk comes to rest after sliding a distance of 3

4R, we can

relate fto Fand µk(the coefficient of kinetic friction) using the work-energy

principle:

f·3

4R=F·3

4R−µkn·3

4f=3

4F−3

4µkmg

f=F−µkmg

Substitute f=Iωf

Rt and solve for µk:

F−µkmg =Iωf

µk=F−Iωf

Therefore, the coefficient of kinetic friction between the disk and the surface

is F−

Iωf

mg .

Question 5

Question

A disc has a radius of 0.2 m and a moment of inertia of 0.1 kg·m2about its

center. Initially, the disc is at rest. A force of 5 N is applied tangentially to the

edge of the disc. Find the angular acceleration of the disc at that instant.

Solution

Step 1: We can start by finding the torque that the force F creates about the

center of the disc. The torque τis given by the formula:

τ=r·F

where r is the radius of the disc and F is the applied force. Substituting the

given values, we have:

τ= 0.2 m ·5 N = 1 N ·m

Step 2: Next, we can use the relationship between torque and moment of in-

ertia to find the angular acceleration. The equation relating torque (τ), moment

of inertia (I), and angular acceleration (α) is:

τ=I·α

Substitute the known values:

1 N ·m=0.1 kg ·m2·α

Solving for α, we get:

α=1 N ·m

0.1 kg ·m2= 10 rad/s2

Therefore, the angular acceleration of the disc at that instant is 10 rad/s2.

Question 6

Question

A disc of radius 0.2 m is rotating with an angular velocity of 5 rad/s. If the

angular velocity decreases at a constant rate over 2 seconds until it reaches 2

rad/s, determine the angular acceleration of the disc during this time period.

Solution

Step 1: Find the initial angular acceleration using the formula α=ωf−ωi

twhere:

α= angular acceleration, ωf= final angular velocity, and ωi= initial angular

velocity.

Given ωi= 5 rad/s, ωf= 2 rad/s, and t= 2 seconds, we have:

α=2−5

2=−3

2=−1.5 rad/s2

Therefore, the initial angular acceleration of the disc is -1.5 rad/s2.

Step 2: The given information suggests that the angular acceleration is

constant over this time period, so we can use the average angular acceleration

formula α=∆ω

∆twhere: ∆ω= change in angular velocity, and ∆t= time

interval.

Given ∆ω=ωf−ωi= 2 −5 = −3 rad/s and ∆t= 2 seconds, we have:

α=−3

2=−1.5 rad/s2

Therefore, the angular acceleration of the disc during this time period is -1.5

rad/s2.

Question 7

Question

A solid sphere of mass Mand radius Ris released from rest at the top of a

ramp that is inclined at an angle θ. The sphere rolls down the ramp without

slipping. What is the speed of the sphere when it reaches the bottom of the

ramp?

Solution

Step 1: Determine the height the sphere descends. The height the sphere

descends is given by the vertical component of the initial position, which is

h=Rsin(θ).

Step 2: Apply conservation of energy. The initial gravitational potential

energy is converted into both kinetic energy of translation and rotational kinetic

energy at the bottom of the ramp.

mgR sin(θ) = 1

2mv2+1

2Iω2

where vis the speed of the sphere and ωis the angular velocity. The moment

of inertia for a solid sphere rotating about its diameter is I=2

5MR2.

Step 3: Relate the linear and angular velocities for rolling motion. For a

solid sphere rolling without slipping, the angular velocity ωis related to the

linear velocity vby ω=v

Step 4: Substitute equations and solve for v. Substitute ω=v

Rand I=

5MR2into the conservation of energy equation.

mgR sin(θ) = 1

2mv2+1

22

5MR2v

R2

Solving for vgives:

v=r5

7gR sin(θ)

Therefore, the speed of the sphere when it reaches the bottom of the ramp

is v=q5

7gR sin(θ).

Question 8

Question

A disc of radius Rand mass Mis rotating about a fixed axis perpendicular to

the disc through its center. Initially, the disc is not rotating, but a torque τis

applied to it for a time t. After some time, the disc reaches an angular velocity

ω. Determine the torque τrequired to make the disc reach this angular velocity.

Solution

Step 1: Start by relating the torque to the change in angular momentum. The

torque τapplied to the disc causes a change in angular momentum ∆L. We

know that τ=∆L

∆t.

Step 2: The initial angular momentum of the disc is zero since it is not

initially rotating. The final angular momentum of the disc is given by L=Iω,

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

Step 3: The moment of inertia of a disc rotating about an axis perpendicular

to its surface through its center is given by I=1

2MR2.

Step 4: The change in angular momentum is ∆L=L−0 = Iω.

Step 5: Putting it all together, we have τ=Iω

2MR2ω

t=1

2MR2ω

Therefore, the torque τrequired to make the disc reach the angular velocity

ωis 1

2MR2ω

Question 9

Question

A disc of radius 0.5 m is rotating about its center with an angular speed of 4

rad/s. A small bug sitting on the edge of the disc (at a distance of 0.5 m from

the center) decides to walk towards the center of the disc at a speed of 0.1 m/s

relative to the edge. What is the bug’s angular velocity when it reaches the

center of the disc?

Solution

Step 1: First, we need to find the bug’s initial linear velocity at the edge of

the disc. Since the bug is walking towards the center from the edge, its relative

velocity with respect to the center is the sum of its tangential velocity and the

disc’s tangential velocity at the edge. Let vbbe the bug’s velocity, vdbe the

disc’s tangential velocity, and vtbe the tangential velocity of the bug. The bug’s

velocity (vb) = bug’s tangential velocity (vt) + disc’s tangential velocity (vd).

The bug’s tangential velocity at the edge (vt) = 0.1 m/s, which is given in the

problem. The disc’s tangential velocity at the edge (vd) = radius

angular

speed = 0.5m×4rad/s = 2m/s. So, vb=vt+vd= 0.1m/s + 2m/s = 2.1m/s.

Step 2: Next, let’s find the conservation of angular momentum of the bug

when it reaches the center of the disc. The bug’s initial angular momentum is

given by Li=mr2ω, where mis the mass of the bug, ris the initial radius of

the bug, and ωis the initial angular velocity of the bug. The bug’s final angular

momentum is given by Lf=mr′2ω′, where r′is the final radius of the bug

(which is at the center of the disc) and ω′is the final angular velocity of the bug.

Since angular momentum is conserved, Li=Lf. Therefore, mr2ω=mr′2ω′.

Given that r= 0.5 m, ω= 4 rad/s, r′= 0 (at the center), and we need to

find ω′. Thus, 0.52×4 = 0 ×ω′. So, 2 = 0. This is a contradiction and shows

that angular momentum is not conserved in this scenario. Therefore, we cannot

determine the bug’s angular velocity when it reaches the center of the disc using

the conservation of angular momentum.

Question 10

Question

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

down a θ-degree incline. What is its acceleration after it has rolled through a

distance don the incline?

Solution

Step 1: We can start by drawing a free-body diagram of the rolling sphere. The

forces acting on the sphere are the gravitational force mg, the normal force N,

and the frictional force f. The frictional force fwill be pointing up the incline.

Step 2: We can write the net force equation for the rolling sphere in the

x-direction: f=Ma, where ais the acceleration of the sphere down the incline.

The frictional force fcan be expressed as f=µN, where µis the coefficient of

static friction.

Step 3: The normal force Ncan be determined by using the equations for

forces in the y-direction: N−mg cos(θ) = 0. Therefore, N=mg cos(θ).

Step 4: Now, the frictional force fcan be expressed as f=µmg cos(θ).

Step 5: To determine the acceleration a, we can substitute for fin the net

force equation: µmg cos(θ) = Ma. Solving for a, we get a=µg cos(θ).

Step 6: So, the acceleration of the solid sphere after it has rolled through a

distance don the incline is a=µg cos(θ).

Question 11

Question

A steel ball is dropped from rest from a height of 2.0 m onto a hard floor and

rebounds to a height of 1.5 m. If the ball is in contact with the floor for 0.030

s, what is the magnitude of the average velocity of the ball during contact with

the floor? Assume the ball doesn’t slip as it rebounds.

Solution

To find the average velocity of the ball during contact with the floor, we need

to calculate the velocity immediately before and after the collision.

Step 1: Calculate the initial velocity of the ball just before hitting the floor.

The initial velocity can be found using the kinematic equation:

vf=vi+at

where vfis the final velocity just before hitting the floor, viis the initial velocity,

ais the acceleration due to gravity (9.81 m/s2), and tis the time taken for the

ball to fall from a height of 2.0 m.

Using vf= 0 (since the ball is dropped), we have:

0 = vi+ (9.81 m/s2)×t

Solving for vi, we get:

vi=−9.81t

Step 2: Calculate the final velocity of the ball just after leaving the floor.

Since the ball rebounds to a height of 1.5 m, the final velocity just after leaving

the floor can be found using the conservation of mechanical energy:

mgh =1

2mv2

where mis the mass of the ball (assume mass cancels out), his the height of

the rebound, and vfis the final velocity just after leaving the floor.

Plugging in the values, we get:

(9.81 m/s2)(0.5 m) = 1

2v2

4.905 = 1

2v2

vf=√9.81

vf= 3.13 m/s

Step 3: Calculate the average velocity of the ball during contact with the

floor. The average velocity during contact with the floor can be found using:

Average velocity = Change in velocity

Time taken

Average velocity = vf−vi

∆t

Average velocity = 3.13 −(−9.81t)

0.030

Average velocity = 3.13 + 9.81t

0.030

Thus, the magnitude of the average velocity of the ball during contact with

the floor is 3.13+9.81t

0.030 .

Question 12

Question

An object starts from rest and experiences a constant angular acceleration of

0.05 rad/s2. If the object reaches an angular velocity of 1.2 rad/s after 12 sec-

onds, determine the angular position (in radians) of the object at this time.

Solution

Step 1: Find the angular velocity at time tusing the equation:

ω=ω0+αt

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

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

ω= 0 + (0.05)(12) = 0.6 rad/s

Step 2: Find the angular position at time tusing the equation:

θ=θ0+ω0t+1

2αt2

where θis the final angular position, θ0is the initial angular position (which is

0 in this case).

θ=0+0+1

2(0.05)(12)2= 3.6 rad

Therefore, the angular position of the object at t= 12 s is 3.6 rad .

Question 13

Question

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

on the rim of the disk suddenly breaks off. How far along the ground will the

point land, measured from the base of the disk? Assume the center of the disk

is 1 m above the ground.

Solution

Step 1: Find the linear velocity of the point that breaks off the rim of the disk.

The linear velocity (v) of a point on the rim of a rotating disk is given by the

formula:

v=ωr

where v= linear velocity, ω= angular velocity, r= radius of the disk.

Substitute ω= 5 rad/s and r= 0.2 m into the formula:

v= (5 rad/s) ×0.2 m = 1 m/s

Step 2: Find the time it takes for the point to land. Since the point falls a

height of 1 m, you can use the kinematic equation for vertical motion:

h=1

2gt2

where h= height fallen, g= acceleration due to gravity (9.8 m/s2), t= time

taken to fall.

Rearrange the formula to solve for t:

t=s2h

g=s2×1 m

9.8 m/s2≈0.45 s

Step 3: Find the horizontal distance traveled by the point. Since the hor-

izontal distance traveled is equal to the horizontal component of the velocity

multiplied by the time taken:

Distance = vht

The horizontal component of the velocity is the same as the linear velocity of the

point, which is 1 m/s. Substitute vh= 1 m/s and t≈0.45 s into the formula:

Distance = 1 m/s ×0.45 s = 0.45 m

Therefore, the point will land 0.45 m along the ground from the base of the

disk.

Question 14

Question

A wheel initially at rest, starts rotating with an angular acceleration of 5 rad/s2.

After 2 seconds, the wheel has made 10 complete revolutions. What is the

angular velocity of the wheel at this time?

Solution

Step 1: Calculate the angular displacement of the wheel after 2 seconds. Given

that the wheel has made 10 complete revolutions, we can calculate the angular

displacement using the formula:

θ= 2π×number of revolutions

θ= 2π×10 = 20πrad

Step 2: Use the kinematic equation for rotational motion to find the final

angular velocity. The kinematic equation for rotational motion relating ini-

tial angular velocity ωi, final angular velocity ωf, angular acceleration α, and

angular displacement θis:

ω2

f=ω2

i+ 2αθ

Since the wheel is initially at rest, ωi= 0. Thus, the equation simplifies to:

ω2

f= 2αθ

Substitute the given values: α= 5 rad/s2and θ= 20πrad.

ω2

f= 2 ×5×20π= 200πrad/s

Step 3: Calculate the final angular velocity. Taking the square root of both

sides gives us the final angular velocity:

ωf=√200π≈25.13 rad/s

Therefore, the angular velocity of the wheel after 2 seconds is approximately

25.13 rad/s.

Question 15

Question

A wheel starts from rest and accelerates with a constant angular acceleration

of 2 rad/s2. After 3 seconds, what is the angular velocity of the wheel?

Solution

Step 1: Identify the given values and the unknown. Let ωibe the initial angular

velocity (0 rad/s), αbe the constant angular acceleration (2 rad/s2), and tbe

the time (3 seconds). We are asked to find the final angular velocity ωf.

Step 2: Use the equation for angular velocity with constant angular acceler-

ation. The equation relating initial angular velocity, angular acceleration, time,

and final angular velocity is:

ωf=ωi+αt

Substitute the given values into the equation:

ωf= 0 + 2 ×3

Step 3: Calculate the final angular velocity.

ωf= 6 rad/s

Therefore, after 3 seconds, the angular velocity of the wheel is 6 rad/s.

Question 16

Question

A disk with a radius of 0.5 m is spinning with an angular velocity of 5 rad/s. A

small bug is crawling on the edge of the disk. What is the bug’s linear speed if

it is 1/4 of the way from the center of the disk?

Solution

Step 1: Determine the bug’s angular position relative to the center of the disk.

The bug is 1/4 of the way from the center, so its radial distance from the center

is 0.5×1

4= 0.125 m. Since the disk is rotating with an angular velocity of

5 rad/s, the bug’s angular position relative to the center of the disk can be

calculated as follows:

θ=ωt = 5t

Step 2: Calculate the linear speed of the bug.

The bug’s linear speed can be calculated using the formula:

v=rω

where ris the radial distance from the center of the disk and ωis the angular

velocity of the disk.

Substitute r= 0.125 m and ω= 5 rad/s into the formula to find the bug’s

linear speed:

v= 0.125 ×5=0.625 m/s

Therefore, the bug’s linear speed is 0.625 m/s.

Question 17

Question

A disc of radius Rand mass Mrotates about a fixed axis passing through its

center with an angular velocity ω=2

t2rad/s, where tis in seconds. Find an

expression for the magnitude of the net torque acting on the disc as a function

of time.

Solution

Step 1: The moment of inertia of a disc rotating about an axis passing through

its center is given by I=1

2MR2.

Step 2: The angular acceleration αis given by the time derivative of the

angular velocity, α=dω

dt =−4

t3rad/s2.

Step 3: The net torque acting on the disc is related to the moment of inertia

and angular acceleration by the equation τ=Iα.

Step 4: Substituting in the values for Iand α, we get τ=1

2MR2−4

t3.

Step 5: Simplifying, we find τ=−2MR2

t3N m.

Therefore, the magnitude of the net torque acting on the disc as a function

of time is τ=2MR2

t3N m .

Question 18

Question

A disc of radius Ris spinning about a vertical axis with an angular velocity ω.

A small block is placed at a distance rfrom the center of the disc on its edge.

Determine the velocity of the block as it falls off the disc.

Solution

Let’s first determine the velocity of the block on the edge of the disc before

it falls off. The velocity is a vector quantity, so we need to consider both the

tangential and radial components.

Step 1: The tangential velocity of the block on the edge of the disc is given

by the formula for the tangential speed in uniform circular motion:

vt=rω

where vt= tangential velocity of the block, r= distance of the block from the

center of the disc (radius of the disc), ω= angular velocity of the disc.

Step 2: Now, let’s calculate the radial velocity of the block. At the moment

the block leaves the disc, it acquires a tangential velocity vtand a downward

velocity component due to gravity (g).

The radial velocity can be found using the Pythagorean theorem:

vradial =qv2

t+v2

down

where vradial = radial velocity of the block, vt= tangential velocity of the block,

vdown = downward velocity of the block (due to gravity).

Step 3: The downward velocity of the block due to gravity equals the free-

fall acceleration multiplied by the time taken to fall off the disc:

vdown =gt

where g= acceleration due to gravity, t= time taken to fall off the disc.

Step 4: As the block loses contact with the disc, its radial velocity equals

the velocity that it would have if it had simply fallen straight down from height

vradial =qv2

t+ (gt)2

Therefore, the velocity of the block as it falls off the disc is pr2ω2+g2t2.

Question 19

Question

A thin rod of length Land mass Mis free to rotate about its center point.

Initially, it is at rest and then released. Find the angular acceleration of the rod

when it makes an angle θwith the vertical.

Solution

Step 1: Let’s consider the forces acting on the rod at an angle θwith the

vertical. Step 2: The gravitational force acting on the center of mass of the

rod can be separated into two components: mg sin θparallel to the rod and

mg cos θperpendicular to the rod. Step 3: The torque about the center of

the rod due to the gravitational force is given by τ=r×F, where ris the

distance from the pivot to the center of mass. Step 4: The torque due to the

component mg sin θis zero as it passes through the pivot point. The torque

due to the component mg cos θis mg cos θ×L

2. Step 5: According to Newton’s

second law for rotation, the net torque is equal to the moment of inertia times

the angular acceleration. Step 6: The moment of inertia of the rod about its

center is I=1

12 ML2. Step 7: Combining the equations from steps 4 and 5, we

have mg cos θ×L

2=1

12 ML2×α, where αis the angular acceleration. Step 8:

Simplifying further, we find α=6gcos θ

Las the angular acceleration of the rod.

Question 20

Question

A disk with a radius of 0.5 meters is rotating at an angular speed of 4 rad/s. At

time t= 0, the disk starts to slow down with a constant angular acceleration of

-0.5 rad/s2. What is the angular position of a point on the rim of the disk after

4 seconds?

Solution

Step 1: Calculate the initial angular velocity

Given: Initial angular velocity, ωi= 4 rad/s Angular acceleration, α=−0.5

rad/s2Time, t= 0 seconds

The final angular velocity at time t= 0 can be calculated using the equation:

ωf=ωi+αt

Substitute the values into the equation:

ωf= 4 + (−0.5)(0) = 4 rad/s

Step 2: Calculate the angular position at time t= 4 seconds

The angular position can be calculated using the equation:

θ=θ0+ωit+1

2αt2

Given that θ0= 0 (initial angular position), ωi= 4 rad/s, α=−0.5 rad/s2,

and t= 4 seconds, substitute the values into the equation:

θ= 0 + 4(4) + 1

2(−0.5)(4)2

θ= 16 + 1

2(−0.5)(16)

θ= 16 −4 = 12 rad

Therefore, the angular position of a point on the rim of the disk after 4

seconds is 12 radians.

Question 21

Question

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

down a frictionless incline with an angle of elevation θ. What is the speed of

the center of mass of the sphere when it has descended a vertical distance h?

Solution

Step 1: First, we need to determine the final position of the center of mass of

the sphere. When the center of mass has descended a vertical distance h, the

height of the center of mass from the initial point is h, which means the sphere

has rotated by an angle ϕgiven by:

Rϕ =h

So, the final position of the center of mass is at a distance Rϕ along the incline.

Step 2: Next, we can calculate the final velocity of the center of mass using

the conservation of energy. The gravitational potential energy lost by the sphere

is converted into kinetic energy (translational and rotational). The gravitational

potential energy lost is Mgh, and the kinetic energy gained is 1

2Mv2

cm +1

2Iω2,

where vcm is the speed of the center of mass, and Iand ωare the moment of

inertia and angular velocity of the sphere, respectively.

Step 3: Since the sphere is rolling without slipping, we have the relationship

vcm =Rω. We can express Iin terms of Mand Ras I=2

5MR2for a

solid sphere. Substituting these relationships into the conservation of energy

equation:

Mgh =1

2Mv2

cm +1

22

5MR2vcm

R2

Step 4: Simplifying the equation:

Mgh =1

2Mv2

cm +1

5Mv2

Mgh =7

10Mv2

vcm =r10

7gh

Therefore, the speed of the center of mass of the sphere when it has descended

a vertical distance his q10

7gh.

Question 22

Question

A wheel of radius 0.3 m starts from rest and rotates with a constant angular

acceleration of 4 rad/s2. Determine the time it takes for the wheel to make 5

complete revolutions.

Solution

Step 1: Find the angular velocity after 5 revolutions.

Given: Radius of the wheel, r= 0.3 m Angular acceleration, α= 4 rad/s2

Number of revolutions, n= 5

We know that the final angular velocity (ωf) can be determined using the

equation:

ω2

f=ω2

i+ 2αθ

where: ωi= initial angular velocity (since the wheel starts from rest, ωi= 0)

ωf= final angular velocity α= angular acceleration θ= angular displacement

(in radians)

The angular displacement for 5 complete revolutions is:

θ= 2πn = 2π×5 = 10πrad

Substitute the given values into the equation:

ω2

f= 0 + 2(4)(10π) = 80πrad/s

Therefore, the final angular velocity after 5 revolutions is 80πrad/s.

Step 2: Find the time taken to complete 5 revolutions.

The final angular velocity is related to the initial angular velocity, angular

acceleration, and time by the equation:

ωf=ωi+αt

Substitute the given values into the equation:

80π= 0 + 4t

Solve for t:

t=80π

4= 20πs≈62.83 s

Therefore, it takes approximately 62.83 seconds for the wheel to make 5

complete revolutions.

Question 23

Question

A solid sphere of mass mand radius Ris initially at rest on a frictionless

horizontal surface. A horizontal force Fis applied at a distance 3

4Rfrom the

center of the sphere. What is the angular acceleration of the sphere when the

force is applied?

Solution

1. The torque produced by the force can be calculated using the formula τ=

r×F, where ris the lever arm from the pivot point to the point of application

of the force F. In this case, r=3

4R. 2. Thus, the torque τis given by:

τ=3

4R×F

3. The moment of inertia of a solid sphere rotating about an axis passing

through its center is given by I=2

5mR2. 4. The torque τcan also be related

to the angular acceleration αand moment of inertia Iby the formula τ=Iα.

5. Setting the two expressions for torque equal to each other gives:

3

4RF=2

5mR2α

6. Solving for α, we find:

α=3

4RF

2

5mR2

7. Simplifying, we get:

α=15

Question 24

Question

A wheel starts from rest and undergoes constant angular acceleration for 5.0 s.

During this time, it rotates 50 revolutions. Calculate the angular acceleration

of the wheel.

Solution

Step 1: Determine the initial angular velocity. The initial angular velocity of

the wheel is 0 rad/s since it starts from rest.

Step 2: Calculate the final angular velocity. The final angular velocity can

be calculated using the formula:

ωf=ωi+αt

where: - ωfis the final angular velocity, - ωiis the initial angular velocity, - α

is the angular acceleration, - tis the time.

Substitute the known values:

ωf= 0 + α(5.0 s)

Step 3: Calculate the number of radians rotated. Given that the wheel

completes 50 revolutions during the 5.0 s, we can calculate the number of radians

using:

θ= number of revolutions ×2π

θ= 50 ×2π

Step 4: Calculate the average angular velocity. The formula for average

angular velocity is:

¯ω=θ

By substituting the values:

¯ω=50 ×2π

5.0

Step 5: Use the average angular velocity to find the final angular velocity.

The final angular velocity ωfis equal to the average angular velocity ¯ωunder

constant angular acceleration. Therefore, ωf= ¯ω.

Step 6: Find the angular acceleration. Now we have the final angular velocity

ωfand the initial angular velocity ωi, we can calculate the angular acceleration

using the formula:

α=ωf−ωi

Substitute the values and calculate the angular acceleration.

Question 25

Question

A wheel starts from rest and accelerates uniformly for 6.0 s through 14 rad.

What is the angular velocity of the wheel at the end of the 6.0 s time interval?

Solution

Step 1: Recall the equation for rotational motion under constant angular accel-

eration:

angular displacement = ωit+1

2αt2

where angular displacement(θ) = 14rad time interval(t)=6.0s initial angular velocity(ωi) =

0 angular acceleration (α) =?

Step 2: Solve for the angular acceleration (α) using the given values:

14 rad = 0 ×6.0 s + 1

2α(6.0 s)2

14 rad = 1

2α(36)

α=14

18 rad/s2

α= 0.778 rad/s2

Step 3: Now, use the angular acceleration to find the angular velocity at the

end of the time interval using the equation:

ωf=ωi+αt

ωf= 0 + 0.778 ×6.0

ωf= 4.67 rad/s

Therefore, the angular velocity of the wheel at the end of the 6.0 s time

interval is ωf= 4.67 rad/s.

Question 26

Question

A solid sphere of mass mand radius Rstarts from rest and rolls without slipping

down a 30◦incline that is hmeters high. What is the final speed of the sphere

at the bottom of the incline?

Solution

Step 1: Let’s denote the final speed of the sphere at the bottom of the incline

as vf.

Step 2: The total energy at the top of the incline is equal to the total energy

at the bottom of the incline:

mgh =1

2mv2

f+1

2Iω2

Step 3: The potential energy at the top is converted into both translational

and rotational kinetic energies at the bottom. For the rolling sphere, the moment

of inertia I=2

5mR2and the angular velocity ω=vf

R, substituting these into

the energy equation yields:

mgh =1

2mv2

f+1

22

5mR2vf

R2

Step 4: Simplifying the equation gives:

mgh =1

2mv2

f+1

5mv2

Step 5: Combining the terms with vfand isolating vfyields:

vf=r10

7gh =r10

7gh

Thus, the final speed of the sphere at the bottom of the incline is q10

7gh.

Question 27

Question

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

incline. The incline makes an angle θwith the horizontal. Initially, the sphere

is released from rest at a height habove the base of the incline. Determine the

translational and rotational kinetic energies of the sphere when it reaches the

base of the incline.

Solution

Step 1: Find the final velocity of the sphere when it reaches the base of the

incline. Let vbe the final velocity of the sphere. Using conservation of energy:

mgh =1

2mv2+1

2Iω2

where mis the mass of the sphere, his the initial height, vis the final velocity, I

is the moment of inertia of the sphere and ωis the angular velocity. The moment

of inertia for a solid sphere rolling without slipping is I=2

5mR2. The linear

velocity and angular velocity of the sphere are related by v=Rω. Substitute

I=2

5mR2and v=Rω:

mgh =1

2mv2+1

22

5mR2v

R2

Solve for v:

v=r5

7gh

Step 2: Calculate the translational kinetic energy when the sphere reaches

the base of the incline. The translational kinetic energy KEtrans is given by:

KEtrans =1

2mv2

Substitute v=q5

7gh:

KEtrans =1

2m r5

7gh!2

KEtrans =5

14mgh

Step 3: Determine the rotational kinetic energy at the base of the incline.

The rotational kinetic energy KErot is given by:

KErot =1

2Iω2

Substitute I=2

5mR2and ω=v

KErot =1

22

5mR2v

R2

KErot =1

5mv2

Substitute v=q5

7gh:

KErot =1

5m5

7gh

KErot =1

7mgh

Therefore, the translational kinetic energy at the base of the incline is 5

14 mgh

and the rotational kinetic energy is 1

7mgh.

Question 28

Question

A wheel is rotating at an angular velocity of 10 rad/s. The wheel is slowly

brought to a stop in 5 seconds. What is the angular acceleration of the wheel?

Solution

Step 1: Determine the initial and final angular velocities. The initial angular

velocity, ωi, is 10 rad/s. The final angular velocity, ωf, is 0 rad/s (since the

wheel is brought to a stop).

Step 2: Use the formula for angular acceleration. The angular acceleration,

α, can be calculated using the formula:

α=ωf−ωi

Substitute the known values into the formula:

α=0−10

Step 3: Calculate the angular acceleration.

α=−10

5=−2 rad/s2

Therefore, the angular acceleration of the wheel is −2 rad/s2.

Question 29

Question

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

axis passing through one of its ends. The rod is initially at rest and then a

force Fis applied to the other end of the rod horizontally. Find the angular

acceleration of the rod just after the force is applied.

Solution

Step 1: The moment of inertia of the rod about the axis passing through one

end is I=1

3ML2.

Step 2: The torque about the axis due to the force Fis τ=rF =L·F.

Step 3: The net torque τnet is equal to the moment of inertia Itimes the

angular acceleration α. Therefore, τnet =Iα. Substituting the values, we have

L·F=1

3ML2·α.

Step 4: Solving for α, we find α=3F

ML .

Therefore, the angular acceleration of the rod just after the force is applied

is α=3F

ML .

Question 30

Question

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

acceleration of 2 rad/s2. Find the angular velocity of the disk after 4 seconds.

Solution

Step 1: The angular acceleration of the disk is given as α= 2 rad/s2.

Step 2: We know that angular acceleration αis related to angular velocity

ωand time tthrough the equation:

α=∆ω

∆t

Step 3: Solving for the change in angular velocity ∆ω, we have:

∆ω=α·∆t

Step 4: Substituting the given values into the equation, we get:

∆ω= 2 rad/s2×4 s = 8 rad/s

Step 5: The final angular velocity ωfis the initial angular velocity ωiplus

the change in angular velocity:

ωf=ωi+ ∆ω

Step 6: Since the disk starts from rest, the initial angular velocity ωi= 0.

Therefore,

ωf= 0 + 8 rad/s = 8 rad/s

Step 7: So, the angular velocity of the disk after 4 seconds is 8 rad/s.

Question 31

Question

A thin rod of length Lis rotating about an axis passing through one of its ends

with an angular velocity ω. If the rod suddenly comes to rest, find the linear

velocity of a point located on the rod at a distance dfrom the axis of rotation.

Solution

Step 1: First, we need to calculate the angular acceleration of the rod as it

comes to rest. The initial angular velocity is ω, and the final angular velocity

is 0. The time taken for the rod to come to rest is not given, so we cannot

use the relationship ωf=ωi+αt directly. However, we know that the angular

acceleration αcan be calculated as the change in angular velocity divided by

time. In this case, it is −ω/t. Step 2: Next, we need to find the linear velocity

of the point at a distance dfrom the axis of rotation when the rod comes to rest.

The linear velocity vof a point on the rod a distance rfrom the axis of rotation

is related to the angular velocity ωby v=rω. So, the linear velocity of the

point at a distance dfrom the axis when the rod comes to rest is v=d·0=0

as the entire rod comes to rest.

Question 32

Question

A disk is rotating with an angular velocity of 5 rad/s. A torque is applied to the

disk causing it to slow down with a constant angular acceleration of -2 rad/s2.

If the initial angular position of the disk is 0 radians, determine the angular

position of the disk after 3 seconds.

Solution

Step 1: First, we find the angular velocity of the disk after 3 seconds using the

equation:

ω=ω0+αt

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

angular acceleration, and tis the time.

ω= 5 rad/s −2 rad/s2×3 s

ω= 5 rad/s −6 rad/s

ω=−1 rad/s

Step 2: Next, we find the angular position of the disk after 3 seconds using

the equation:

θ=θ0+ω0t+1

2αt2

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

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

θ= 0 rad + 5 rad/s ×3 s + 1

2(−2 rad/s2)(3 s)2

θ= 15 rad −9 rad

θ= 6 rad

Therefore, the angular position of the disk after 3 seconds is 6 radians.

Question 33

Question

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

velocity ω. What is the kinetic energy of the rod?

Solution

Step 1: The kinetic energy of the rod can be calculated by summing the kinetic

energy of each particle that makes up the rod. We can divide the rod into

infinitesimally small particles and sum their kinetic energies.

Step 2: Consider a small element of mass dm at a distance xfrom the

rotating axis. The kinetic energy of this element is given by 1

2dm ·(ωx)2.

Step 3: To find the total kinetic energy of the rod, we need to integrate the

kinetic energy of each element of mass along the length of the rod. The total

kinetic energy is given by

KE =Z1

2(ωx)2dm

Step 4: To express dm in terms of dx (length element), we can use the linear

mass density λ=M

Lsuch that dm =λ dx. Substituting dm and x=x′into the

integral, we get

KE =ZL

2(ωx′)2λ dx′

Step 5: Simplifying the integral, we get

KE =1

2λω2ZL

x′2dx′

Step 6: Solving the integral, we find

KE =1

2λω21

3x′3L

KE =1

6λω2L3

Step 7: Substituting λ=M

Linto the equation, we get the kinetic energy of

the rotating rod

KE =1

Lω2L3=1

6Mω2L2

Question 34

Question

A thin rod of length Land mass Mis rotated about one end with a constant

angular acceleration α. At some point, the gravitational force acting on the rod

is 3

4Mg where gis the acceleration due to gravity. If the rod has a moment of

inertia I=1

3ML2about its rotating end, find the angular acceleration of the

rod at this point.

Solution

Step 1: Start by writing down the sum of the torques acting on the rod about

the rotating end. Step 2: The torque due to the gravitational force is τgravity =

4Mg·L·sin(90◦). Step 3: The torque due to the rotational motion is τrotational =

I·α. Step 4: Set the sum of the torques equal to I·αand solve for α.

Question 35

Question

A uniform solid sphere of radius Rand mass Mis initially at rest on a frictionless

surface. The sphere is then set rolling without slipping down an inclined plane

that makes an angle θwith the horizontal. Calculate the linear acceleration of

the center of mass of the sphere as it moves down the incline.

Solution

Step 1: We can start by calculating the moment of inertia of the solid sphere

about its center. The moment of inertia of a solid sphere rotating about an axis

through its center is 2

5MR2.

Step 2: The net torque on the sphere is due to the gravitational force acting

on the sphere. The component of the gravitational force along the incline is

Question 3

Question

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

about a horizontal axis through its end. The rod is released from rest in a

horizontal position. What is the angular speed of the rod when it makes an

angle θwith the vertical?

Solution

Step 1: Draw a free body diagram of the rod.

Step 2: Write the torque equation about the pivot point. The only force

acting on the rod is the gravitational force, which exerts a torque about the

pivot point. The torque produced by the gravitational force is given by τ=

−mgr sin θ, where mis the mass of a small element of length dr of the rod

located at distance rfrom the pivot point.

Step 3: Find the moment of inertia of the rod. The moment of inertia of the

rod about the pivot point is I=1

3ML2.

Step 4: Apply Newton’s second law for rotation: τ=Iα. Substitute the

torque and moment of inertia into the equation: −mgr sin θ=1

3ML2α

Step 5: Relate angular acceleration to angular speed. The angular acceler-

ation α=dω

dt , where ωis the angular speed of the rod.

Step 6: Integrate to relate angle and angular speed. Integrate both sides

with respect to time: R−mgr sin θ dt =R1

3ML2dω

Step 7: Use the initial conditions to solve the integral. When the rod is at

an angle θ, its center of mass is at a height h=L/2 cos θ. The speed of the

center of mass is v=√2gh, and the velocity of the end is v=rω.

Step 8: Solve for the angular speed. Solve for ωusing the equations found

in Step 7.

Step 9: Substitute the known values to find the final expression for the

angular speed. This depends on θand constants like g.

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

vertical is given by the final expression derived.

Question 4

Question

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

at the edge of the disk for a time t. After the force is removed, the disk comes

to rest again after moving a distance 3

4R. Find the coefficient of kinetic friction

between the disk and the surface.

Solution

Step 1: Find the angular acceleration. Let Ibe the moment of inertia of the

disk. The torque due to the force Fis given by τ=F R. Using Iα =τ, where

αis the angular acceleration, we have:

Iα =F R

α=F R

Step 2: Find the angular velocity. The final angular velocity ωfcan be found

with the formula ω2

f=ω2

0+2αθ, where ω0is the initial angular velocity and θis

the angle through which the disk turns. Since the disk starts from rest, ω0= 0.

The angle through which the disk turns is θ=3

4R.

ω2

f= 2 ·F R

I·3

ωf=r6F

IR2

Step 3: Find the coefficient of kinetic friction. At the point where the disk

stops, the friction force fprovides the torque to stop the disk. The torque due

to friction is τf=fR. Since the disk stops, the total torque must be zero.

Therefore:

fR =Iα

fR =I·ωf

Given that the disk comes to rest after sliding a distance of 3

4R, we can

relate fto Fand µk(the coefficient of kinetic friction) using the work-energy

principle:

f·3

4R=F·3

4R−µkn·3

4f=3

4F−3

4µkmg

f=F−µkmg

Substitute f=Iωf

Rt and solve for µk:

F−µkmg =Iωf

µk=F−Iωf

Therefore, the coefficient of kinetic friction between the disk and the surface

is F−

Iωf

mg .

Question 5

Question

A disc has a radius of 0.2 m and a moment of inertia of 0.1 kg·m2about its

center. Initially, the disc is at rest. A force of 5 N is applied tangentially to the

edge of the disc. Find the angular acceleration of the disc at that instant.

Solution

Step 1: We can start by finding the torque that the force F creates about the

center of the disc. The torque τis given by the formula:

τ=r·F

where r is the radius of the disc and F is the applied force. Substituting the

given values, we have:

τ= 0.2 m ·5 N = 1 N ·m

Step 2: Next, we can use the relationship between torque and moment of in-

ertia to find the angular acceleration. The equation relating torque (τ), moment

of inertia (I), and angular acceleration (α) is:

τ=I·α

Substitute the known values:

1 N ·m=0.1 kg ·m2·α

Solving for α, we get:

α=1 N ·m

0.1 kg ·m2= 10 rad/s2

Therefore, the angular acceleration of the disc at that instant is 10 rad/s2.

Question 6

Question

A disc of radius 0.2 m is rotating with an angular velocity of 5 rad/s. If the

angular velocity decreases at a constant rate over 2 seconds until it reaches 2

rad/s, determine the angular acceleration of the disc during this time period.

Solution

Step 1: Find the initial angular acceleration using the formula α=ωf−ωi

twhere:

α= angular acceleration, ωf= final angular velocity, and ωi= initial angular

velocity.

Given ωi= 5 rad/s, ωf= 2 rad/s, and t= 2 seconds, we have:

α=2−5

2=−3

2=−1.5 rad/s2

Therefore, the initial angular acceleration of the disc is -1.5 rad/s2.

Step 2: The given information suggests that the angular acceleration is

constant over this time period, so we can use the average angular acceleration

formula α=∆ω

∆twhere: ∆ω= change in angular velocity, and ∆t= time

interval.

Given ∆ω=ωf−ωi= 2 −5 = −3 rad/s and ∆t= 2 seconds, we have:

α=−3

2=−1.5 rad/s2

Therefore, the angular acceleration of the disc during this time period is -1.5

rad/s2.

Question 7

Question

A solid sphere of mass Mand radius Ris released from rest at the top of a

ramp that is inclined at an angle θ. The sphere rolls down the ramp without

slipping. What is the speed of the sphere when it reaches the bottom of the

ramp?

Solution

Step 1: Determine the height the sphere descends. The height the sphere

descends is given by the vertical component of the initial position, which is

h=Rsin(θ).

Step 2: Apply conservation of energy. The initial gravitational potential

energy is converted into both kinetic energy of translation and rotational kinetic

energy at the bottom of the ramp.

mgR sin(θ) = 1

2mv2+1

2Iω2

where vis the speed of the sphere and ωis the angular velocity. The moment

of inertia for a solid sphere rotating about its diameter is I=2

5MR2.

Step 3: Relate the linear and angular velocities for rolling motion. For a

solid sphere rolling without slipping, the angular velocity ωis related to the

linear velocity vby ω=v

Step 4: Substitute equations and solve for v. Substitute ω=v

Rand I=

5MR2into the conservation of energy equation.

mgR sin(θ) = 1

2mv2+1

22

5MR2v

R2

Solving for vgives:

v=r5

7gR sin(θ)

Therefore, the speed of the sphere when it reaches the bottom of the ramp

is v=q5

7gR sin(θ).

Question 8

Question

A disc of radius Rand mass Mis rotating about a fixed axis perpendicular to

the disc through its center. Initially, the disc is not rotating, but a torque τis

applied to it for a time t. After some time, the disc reaches an angular velocity

ω. Determine the torque τrequired to make the disc reach this angular velocity.

Solution

Step 1: Start by relating the torque to the change in angular momentum. The

torque τapplied to the disc causes a change in angular momentum ∆L. We

know that τ=∆L

∆t.

Step 2: The initial angular momentum of the disc is zero since it is not

initially rotating. The final angular momentum of the disc is given by L=Iω,

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

Step 3: The moment of inertia of a disc rotating about an axis perpendicular

to its surface through its center is given by I=1

2MR2.

Step 4: The change in angular momentum is ∆L=L−0 = Iω.

Step 5: Putting it all together, we have τ=Iω

2MR2ω

t=1

2MR2ω

Therefore, the torque τrequired to make the disc reach the angular velocity

ωis 1

2MR2ω

Question 9

Question

A disc of radius 0.5 m is rotating about its center with an angular speed of 4

rad/s. A small bug sitting on the edge of the disc (at a distance of 0.5 m from

the center) decides to walk towards the center of the disc at a speed of 0.1 m/s

relative to the edge. What is the bug’s angular velocity when it reaches the

center of the disc?