PHYS 305 - INTRODUCTION TO MODERN PHYSICS Rotational kinematics Question Bank Set 8 | PHYS 305

PHYS 305 - INTRODUCTION TO

MODERN PHYSICS - Rotational

kinematics

Question Bank - Set 8

Liberty University

Question 1

Question

A solid cylinder with radius Rand mass Mis initially at rest. A force Fis

applied tangentially to the edge of the cylinder causing it to rotate. If the force

is applied for a time t, what is the final angular velocity of the cylinder?

Solution

Step 1: Find the torque applied to the cylinder. The torque (τ) applied to the

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

τ=F r

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

inertia of a solid cylinder rotating about its axis is I=1

2MR2.

Step 3: Apply Newton’s second law for rotation. By Newton’s second law

for rotation, the net torque applied to the cylinder is equal to the change in

angular momentum.

τ=Iα

where αis the angular acceleration.

Step 4: Find the angular acceleration of the cylinder. Substitute the expres-

sions for torque and moment of inertia into the equation above.

F r =1

2MR2α

α=2F r

MR2

Step 5: Find the final angular velocity. The final angular velocity (ωf) can

be found using the equation for angular velocity in terms of angular acceleration

and time.

ωf=αt

Substitute the expression for angular acceleration into the equation above, we

get

ωf=2F r

MR2t

Question 2

Question

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

down an incline plane of height h. What is the final velocity of the sphere at

the bottom of the incline?

Solution

Step 1: Find the moment of inertia of the sphere rolling down the incline. The

moment of inertia of a solid sphere about its center is 2

5MR2.

Step 2: Calculate the potential energy at the top of the incline and the

kinetic energy at the bottom. The potential energy at the top of the incline is

given by P E =Mgh, where his the height of the incline. The kinetic energy

at the bottom of the incline is given by KE =1

2Mv2+1

2Iω2, where vis the

linear velocity at the bottom of the incline and ωis the angular velocity.

Step 3: Apply conservation of energy to find the final velocity. Since there

is no external torque acting on the sphere, the total mechanical energy remains

constant. Therefore, we have:

P E =KE

Mgh =1

2Mv2+1

2Iv

R2

Mgh =1

2Mv2+1

22

5MR2v

R2

Step 4: Solve for the final velocity. Solving the equation from Step 3 for v,

we get:

v=r10

7gh

Question 3

Question

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

acceleration α. At the instant when the rod makes an angle θwith the vertical,

find the tangential acceleration of a point located a distance xfrom the pivot

point.

Solution

Step 1: Find the angular velocity ωof the rod at the instant it makes an angle θ

with the vertical. The angular velocity ωcan be calculated using the equation:

ω2= 2αθ

where θis in radians.

Step 2: Find the velocity vof a point on the rod a distance xfrom the pivot.

The velocity vof a point on the rod can be determined using the equation:

v=ωx

Step 3: Find the tangential acceleration atof the point. The tangential

acceleration atcan be calculated using the equation:

at=αx

Question 4

Question

A solid sphere of radius Rrolls without slipping down an inclined plane of

angle θ. The sphere starts from rest at the top of the plane. What is the linear

acceleration of the center of the sphere when it reaches the bottom of the plane?

Solution

Let’s assume that the moment of inertia of the sphere about its center of mass

is I=2

5mR2, where mis the mass of the sphere.

Step 1: Find the acceleration of the sphere down the incline.

The net torque acting on the sphere is due to the component of the weight

parallel to the incline. Therefore:

τ=Iα =2

5mR2(α)

The torque is also equal to the force causing the angular acceleration multiplied

by the radius R:τ=mgR sin(θ). This gives us:

mgR sin(θ) = 2

5mR2(α)

Solving for α:

α=5gsin(θ)

Step 2: Find the linear acceleration aof the center of the sphere.

The linear acceleration of the center of mass is given by:

a=Rα

Substitute the value of αwe found in Step 1:

a=R5gsin(θ)

2R

a=5

2gsin(θ)

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

the bottom of the plane is 5

2gsin(θ).

Question 5

Question

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

for 10 seconds. If the angular acceleration of the disk is 0.5 rad/s2, determine

the angular velocity of the disk after 10 seconds.

Solution

Step 1: Since the disk starts from rest and accelerates uniformly, we can use the

following kinematic equation for rotational motion:

ω=ω0+αt

where:

ω= final angular velocity

ω0= initial angular velocity (which is 0 rad/s since the disk starts from rest)

α= angular acceleration

t= time

Step 2: Substituting the given values into the equation:

ω= 0 + 0.5×10

ω= 5 rad/s

Step 3: Therefore, the angular velocity of the disk after 10 seconds is 5 rad/s.

Question 6

Question

A disk of radius 0.5 m and mass 2 kg is rotating about a fixed axis passing

through its center with an angular velocity of 5 rad/s. A small object with

mass 0.1 kg is placed on the rim of the disk. What is the angular velocity of

the disk after the object falls off? Assume the object falls off without imparting

any torque to the disk.

Solution

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

angular momentum of the system is given by the sum of the angular momentum

of the disk and the angular momentum of the object:

Linitial =Idiskωinitial +mobjectrωinitial

For a disk rotating about its center, the moment of inertia is Idisk =1

2mr2.

Substitute the given values into the equation:

Linitial =1

2×2×0.52×5+0.1×0.5×5

Step 2: Calculate the final angular momentum of the system after the object

falls off. When the object falls off, the angular momentum of the system remains

conserved. Thus, the final angular momentum is given by:

Lfinal =Idiskωfinal

Step 3: Set the initial and final angular momenta equal to each other to find

the final angular velocity. Since angular momentum is conserved, we have:

Linitial =Lfinal

2×2×0.52×5+0.1×0.5×5 = 1

2×2×0.52×ωfinal

Solve this equation for ωfinal to find the angular velocity of the disk after the

object falls off.

Question 7

Question

A thin uniform rod of length Land mass Mis initially at rest on a frictionless

horizontal surface. One end of the rod is then struck horizontally with an

impulse J. Determine the angular velocity of the rod just after the impulse.

Solution

1. After the impulse is applied, the rod will rotate around its pivot point. Let

the angular velocity of the rod just after the impulse be ω. The linear velocity

of the end of the rod being struck is v=ω·L

2. By applying the principle of conservation of angular momentum, we

have: initial angular momentum = final angular momentum. The initial angular

momentum is zero since the rod is initially at rest. The final angular momentum

is Lfinal =I·ω, where Iis the moment of inertia of the rod.

3. The moment of inertia of a rod rotating about one end is I=1

3Ml2.

Substituting this into the equation, we get: 0 = 1

3Ml2·ω.

4. Solving for ω, we find: ω= 0.

Therefore, the angular velocity of the rod just after the impulse is ω= 0.

Question 8

Question

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

is placed on the rim of the disc and released from rest. Calculate the angular

velocity of the particle when it has fallen through a vertical distance of 1 m.

Assume that the disc is frictionless.

Solution

Step 1: First, we need to find the initial angular velocity of the particle at

rest on the rim of the rotating disc. We can use the conservation of angular

momentum.

Step 2: The initial angular momentum of the particle on the rotating disc is

equal to the final angular momentum of the particle at the bottom. Therefore,

we have:

I1ω1=I2ω2

where I1is the moment of inertia of the disc, I2is the moment of inertia of the

particle, and ω1and ω2are the initial and final angular velocities, respectively.

Step 3: The moment of inertia of the disc is given by Idisc =1

2mr2, where m

is the mass of the disc and ris the radius. The moment of inertia of the particle

about the axis passing through its center is Iparticle =mr2. Therefore, we have:

2mr2·4 = mr2·ω2

Step 4: Solving for ω2, we find:

ω2= 2 rad/s

Step 5: Next, we need to find the angular velocity at the bottom of the disc

after the particle has fallen a distance of 1 m. We can use the conservation of

energy.

Step 6: The change in gravitational potential energy of the particle at the

top of the disc to the bottom is equal to the change in rotational kinetic energy.

That is:

mgh =1

2Iω2

where mis the mass of the particle, gis the acceleration due to gravity, h

is the height fallen, Iis the moment of inertia of the system (particle + disc),

and ωis the final angular velocity.

Step 7: Substituting the values h= 1 m and I=3

2mr2into the above

equation, we find:

mg =3

4mω2

Step 8: Solving for ω, we get:

ω=r4g

3≈2.05 rad/s

Therefore, the angular velocity of the particle when it has fallen through a

vertical distance of 1 m is approximately 2.05 rad/s.

Question 9

Question

A solid cylindrical disk of radius Rand mass Mis freely pivoted about its center

and then set spinning with an initial angular velocity ω0. A small block of mass

mis placed on the edge of the disk and starts sliding towards the center. At

what distance rfrom the center of the disk does the block lose contact with the

disk?

Solution

Step 1: Utilize the conservation of angular momentum to determine the angular

velocity of the block when it loses contact with the disk. Since the system is

isolated (no external torques acting), the total angular momentum is conserved.

Initially, the angular momentum of the disk is Linitial =Idiskω0, and when the

block loses contact, the angular momentum is Lfinal = (Idisk +mr2)ωblock.

⇒Idiskω0= (Idisk +mr2)ωblock

Step 2: Express the moments of inertia in terms of the given quantities. The

moment of inertia of a disk about its center is Idisk =1

2MR2, and for the block

about the edge, it is mr2.

Step 3: Solve for the angular velocity of the block when it loses contact.

Substitute the expressions for Idisk and Iblock into the conservation equation to

get:

2MR2ω0=1

2MR2+mr2ωblock

Now, solve for ωblock:

ωblock =MR2ω0

MR2+ 2mr2

Step 4: Use the relationship between linear and angular velocity to find the

velocity of the block when it loses contact. The linear velocity of a point on the

rotating disk is given by v=ωr. At the point where the block loses contact,

the linear velocity of the edge of the disk is equal to the velocity of the block:

ωblockr=ω0R

Substitute the expression for ωblock obtained earlier:

MR2ω0

MR2+ 2mr2r=ω0R

⇒r=2mR3

MR2+ 2mr2

Therefore, the block loses contact with the disk at a distance rfrom the

center given by r=2mR3

MR2+2mr2.

Question 10

Question

A thin 2.0 m long rod is pivoted at one end and swings in a vertical plane. The

rod is released from rest when it makes an angle of 60◦with the vertical. What

is the angular velocity of the rod when it is vertical?

Solution

Step 1: Let’s first find the potential energy of the rod when it is at the initial

position. The potential energy Uiat an angle θiwith the vertical is given by:

Ui=mghi

where: m= mass of the rod g= acceleration due to gravity hi= height at

angle θi

Step 2: At the initial position, when the rod makes an angle of 60◦with the

vertical, the height can be calculated as:

hi= 2.0×sin(60◦)

Step 3: The potential energy at the initial position is:

Ui=mghi

Step 4: Now, let’s find the potential energy of the rod when it is vertical.

The potential energy Ufat the vertical position is:

Uf=mghf

where: hf= 0 (at the vertical position)

Step 5: The kinetic energy Kfof the rod at the vertical position is equal to

the total initial potential energy Uisince the rod began from rest.

Kf=Ui

Step 6: The kinetic energy Kfat the vertical position is given by:

Kf=1

2Iω2

where: I= moment of inertia of the rod about the pivot ω= angular velocity

of the rod at the vertical position

Step 7: The moment of inertia Iof the rod about the pivot is:

I=1

3mL2

where: L= length of the rod

Step 8: Setting the two expressions for kinetic energy equal, we have:

2Iω2=mghi

Step 9: Solving for ωgives:

ω=r2ghi

Step 10: Substitute the values of m,g,hi, and Linto the equation and

calculate the angular velocity.

Question 11

Question

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

small particle is then placed on the disk at a distance of 0.3 m from the center.

What is the angular velocity of the particle with respect to the disk just after

it is placed?

Solution

Let’s denote the angular velocity of the particle with respect to the disk as ωp.

We know that the angular velocity of the disk is 4 rad/s. The angular velocity

of the particle with respect to the disk can be found using the formula:

ωp=ωdisk +ωparticle

where ωparticle is the angular velocity of the particle in an inertial reference

frame.

Step 1: Calculate the angular velocity of the particle in the inertial reference

frame.

The linear velocity of the particle can be calculated as v=rω, where r= 0.3

m and ω= 4 rad/s.

So, v= 0.3×4=1.2 m/s.

Now, the angular velocity of the particle in the inertial reference frame is

ωparticle =v

r=1.2

0.3= 4 rad/s.

Step 2: Find the angular velocity of the particle with respect to the disk.

Using the formula ωp=ωdisk +ωparticle, we have

ωp= 4 + 4 = 8 rad/s

Therefore, the angular velocity of the particle with respect to the disk just

after it is placed is 8 rad/s.

Question 12

Question

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

acceleration. After 3 seconds, the wheel has made 15 complete revolutions.

What is the angular acceleration of the wheel?

Solution

Step 1: Find the angular velocity of the wheel after 3 seconds. Given that

the wheel starts at rest, the initial angular velocity ω0= 0. The final angular

velocity ωcan be calculated using the equation:

ω=ω0+αt

where αis the angular acceleration and tis the time taken.

ω= 0 + α×3

ω= 3α

Step 2: Convert the number of revolutions to radians. The angle covered by

15 complete revolutions can be calculated as:

θ= 2π×15 = 30πradians

Step 3: Find the final angular velocity using angle covered. The final angular

velocity ωcan also be calculated using the formula:

ω2=ω2

0+ 2αθ

Plugging in the known values:

(3α)2= 0 + 2α×30π

9α2= 60πα

α=60π

9=20π

3≈20.94 rad/s2

Therefore, the angular acceleration of the wheel is 20π

3rad/s2.

Question 13

Question

A disk is rotating counterclockwise at a constant angular acceleration of 1.5

rad/s2. At time t= 0, the angular velocity of the disk is 3 rad/s and the radius

of the disk is 0.5 m. What is the angular velocity of a point on the disk that is

0.25 m from the center of rotation after 2 seconds?

Solution

Step 1: Find the angular velocity of the disk after 2 seconds using the equation

ωf=ωi+αt. Given: ωi= 3 rad/s, α= 1.5 rad/s2,t= 2 s.

Plugging the values into the equation:

ωf= 3 + 1.5×2 = 3 + 3 = 6 rad/s.

Step 2: Find the linear velocity of the point on the disk using v=rω. Given:

r= 0.25 m, ω= 6 rad/s.

Plugging the values into the equation:

v= 0.25 ×6 = 1.5 m/s.

Step 3: Find the angular velocity of the point on the disk using the equation

ω=v/r. Given: v= 1.5 m/s, r= 0.25 m.

Plugging the values into the equation:

ω=1.5

0.25 = 6 rad/s.

Therefore, the angular velocity of a point on the disk that is 0.25 m from

the center of rotation after 2 seconds is 6 rad/s.

Question 14

Question

A thin rod of length Lrotates about one end, with an angular speed of ω. At

a certain instant, a small object of mass mis attached to the free end of the

rod. Determine the angular speed of the rod-object system after the object is

attached.

Solution

1. The initial angular momentum of the rod alone is given by:

Linitial =Irodω

where Irod is the moment of inertia of the rod about its end. We can approximate

the rod as a uniform rod rotating about its end, so Irod =1

3mL2.

2. The final angular momentum of the system (rod-object) is the sum of the

angular momentum of the rod and the angular momentum of the object:

Lfinal =Irodωfinal +mL2ωfinal

where ωfinal is the angular speed of the system after the object is attached.

3. Since angular momentum is conserved, Linitial =Lfinal. Substituting the

expressions for Linitial and Lfinal:

Irodω=Irodωfinal +mL2ωfinal

4. Solving for ωfinal:

3mL2ω=1

3mL2+mL2ωfinal

5. Simplifying the equation gives:

3ω=4

3ωfinal

6. Therefore, the angular speed of the rod-object system after the object is

attached is:

ωfinal =1

4ω

Question 15

Question

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

speed ω. Find the magnitude of the total angular momentum of the rod about

the opposite end.

Solution

Step 1: The angular momentum of a point mass mmoving at a distance rfrom

a fixed axis with angular speed ωis given by L=m·r2·ω. For the entire rod,

we need to integrate this expression over the length of the rod.

Step 2: Consider a small element of length dx at a distance xfrom the axis

of rotation. The mass of this element is dm =M

L·dx.

Step 3: The angular momentum dL of this element is dL =M

L·x2·ω·dx.

Step 4: The total angular momentum Ltotal of the rod is obtained by inte-

grating dL from 0 to L:

Ltotal =ZL

L·x2·ω·dx

Step 5: After integrating, we get:

Ltotal =Mω

LZL

x2·dx

Step 6: Solving the integral, we find:

Ltotal =Mω

L·1

3x3L

Step 7: Substituting the limits of integration and simplifying, we get:

Ltotal =Mω

L·1

3L3−03

Step 8: Finally, we have:

Ltotal =MωL2

Question 16

Question

A disc of radius Ris initially at rest. It starts rotating about a fixed axis passing

through its center with a constant angular acceleration α. At what time will

the tangential speed of a point on the rim reach half of the maximal value?

Solution

Let ωbe the angular velocity of the disc and vbe the tangential speed of a

point on the rim.

Step 1: The relationship between the angular velocity ω, tangential speed

v, and radius Ris given by v=Rω.

Step 2: The angular acceleration αis related to the angular velocity ωas

α=dω

dt .

Step 3: The time derivative of vwith respect to time tis a=Rα.

Step 4: From the given information, when v=1

2Rωmax, we can write

2Rαt =1

2Rωmax.

Step 5: Solving for t, we find t=ωmax

α.

Therefore, the tangential speed of a point on the rim will reach half of the

maximal value at a time t=ωmax

α.

Question 17

Question

A disk starts from rest and accelerates with a constant angular acceleration α.

What is the time it takes to reach an angular velocity ω?

Solution

Step 1: The final angular velocity ωcan be related to the initial angular velocity

ω0, angular acceleration α, and time tusing the equation:

ω=ω0+αt

Step 2: Since the disk starts from rest, the initial angular velocity ω0= 0.

Therefore, the equation simplifies to:

ω=αt

Step 3: Rearranging the equation gives:

t=ω

Step 4: This shows that the time it takes for the disk to reach an angular

velocity ωis ω

α.

Question 18

Question

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

acceleration of 2 rad/s2for 4 s. Determine the angular velocity of the wheel at

t= 4 s.

Solution

Step 1: We can use the equation for rotational kinematics:

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.

Step 2: Substituting the given values, we get:

ω= 0 + (2 rad/s2)(4 s)

Step 3: Calculating the final angular velocity:

ω= 8 rad/s

Step 4: Therefore, the angular velocity of the wheel at t= 4 s is 8 rad/s.

Question 19

Question

A thin rod of length Land mass Mis pivoted at one end and is initially at rest.

A small object of mass mcollides with the free end of the rod and sticks to it.

The collision is perfectly inelastic. Find the angular velocity of the system just

after the collision.

Solution

Let’s denote the initial angular velocity of the system as ωiand the final angular

velocity after the collision as ωf.

Step 1: First, we need to calculate the moment of inertia of the rod-object

system. The moment of inertia of the thin rod about its pivot point is Irod =

3ML2. The moment of inertia of the point mass about the pivot point is

Iobject =mL2. Therefore, the total moment of inertia of the system is Itotal =

Irod +Iobject.

Step 2: Next, we can apply the principle of conservation of angular momen-

tum. Since no external torque acts on the system, the initial angular momen-

tum is equal to the final angular momentum. The initial angular momentum is

Linitial =Itotal ·ωi. After the collision, the system becomes one rigid body ro-

tating about the pivot with angular velocity ωf, so the final angular momentum

is Lfinal =Itotal ·ωf.

Step 3: Setting the initial angular momentum equal to the final angular

momentum and solving for ωf, we get:

Itotal ·ωi=Itotal ·ωf

3ML2·ωi= (1

3ML2+mL2)·ωf

3Mωi=1

3Mωf+mωf

3Mωi=1

3M+mωf

ωf=Mωi

M+ 3m

Therefore, the angular velocity of the system just after the collision is ωf=

Mωi

M+3m.

Question 20

Question

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

bug lands on the edge of the disk and starts walking towards the center at a

constant speed of 0.1 m/s relative to the disk. What is the bug’s acceleration

(magnitude and direction) relative to the disk when the bug is 0.1 m from the

center of the disk?

Solution

Step 1: First, we need to find the bug’s velocity (magnitude and direction)

relative to the disk. Since the bug is walking towards the center of the disk, we

can split its velocity into two components: tangential and radial.

The tangential component of the bug’s velocity vtis equal to the disk’s

angular velocity ωtimes the distance from the bug to the center of the disk r.

Therefore,

vt=ω×r= 5 rad/s ×0.1 m = 0.5 m/s

Step 2: To find the bug’s radial velocity vr(magnitude and direction), we

need to consider the bug’s motion towards the center of the disk. Since the

bug’s velocity relative to the disk is 0.1 m/s, the magnitude of the bug’s radial

velocity is 0.1 m/s.

The bug’s radial velocity is directed towards the center of the disk, opposite

to its motion.

Step 3: Now, we can determine the bug’s total velocity relative to the disk

using the components found in Step 1 and Step 2.

The bug’s total velocity vrelative to the disk is given by the vector sum of

the tangential and radial velocities:

v=qv2

t+v2

r=p0.52+ 0.12=√0.26 ≈0.51 m/s

Step 4: Finally, we can find the bug’s acceleration (magnitude and direction)

relative to the disk by considering the change in the bug’s velocity as it moves

towards the center.

The bug’s acceleration arelative to the disk is directed radially towards the

center of the disk. The magnitude of the bug’s acceleration can be calculated

using the formula for centripetal acceleration:

a=v2

r=0.512

0.1=0.2601

0.1= 2.601 m/s2

Therefore, the bug’s acceleration relative to the disk when the bug is 0.1

m from the center of the disk is approximately 2.601 m/s2directed radially

towards the center of the disk.

Question 21

Question

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

angular acceleration of 2 rad/s2. What is the magnitude of the total distance

covered by a point on the rim of the disk after 5 seconds?

Solution

Step 1: Find the angular velocity of the disk after 5 seconds using the equation

ω=ω0+αt, where ω0is the initial angular velocity (0 in this case), αis the

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

ω= 0 + 2 ×5 = 10 rad/s

Step 2: Find the total angle rotated by the disk after 5 seconds using the

equation θ=1

2αt2, where αis the angular acceleration (2 rad/s2) and tis the

time (5 s).

θ=1

2×2×(5)2= 25 rad

Step 3: Find the total distance covered by a point on the rim of the disk.

Since the point travels along the circumference of the disk, the distance is given

by d=rθ, where ris the radius of the disk (0.3 m) and θis the total angle

covered (25 rad).

d= 0.3×25 = 7.5 m

Therefore, the magnitude of the total distance covered by a point on the rim

of the disk after 5 seconds is 7.5 meters.

Question 22

Question

A solid cylinder with mass Mand radius Ris rolling without slipping down an

incline of angle θ. The moment of inertia of the cylinder about its center of

mass is I=1

2MR2. If the cylinder starts from rest at the top of the incline,

determine the speed of the center of mass of the cylinder when it reaches the

bottom of the incline. Ignore friction.

Solution

Step 1: Calculate the gravitational potential energy at the top and its kinetic

energy at the bottom.

The change in gravitational potential energy is converted into the kinetic energy

at the bottom of the incline.

∆KE = ∆P E

Step 2: Calculate the change in height.

The change in height is equal to the vertical distance the center of mass travels

from the top to the bottom of the incline.

h=R(1 −cos(θ))

Step 3: Calculate the speed of the center of mass at the bottom.

Using the principle of conservation of energy, we can equate the change in po-

tential energy to the change in kinetic energy.

∆KE = ∆P E

KEfinal −KEinitial =P Efinal −P Einitial

2Mv2

final −0=0−M gR(1 −cos(θ))

2Mv2

final =MgR(1 −cos(θ))

vfinal =p2gR(1 −cos(θ))

Therefore, the speed of the center of mass of the cylinder when it reaches

the bottom of the incline is vfinal =p2gR(1 −cos(θ)).

Question 23

Question

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

angular acceleration of 2 rad/s2.

(a) What is the angular velocity of the disk after 3 seconds?

(b) How many revolutions has the disk gone through after 3 seconds?

Solution

(a) Step 1: Use the formula for angular velocity under constant acceleration:

ωf=ωi+αt

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

is 0 for this case since the disk starts from rest), - αis the angular acceleration,

and - tis the time. Step 2: Substitute the given values into the formula:

ωf= 0 + 2 ×3 = 6 rad/s

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

(b) Step 1: Use the formula for angular displacement under constant angular

acceleration:

θ=ωit+1

2αt2

where - θis the angular displacement, - ωiis the initial angular velocity (0), - α

is the angular acceleration, - tis the time. Step 2: Substitute the given values

into the formula:

θ= 0 + 1

2×2×32= 9 rad

Step 3: Calculate the number of revolutions:

Number of revolutions = θ

2π=9

2π≈1.43 revolutions

Therefore, the disk has gone through approximately 1.43 revolutions after 3

seconds.

Question 24

Question

A uniform disc of radius Ris initially at rest. A constant torque τis applied to

the disc, causing it to rotate with an angular acceleration α. Find the angular

velocity of the disc at the time the angular displacement θis equal to π

Solution

Step 1: Identify the relevant equations. The kinematic equation relating angular

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

can be used: ω2

f=ω2

i+ 2αθ.

Step 2: Find the initial angular velocity. Since the disc is initially at rest,

the initial angular velocity ωi= 0.

Step 3: Determine the final angular velocity. Using the kinematic equation,

we have ω2

f= 0+2απ

2=απ. Therefore, the angular velocity of the disc when

θ=π

2is ω=√απ.

Question 25

Question

A wheel initially at rest undergoes an angular acceleration given by α= 4t−2t2,

where αis in radians per second squared and tis in seconds. If the wheel starts

from rest, determine the angular velocity of the wheel after 3 seconds.

Solution

Step 1: To find the angular velocity of the wheel after 3 seconds, we need to

integrate the given angular acceleration function with respect to time to get the

angular velocity function.

Step 2: Integrate α= 4t−2t2with respect to time to find the angular

velocity function ω(t):

Zα dt =Z(4t−2t2)dt

Step 3: Integrating term by term:

Zα dt =Z4t dt −Z2t2dt

Step 4: Integrating each term:

ω(t)=2t2−2

3t3+C

Step 5: Since the wheel starts from rest, the constant of integration Cis

equal to 0.

Step 6: Substitute t= 3 into the angular velocity function to find the angular

velocity after 3 seconds:

ω(3) = 2(3)2−2

3(3)3= 18 rad/s

Step 7: Therefore, the angular velocity of the wheel after 3 seconds is ω=

18 rad/s.

Question 26

Question

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

surface from rest. It then ascends a 30◦incline. Determine the linear speed of

the sphere when it reaches a height hup the incline.

Solution

Let’s denote the linear speed of the sphere when it reaches a height hup the

incline as v.

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

When the sphere starts rolling without slipping, the initial condition relates

the linear speed (v), angular speed (ω), and radius (R) of the sphere. Since the

sphere rolls without slipping, we have v=Rω. Therefore, the initial angular

speed of the sphere is given by:

ω=v

Step 2: Determine the final angular speed of the sphere.

To determine the final angular speed of the sphere as it reaches height hon

the incline, we will use the conservation of energy. The initial kinetic energy of

the sphere is due to its rotational motion and translation motion:

KEinitial =1

2Iω2+1

2Mv2

The final kinetic energy is due to its increased potential energy at height hup

the incline:

KEfinal =1

2Iω2

f+1

2Mv2

f+Mgh

Since the sphere is rolling without slipping, the moment of inertia of the sphere

about its center of mass is I=2

5MR2. Setting the initial kinetic energy equal

to the final kinetic energy, we have:

22

5MR2v

R2+1

2Mv2=1

22

5MR2ω2

f+1

2Mv2

f+Mgh

Step 3: Determine the linear speed of the sphere at height h.

Now, we can solve for vf(the linear speed of the sphere at height h) in terms

of R,v,h, and other constants. Simplify the expression to find vf:

vf=5

7v2+ 5gh1/2

Question 27

Question

A thin uniform rod of length Land mass Mis pivoted about one end. Initially,

the rod is at rest. A bullet of mass mmoving horizontally with speed vstrikes

the rod at a distance dfrom the pivot point and gets embedded in it. Find the

angular speed of the rod just after the collision.

Solution

1. Let Ibe the moment of inertia of the rod about the pivot point. 2. The initial

angular momentum of the system (rod + bullet) about the pivot point is zero

since the rod is at rest. 3. Just after the collision, the final angular momentum

of the system is equal to the angular momentum of the bullet after it gets

embedded in the rod. 4. The angular momentum of the bullet just after the

collision is mvd. 5. The final angular momentum of the system of the rod and

bullet just after the collision is Iω, where ωis the angular speed of the rod just

after the collision. 6. Therefore, we have mvd =Iω. 7. The moment of inertia

of the rod about the pivot point is 1

3ML2, and by the parallel axis theorem, the

moment of inertia of the rod with the bullet about the pivot point is I+Md2.

8. Substituting I=1

3ML2into the equation gives mvd =1

3ML2+Md2ω.

9. Solving for ωgives ω=mvd

3ML2+Md2).

Question 28

Question

A wheel initially at rest undergoes constant angular acceleration for 10 seconds,

after which it reaches an angular velocity of 120 rad/s. If the wheel’s angular

acceleration is constant during this time, what is the angular acceleration of the

wheel?

Solution

Step 1: We know that the final angular velocity (ωf) is 120 rad/s, the initial

angular velocity (ωi) is 0 rad/s, and the time taken (t) is 10 seconds. We are

asked to find the angular acceleration (α).

Step 2: The angular acceleration can be calculated using the equation:

ωf=ωi+αt

Step 3: Substituting the given values into the equation:

120 = 0 + α×10

Step 4: Solving the equation for α:

α=120

10 = 12 rad/s2

Step 5: Therefore, the angular acceleration of the wheel is 12 rad/s2.

Question 29

Question

A wheel of radius 0.5 m is rotating at an angular speed of 10 rad/s. Initially, a

point on the rim of the wheel is at the horizontal position. Find the horizontal

position of the point after the wheel has made 5 complete revolutions.

Solution

Step 1: Find the angular displacement made by the wheel in 5 complete revo-

lutions.

Given that the angular speed of the wheel is 10 rad/s, we can find the angular

displacement using the formula:

θ=ωt

where θis the angular displacement, ωis the angular speed, and tis the time.

Since the wheel makes 5 complete revolutions, the time taken is:

t=5 revolutions

10 rev/s = 0.5 s

Therefore, the angular displacement is:

θ= 10 rad/s ×0.5 s = 5 rad

Step 2: Find the horizontal position of the point after 5 complete revolutions.

The horizontal position of the point on the rim of the wheel is given by:

x=rcos(θ)

where ris the radius of the wheel and θis the angular displacement. Substitute

r= 0.5 m and θ= 5 rad into the formula:

x= 0.5 m ×cos(5) ≈0.5 m ×cos(3.13) ≈0.5 m ×(−0.999) ≈ −0.4995 m

Therefore, the horizontal position of the point after 5 complete revolutions is

approximately −0.4995 m.

Question 30

Question

A disc of radius 0.1 m is rotating about its central axis with an angular velocity

of 5 rad/s. At a certain moment, its angular velocity decreases uniformly to

2 rad/s in 4 seconds. Find the magnitude of the angular acceleration and the

angle through which the disc rotates during those 4 seconds.

Solution

Step 1: Calculate the angular acceleration using the formula α=∆ω

∆t.

Given: ωi= 5 rad/s, ωf= 2 rad/s,∆t= 4 s

α=2−5

4=−3

4=−0.75 rad/s2

Step 2: Calculate the angle through which the disc rotates using the formula

θ=ωit+1

2αt2.

θ= 5 ×4 + 1

2×(−0.75) ×42

θ= 20 −6 = 14 rad

Therefore, the magnitude of the angular acceleration is 0.75 rad/s2and the

disc rotates through an angle of 14 radians in those 4 seconds.

Question 31

Question

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

2.0 rad/s2. After 3.0 s, what is the angular velocity of the wheel?

Solution

Step 1: Identify the knowns and the unknowns.

The knowns are: - Initial angular velocity, ω0= 0 (since it starts from rest). -

Angular acceleration, α= 2.0 rad/s2. - Time, t= 3.0 s.

The unknown is the final angular velocity, ωafter 3.0 s.

Step 2: Use the equation relating angular velocity, angular acceleration, and

time.

The equation relating angular velocity, angular acceleration, and time is:

ω=ω0+αt

Substitute the known values into the equation:

ω= 0 + (2.0 rad/s2)(3.0 s)

Step 3: Calculate the final angular velocity.

ω= 0 + 6.0 rad/s = 6.0 rad/s

Therefore, after 3.0 s, the angular velocity of the wheel is 6.0 rad/s.

Question 32

Question

A solid sphere of radius Rand mass Mis initially at rest. A constant hori-

zontal force 

Fis applied to the sphere at a distance habove the center of the

sphere. Calculate the angular acceleration of the sphere immediately after the

application of the force.

Solution

Step 1: The torque due to the force 

Fis given by:

τ=r ×

Since the force and displacement are in the same direction, r×

F=rF sin θ=rF

where θ= 90◦. The lever arm r=R+h, so:

τ= (R+h)F

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

5MR2.

Applying Newton’s second law for rotation, we have:

τ=Iα

Substitute the expressions for τ,Iinto the equation above:

(R+h)F=2

5MR2α

Step 3: Solve for the angular acceleration α:

α=(R+h)F

5MR2

α=5(R+h)F

2MR2

Therefore, the angular acceleration of the sphere immediately after the ap-

plication of the force is α=5(R+h)F

2MR2.

Question 33

Question

A thin rod of length Lrotates about one end with an angular velocity ω0. A

small object of mass mslides without friction along the rod. At what angular

velocity will the object be farthest from the rotating end of the rod?

Solution

Step 1: Let’s denote the distance of the object from the rotating end of the rod

as x. From the perpendicular distance from the axis of rotation to the object

can be expressed as L−x.

Step 2: The angular velocity of the object is given by

ω=v

where vis the linear velocity of the object.

Step 3: The linear velocity of the object can be expressed in terms of the

angular velocity of the rod. Taking the derivative of the distance xwith respect

to time t:

v=dx

dt =ω0Lcos(ω0t),

where ω0is the initial angular velocity of the rod.

Step 4: Substituting the expression for vinto the formula for ωgives

ω=ω0Lcos(ω0t)

Step 5: At the point where the object is farthest from the rotating end, the

velocity is perpendicular to the rod. Therefore, cos(ω0t) = 0, which implies that

ω=ω0L

xmax .

Step 6: In order for the object to be located farthest from the rotating end,

we need to maximize x. This happens when the object’s velocity is zero, i.e.,

when cos(ω0t) = 0. Therefore, xmax =L.

Step 7: Finally, substituting xmax =Linto the expression for ωgives ω=

ω0L

L=ω0. Thus, the object will be farthest from the rotating end of the rod

when its angular velocity is ω0.

Question 34

Question

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

acceleration of 0.4 rad/s2. At what time will the tangential velocity of a point

on the rim of the disk be 5 m/s?

Solution

Step 1: Identify the known values and the unknown. Let’s denote the final

tangential velocity as vf, the initial tangential velocity as viwhich is 0 m/s

since the disk starts from rest, the angular acceleration as α, and the time as t.

We are given r(radius) = 0.5 m, α= 0.4 rad/s2, and vf= 5 m/s. We need to

find t.

Step 2: Find the angular velocity at time tusing the equation ωf=ωi+αt.

The initial angular velocity ωiis 0 rad/s, so ωf=αt.

Step 3: Find the tangential velocity at time tusing the equation v=rω.

Substitute ω=αt into the equation above to get v=rαt.

Step 4: Set up and solve the equation v=rαt for t. Given that v= 5 m/s,

r= 0.5 m, and α= 0.4 rad/s2, we have:

5=0.5×0.4×t

Solve for tto find the time at which the tangential velocity is 5 m/s.

Question 35

Question

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

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

3 seconds?

Solution

Step 1: The angular velocity of the disk can be calculated using the formula:

ωf=ω0+αt

where: - ωfis the final angular velocity, - ω0is the initial angular velocity

(which is 0 since the disk starts from rest), and - αis the angular acceleration.

Substitute the given values into the formula:

ωf= 0 + 2 ×3 = 6 rad/s

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

Question 6

Question

A disk of radius 0.5 m and mass 2 kg is rotating about a fixed axis passing

through its center with an angular velocity of 5 rad/s. A small object with

mass 0.1 kg is placed on the rim of the disk. What is the angular velocity of

the disk after the object falls off? Assume the object falls off without imparting

any torque to the disk.

Solution

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

angular momentum of the system is given by the sum of the angular momentum

of the disk and the angular momentum of the object:

Linitial =Idiskωinitial +mobjectrωinitial

For a disk rotating about its center, the moment of inertia is Idisk =1

2mr2.

Substitute the given values into the equation:

Linitial =1

2×2×0.52×5+0.1×0.5×5

Step 2: Calculate the final angular momentum of the system after the object

falls off. When the object falls off, the angular momentum of the system remains

conserved. Thus, the final angular momentum is given by:

Lfinal =Idiskωfinal

Step 3: Set the initial and final angular momenta equal to each other to find

the final angular velocity. Since angular momentum is conserved, we have:

Linitial =Lfinal

2×2×0.52×5+0.1×0.5×5 = 1

2×2×0.52×ωfinal

Solve this equation for ωfinal to find the angular velocity of the disk after the

object falls off.

Question 7

Question

A thin uniform rod of length Land mass Mis initially at rest on a frictionless

horizontal surface. One end of the rod is then struck horizontally with an

impulse J. Determine the angular velocity of the rod just after the impulse.

Solution

1. After the impulse is applied, the rod will rotate around its pivot point. Let

the angular velocity of the rod just after the impulse be ω. The linear velocity

of the end of the rod being struck is v=ω·L

2. By applying the principle of conservation of angular momentum, we

have: initial angular momentum = final angular momentum. The initial angular

momentum is zero since the rod is initially at rest. The final angular momentum

is Lfinal =I·ω, where Iis the moment of inertia of the rod.

3. The moment of inertia of a rod rotating about one end is I=1

3Ml2.

Substituting this into the equation, we get: 0 = 1

3Ml2·ω.

4. Solving for ω, we find: ω= 0.

Therefore, the angular velocity of the rod just after the impulse is ω= 0.

Question 8

Question

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

is placed on the rim of the disc and released from rest. Calculate the angular

velocity of the particle when it has fallen through a vertical distance of 1 m.

Assume that the disc is frictionless.

Solution

Step 1: First, we need to find the initial angular velocity of the particle at

rest on the rim of the rotating disc. We can use the conservation of angular

momentum.

Step 2: The initial angular momentum of the particle on the rotating disc is

equal to the final angular momentum of the particle at the bottom. Therefore,

we have:

I1ω1=I2ω2

where I1is the moment of inertia of the disc, I2is the moment of inertia of the

particle, and ω1and ω2are the initial and final angular velocities, respectively.

Step 3: The moment of inertia of the disc is given by Idisc =1

2mr2, where m

is the mass of the disc and ris the radius. The moment of inertia of the particle

about the axis passing through its center is Iparticle =mr2. Therefore, we have:

2mr2·4 = mr2·ω2

Step 4: Solving for ω2, we find:

ω2= 2 rad/s

Step 5: Next, we need to find the angular velocity at the bottom of the disc

after the particle has fallen a distance of 1 m. We can use the conservation of

energy.

Step 6: The change in gravitational potential energy of the particle at the

top of the disc to the bottom is equal to the change in rotational kinetic energy.

That is:

mgh =1

2Iω2

where mis the mass of the particle, gis the acceleration due to gravity, h

is the height fallen, Iis the moment of inertia of the system (particle + disc),

and ωis the final angular velocity.

Step 7: Substituting the values h= 1 m and I=3

2mr2into the above

equation, we find:

mg =3

4mω2

Step 8: Solving for ω, we get:

ω=r4g

3≈2.05 rad/s

Therefore, the angular velocity of the particle when it has fallen through a

vertical distance of 1 m is approximately 2.05 rad/s.

Question 9

Question

A solid cylindrical disk of radius Rand mass Mis freely pivoted about its center

and then set spinning with an initial angular velocity ω0. A small block of mass

mis placed on the edge of the disk and starts sliding towards the center. At

what distance rfrom the center of the disk does the block lose contact with the

disk?

Solution

Step 1: Utilize the conservation of angular momentum to determine the angular

velocity of the block when it loses contact with the disk. Since the system is

isolated (no external torques acting), the total angular momentum is conserved.

Initially, the angular momentum of the disk is Linitial =Idiskω0, and when the

block loses contact, the angular momentum is Lfinal = (Idisk +mr2)ωblock.

⇒Idiskω0= (Idisk +mr2)ωblock

Step 2: Express the moments of inertia in terms of the given quantities. The

moment of inertia of a disk about its center is Idisk =1

2MR2, and for the block

about the edge, it is mr2.

Step 3: Solve for the angular velocity of the block when it loses contact.

Substitute the expressions for Idisk and Iblock into the conservation equation to

get:

2MR2ω0=1

2MR2+mr2ωblock

Now, solve for ωblock:

ωblock =MR2ω0

MR2+ 2mr2

Step 4: Use the relationship between linear and angular velocity to find the

velocity of the block when it loses contact. The linear velocity of a point on the

rotating disk is given by v=ωr. At the point where the block loses contact,

the linear velocity of the edge of the disk is equal to the velocity of the block:

ωblockr=ω0R

Substitute the expression for ωblock obtained earlier:

MR2ω0

MR2+ 2mr2r=ω0R

⇒r=2mR3

MR2+ 2mr2

Therefore, the block loses contact with the disk at a distance rfrom the

center given by r=2mR3

MR2+2mr2.

Question 10

Question

A thin 2.0 m long rod is pivoted at one end and swings in a vertical plane. The

rod is released from rest when it makes an angle of 60◦with the vertical. What

is the angular velocity of the rod when it is vertical?

Solution

Step 1: Let’s first find the potential energy of the rod when it is at the initial

position. The potential energy Uiat an angle θiwith the vertical is given by:

Ui=mghi

where: m= mass of the rod g= acceleration due to gravity hi= height at

angle θi

Step 2: At the initial position, when the rod makes an angle of 60◦with the

vertical, the height can be calculated as:

hi= 2.0×sin(60◦)

Step 3: The potential energy at the initial position is:

Ui=mghi

Step 4: Now, let’s find the potential energy of the rod when it is vertical.

The potential energy Ufat the vertical position is:

Uf=mghf

where: hf= 0 (at the vertical position)

Step 5: The kinetic energy Kfof the rod at the vertical position is equal to

the total initial potential energy Uisince the rod began from rest.

Kf=Ui

Step 6: The kinetic energy Kfat the vertical position is given by:

Kf=1

2Iω2

where: I= moment of inertia of the rod about the pivot ω= angular velocity

of the rod at the vertical position

Step 7: The moment of inertia Iof the rod about the pivot is:

I=1

3mL2

where: L= length of the rod

Step 8: Setting the two expressions for kinetic energy equal, we have:

2Iω2=mghi

Step 9: Solving for ωgives:

ω=r2ghi

Step 10: Substitute the values of m,g,hi, and Linto the equation and

calculate the angular velocity.

Question 11

Question

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

small particle is then placed on the disk at a distance of 0.3 m from the center.

What is the angular velocity of the particle with respect to the disk just after

it is placed?

Solution

Let’s denote the angular velocity of the particle with respect to the disk as ωp.

We know that the angular velocity of the disk is 4 rad/s. The angular velocity

of the particle with respect to the disk can be found using the formula:

ωp=ωdisk +ωparticle

where ωparticle is the angular velocity of the particle in an inertial reference

frame.

Step 1: Calculate the angular velocity of the particle in the inertial reference

frame.

The linear velocity of the particle can be calculated as v=rω, where r= 0.3

m and ω= 4 rad/s.

So, v= 0.3×4=1.2 m/s.

Now, the angular velocity of the particle in the inertial reference frame is

ωparticle =v

r=1.2

0.3= 4 rad/s.

Step 2: Find the angular velocity of the particle with respect to the disk.

Using the formula ωp=ωdisk +ωparticle, we have

ωp= 4 + 4 = 8 rad/s

Therefore, the angular velocity of the particle with respect to the disk just

after it is placed is 8 rad/s.

Question 12

Question

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

acceleration. After 3 seconds, the wheel has made 15 complete revolutions.

What is the angular acceleration of the wheel?

Solution

Step 1: Find the angular velocity of the wheel after 3 seconds. Given that

the wheel starts at rest, the initial angular velocity ω0= 0. The final angular

velocity ωcan be calculated using the equation:

ω=ω0+αt

where αis the angular acceleration and tis the time taken.

ω= 0 + α×3

ω= 3α

Step 2: Convert the number of revolutions to radians. The angle covered by

15 complete revolutions can be calculated as:

θ= 2π×15 = 30πradians

Step 3: Find the final angular velocity using angle covered. The final angular

velocity ωcan also be calculated using the formula:

ω2=ω2

0+ 2αθ

Plugging in the known values:

(3α)2= 0 + 2α×30π

9α2= 60πα

α=60π

9=20π

3≈20.94 rad/s2

Therefore, the angular acceleration of the wheel is 20π

3rad/s2.

Question 13

Question

A disk is rotating counterclockwise at a constant angular acceleration of 1.5

rad/s2. At time t= 0, the angular velocity of the disk is 3 rad/s and the radius

of the disk is 0.5 m. What is the angular velocity of a point on the disk that is

0.25 m from the center of rotation after 2 seconds?

Solution

Step 1: Find the angular velocity of the disk after 2 seconds using the equation

ωf=ωi+αt. Given: ωi= 3 rad/s, α= 1.5 rad/s2,t= 2 s.

Plugging the values into the equation:

ωf= 3 + 1.5×2 = 3 + 3 = 6 rad/s.

Step 2: Find the linear velocity of the point on the disk using v=rω. Given:

r= 0.25 m, ω= 6 rad/s.

Plugging the values into the equation:

v= 0.25 ×6 = 1.5 m/s.

Step 3: Find the angular velocity of the point on the disk using the equation

ω=v/r. Given: v= 1.5 m/s, r= 0.25 m.

Plugging the values into the equation:

ω=1.5

0.25 = 6 rad/s.

Therefore, the angular velocity of a point on the disk that is 0.25 m from

the center of rotation after 2 seconds is 6 rad/s.

Question 14

Question

A thin rod of length Lrotates about one end, with an angular speed of ω. At

a certain instant, a small object of mass mis attached to the free end of the

rod. Determine the angular speed of the rod-object system after the object is

attached.

Solution

1. The initial angular momentum of the rod alone is given by:

Linitial =Irodω

where Irod is the moment of inertia of the rod about its end. We can approximate

the rod as a uniform rod rotating about its end, so Irod =1

3mL2.

2. The final angular momentum of the system (rod-object) is the sum of the

angular momentum of the rod and the angular momentum of the object:

Lfinal =Irodωfinal +mL2ωfinal

where ωfinal is the angular speed of the system after the object is attached.

3. Since angular momentum is conserved, Linitial =Lfinal. Substituting the

expressions for Linitial and Lfinal:

Irodω=Irodωfinal +mL2ωfinal

4. Solving for ωfinal:

3mL2ω=1

3mL2+mL2ωfinal

5. Simplifying the equation gives:

3ω=4

3ωfinal

6. Therefore, the angular speed of the rod-object system after the object is

attached is:

ωfinal =1

4ω

Question 15

Question

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

speed ω. Find the magnitude of the total angular momentum of the rod about

the opposite end.

Solution

Step 1: The angular momentum of a point mass mmoving at a distance rfrom

a fixed axis with angular speed ωis given by L=m·r2·ω. For the entire rod,

we need to integrate this expression over the length of the rod.

Step 2: Consider a small element of length dx at a distance xfrom the axis

of rotation. The mass of this element is dm =M

L·dx.

Step 3: The angular momentum dL of this element is dL =M

L·x2·ω·dx.

Step 4: The total angular momentum Ltotal of the rod is obtained by inte-

grating dL from 0 to L:

Ltotal =ZL

L·x2·ω·dx

Step 5: After integrating, we get:

Ltotal =Mω

LZL

x2·dx

Step 6: Solving the integral, we find:

Ltotal =Mω

L·1

3x3L

Step 7: Substituting the limits of integration and simplifying, we get:

Ltotal =Mω

L·1

3L3−03

Step 8: Finally, we have:

Ltotal =MωL2

Question 16

Question

A disc of radius Ris initially at rest. It starts rotating about a fixed axis passing

through its center with a constant angular acceleration α. At what time will

the tangential speed of a point on the rim reach half of the maximal value?

Solution

Let ωbe the angular velocity of the disc and vbe the tangential speed of a

point on the rim.

Step 1: The relationship between the angular velocity ω, tangential speed

v, and radius Ris given by v=Rω.

Step 2: The angular acceleration αis related to the angular velocity ωas

α=dω

dt .

Step 3: The time derivative of vwith respect to time tis a=Rα.

Step 4: From the given information, when v=1

2Rωmax, we can write

2Rαt =1

2Rωmax.

Step 5: Solving for t, we find t=ωmax

α.

Therefore, the tangential speed of a point on the rim will reach half of the

maximal value at a time t=ωmax

α.

Question 17

Question

A disk starts from rest and accelerates with a constant angular acceleration α.

What is the time it takes to reach an angular velocity ω?

Solution

Step 1: The final angular velocity ωcan be related to the initial angular velocity

ω0, angular acceleration α, and time tusing the equation:

ω=ω0+αt

Step 2: Since the disk starts from rest, the initial angular velocity ω0= 0.

Therefore, the equation simplifies to:

ω=αt

Step 3: Rearranging the equation gives:

t=ω

Step 4: This shows that the time it takes for the disk to reach an angular

velocity ωis ω

α.

Question 18

Question

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

acceleration of 2 rad/s2for 4 s. Determine the angular velocity of the wheel at

t= 4 s.

Solution

Step 1: We can use the equation for rotational kinematics:

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.

Step 2: Substituting the given values, we get:

ω= 0 + (2 rad/s2)(4 s)

Step 3: Calculating the final angular velocity:

ω= 8 rad/s

Step 4: Therefore, the angular velocity of the wheel at t= 4 s is 8 rad/s.

Question 19

Question

A thin rod of length Land mass Mis pivoted at one end and is initially at rest.

A small object of mass mcollides with the free end of the rod and sticks to it.

The collision is perfectly inelastic. Find the angular velocity of the system just

after the collision.

Solution

Let’s denote the initial angular velocity of the system as ωiand the final angular

velocity after the collision as ωf.

Step 1: First, we need to calculate the moment of inertia of the rod-object

system. The moment of inertia of the thin rod about its pivot point is Irod =

3ML2. The moment of inertia of the point mass about the pivot point is

Iobject =mL2. Therefore, the total moment of inertia of the system is Itotal =

Irod +Iobject.

Step 2: Next, we can apply the principle of conservation of angular momen-

tum. Since no external torque acts on the system, the initial angular momen-

tum is equal to the final angular momentum. The initial angular momentum is

Linitial =Itotal ·ωi. After the collision, the system becomes one rigid body ro-

tating about the pivot with angular velocity ωf, so the final angular momentum

is Lfinal =Itotal ·ωf.

Step 3: Setting the initial angular momentum equal to the final angular

momentum and solving for ωf, we get:

Itotal ·ωi=Itotal ·ωf

3ML2·ωi= (1

3ML2+mL2)·ωf

3Mωi=1

3Mωf+mωf

3Mωi=1

3M+mωf

ωf=Mωi

M+ 3m

Therefore, the angular velocity of the system just after the collision is ωf=

Mωi

M+3m.

Question 20

Question

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

bug lands on the edge of the disk and starts walking towards the center at a

constant speed of 0.1 m/s relative to the disk. What is the bug’s acceleration

(magnitude and direction) relative to the disk when the bug is 0.1 m from the

center of the disk?

Solution

Step 1: First, we need to find the bug’s velocity (magnitude and direction)

relative to the disk. Since the bug is walking towards the center of the disk, we

can split its velocity into two components: tangential and radial.

The tangential component of the bug’s velocity vtis equal to the disk’s

angular velocity ωtimes the distance from the bug to the center of the disk r.

Therefore,

vt=ω×r= 5 rad/s ×0.1 m = 0.5 m/s

Step 2: To find the bug’s radial velocity vr(magnitude and direction), we

need to consider the bug’s motion towards the center of the disk. Since the

bug’s velocity relative to the disk is 0.1 m/s, the magnitude of the bug’s radial

velocity is 0.1 m/s.

The bug’s radial velocity is directed towards the center of the disk, opposite

to its motion.

Step 3: Now, we can determine the bug’s total velocity relative to the disk

using the components found in Step 1 and Step 2.

The bug’s total velocity vrelative to the disk is given by the vector sum of

the tangential and radial velocities:

v=qv2

t+v2

r=p0.52+ 0.12=√0.26 ≈0.51 m/s

Step 4: Finally, we can find the bug’s acceleration (magnitude and direction)

relative to the disk by considering the change in the bug’s velocity as it moves

towards the center.

The bug’s acceleration arelative to the disk is directed radially towards the

center of the disk. The magnitude of the bug’s acceleration can be calculated

using the formula for centripetal acceleration:

a=v2

r=0.512

0.1=0.2601

0.1= 2.601 m/s2

Therefore, the bug’s acceleration relative to the disk when the bug is 0.1

m from the center of the disk is approximately 2.601 m/s2directed radially

towards the center of the disk.

Question 21

Question

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

angular acceleration of 2 rad/s2. What is the magnitude of the total distance

covered by a point on the rim of the disk after 5 seconds?

Solution

Step 1: Find the angular velocity of the disk after 5 seconds using the equation

ω=ω0+αt, where ω0is the initial angular velocity (0 in this case), αis the

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

ω= 0 + 2 ×5 = 10 rad/s

Step 2: Find the total angle rotated by the disk after 5 seconds using the

equation θ=1

2αt2, where αis the angular acceleration (2 rad/s2) and tis the

time (5 s).

θ=1

2×2×(5)2= 25 rad

Step 3: Find the total distance covered by a point on the rim of the disk.

Since the point travels along the circumference of the disk, the distance is given

by d=rθ, where ris the radius of the disk (0.3 m) and θis the total angle

covered (25 rad).

d= 0.3×25 = 7.5 m

Therefore, the magnitude of the total distance covered by a point on the rim

of the disk after 5 seconds is 7.5 meters.

Question 22

Question

A solid cylinder with mass Mand radius Ris rolling without slipping down an

incline of angle θ. The moment of inertia of the cylinder about its center of

mass is I=1

2MR2. If the cylinder starts from rest at the top of the incline,

determine the speed of the center of mass of the cylinder when it reaches the

bottom of the incline. Ignore friction.

Solution

Step 1: Calculate the gravitational potential energy at the top and its kinetic

energy at the bottom.

The change in gravitational potential energy is converted into the kinetic energy

at the bottom of the incline.

∆KE = ∆P E

Step 2: Calculate the change in height.

The change in height is equal to the vertical distance the center of mass travels

from the top to the bottom of the incline.

h=R(1 −cos(θ))

Step 3: Calculate the speed of the center of mass at the bottom.

Using the principle of conservation of energy, we can equate the change in po-

tential energy to the change in kinetic energy.

∆KE = ∆P E

KEfinal −KEinitial =P Efinal −P Einitial

2Mv2

final −0=0−M gR(1 −cos(θ))

2Mv2

final =MgR(1 −cos(θ))

vfinal =p2gR(1 −cos(θ))

Therefore, the speed of the center of mass of the cylinder when it reaches

the bottom of the incline is vfinal =p2gR(1 −cos(θ)).

Question 23

Question

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

angular acceleration of 2 rad/s2.

(a) What is the angular velocity of the disk after 3 seconds?

(b) How many revolutions has the disk gone through after 3 seconds?

Solution

(a) Step 1: Use the formula for angular velocity under constant acceleration:

ωf=ωi+αt

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

is 0 for this case since the disk starts from rest), - αis the angular acceleration,

and - tis the time. Step 2: Substitute the given values into the formula:

ωf= 0 + 2 ×3 = 6 rad/s

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

(b) Step 1: Use the formula for angular displacement under constant angular

acceleration:

θ=ωit+1

2αt2

where - θis the angular displacement, - ωiis the initial angular velocity (0), - α

is the angular acceleration, - tis the time. Step 2: Substitute the given values

into the formula:

θ= 0 + 1

2×2×32= 9 rad

Step 3: Calculate the number of revolutions:

Number of revolutions = θ

2π=9

2π≈1.43 revolutions

Therefore, the disk has gone through approximately 1.43 revolutions after 3

seconds.

Question 24

Question

A uniform disc of radius Ris initially at rest. A constant torque τis applied to

the disc, causing it to rotate with an angular acceleration α. Find the angular

velocity of the disc at the time the angular displacement θis equal to π

Solution

Step 1: Identify the relevant equations. The kinematic equation relating angular

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

can be used: ω2

f=ω2

i+ 2αθ.

Step 2: Find the initial angular velocity. Since the disc is initially at rest,

the initial angular velocity ωi= 0.

Step 3: Determine the final angular velocity. Using the kinematic equation,

we have ω2

f= 0+2απ

2=απ. Therefore, the angular velocity of the disc when

θ=π

2is ω=√απ.

Question 25

Question

A wheel initially at rest undergoes an angular acceleration given by α= 4t−2t2,

where αis in radians per second squared and tis in seconds. If the wheel starts

from rest, determine the angular velocity of the wheel after 3 seconds.

Solution

Step 1: To find the angular velocity of the wheel after 3 seconds, we need to

integrate the given angular acceleration function with respect to time to get the

angular velocity function.

Step 2: Integrate α= 4t−2t2with respect to time to find the angular

velocity function ω(t):

Zα dt =Z(4t−2t2)dt

Step 3: Integrating term by term:

Zα dt =Z4t dt −Z2t2dt

Step 4: Integrating each term:

ω(t)=2t2−2

3t3+C

Step 5: Since the wheel starts from rest, the constant of integration Cis

equal to 0.

Step 6: Substitute t= 3 into the angular velocity function to find the angular

velocity after 3 seconds:

ω(3) = 2(3)2−2

3(3)3= 18 rad/s

Step 7: Therefore, the angular velocity of the wheel after 3 seconds is ω=

18 rad/s.

Question 26

Question

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

surface from rest. It then ascends a 30◦incline. Determine the linear speed of

the sphere when it reaches a height hup the incline.

Solution

Let’s denote the linear speed of the sphere when it reaches a height hup the

incline as v.

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

When the sphere starts rolling without slipping, the initial condition relates

the linear speed (v), angular speed (ω), and radius (R) of the sphere. Since the

sphere rolls without slipping, we have v=Rω. Therefore, the initial angular

speed of the sphere is given by:

ω=v

Step 2: Determine the final angular speed of the sphere.

To determine the final angular speed of the sphere as it reaches height hon

the incline, we will use the conservation of energy. The initial kinetic energy of

the sphere is due to its rotational motion and translation motion:

KEinitial =1

2Iω2+1

2Mv2

The final kinetic energy is due to its increased potential energy at height hup

the incline:

KEfinal =1

2Iω2

f+1

2Mv2

f+Mgh

Since the sphere is rolling without slipping, the moment of inertia of the sphere

about its center of mass is I=2

5MR2. Setting the initial kinetic energy equal

to the final kinetic energy, we have:

22

5MR2v

R2+1

2Mv2=1

22

5MR2ω2

f+1

2Mv2

f+Mgh

Step 3: Determine the linear speed of the sphere at height h.

Now, we can solve for vf(the linear speed of the sphere at height h) in terms

of R,v,h, and other constants. Simplify the expression to find vf:

vf=5

7v2+ 5gh1/2

Question 27

Question

A thin uniform rod of length Land mass Mis pivoted about one end. Initially,

the rod is at rest. A bullet of mass mmoving horizontally with speed vstrikes

the rod at a distance dfrom the pivot point and gets embedded in it. Find the

angular speed of the rod just after the collision.