PHYS 305 - INTRODUCTION TO
MODERN PHYSICS - Rotational
kinematics
Question Bank - Set 5
Liberty University
Question 1
Question
A flywheel starts from rest and accelerates with a constant angular acceleration
of 0.05 rad/s2for 10 seconds. If the flywheel has a radius of 0.5 meters, determine
the angular velocity of the flywheel at the end of the 10 seconds.
Solution
Step 1: Identify the known values and the angular acceleration equation. The
given values are: Angular acceleration, α= 0.05 rad/s2Time, t= 10 s Initial
angular velocity, ω0= 0 (as the flywheel starts from rest) Radius of the flywheel,
r= 0.5 m
The angular acceleration equation relates the final angular velocity ωto the
initial angular velocity ω0, angular acceleration α, and time t:
ω=ω0+αt
Step 2: Calculate the final angular velocity. Plugging in the values into the
angular acceleration equation:
ω= 0 + 0.05 ×10
ω= 0.5 rad/s
Therefore, the angular velocity of the flywheel at the end of 10 seconds is
0.5 rad/s.
Question 2
Question
A wheel starts at 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 variables and the unknown we are looking for. Let
ω0= 0 rad/s be the initial angular velocity of the wheel, α= 2 rad/s2be the
angular acceleration of the wheel, and t= 3 s be the time at which we want to
find the angular velocity of the wheel.
Step 2: Use the equation for angular motion under constant acceleration.
The equation for angular motion under constant acceleration is:
ω=ω0+αt
where: ωis the final angular velocity, ω0is the initial angular velocity, αis the
angular acceleration, and tis the time.
Step 3: Substitute the values into the equation.
ω= 0 + 2 ×3
Step 4: Calculate the final angular velocity.
ω= 6 rad/s
Therefore, after 3 seconds, the angular velocity of the wheel is 6 rad/s.
Question 3
Question
A uniform circular disc of radius Rand mass Mrotates about its axis passing
through its center with an angular velocity ω. A small ball is placed on top of
the disc at a distance rfrom the center (as shown in the figure below).
r
R
Center
Ball
ω
If the disc suddenly comes to a stop, find the initial angular speed of the
ball relative to the the disc just before the disc stops rotating.
2
Solution
Step 1: Let vbe the velocity of the ball relative to the disc just before the disc
stops. This velocity can be found using the relative circular motion equation:
v=ωR
Step 2: The initial angular momentum of the ball about the center of the
disc is given by:
Li=m·(v·r)
Step 3: For the ball-disc system, angular momentum is conserved. The final
angular momentum of the ball is zero since it is no longer rotating. Therefore,
the change in angular momentum is given by:
∆L=Li
Step 4: The change in angular momentum is also given by:
∆L=Idisc ·ω
Step 5: Substituting the expressions for Li,ω, and Idisc =1
2MR2into the
above equation, we have:
m·(v·r) = 1
2MR2·ω
Step 6: Substituting v=ωR into the above equation:
m·ωR ·r=1
2MR2·ω
Step 7: Solving for ωgives:
ω=2m
M·r
R
Hence, the initial angular speed of the ball relative to the disc just before
the disc stops rotating is 2m
M·r
R.
Question 4
Question
A solid sphere with radius Rand moment of inertia Iabout its center of mass
starts from rest and rolls without slipping down a 30-degree incline. The sphere’s
mass is M. Find the final velocity of the sphere when it reaches the bottom of
the incline.
3
Solution
Step 1: Determine the acceleration of the sphere down the incline. The net
torque about the center of mass of the sphere is equal to the moment of inertia
times the angular acceleration. The torque is due to the gravitational force
component parallel to the incline, so we have:
τ=2
5MR2α=MgR sin(30◦)R
2
5R2α=gR sin(30◦)
α=5gsin(30◦)
2R
acm =Rα =5gsin(30◦)
2
Step 2: Determine the linear acceleration of the center of mass. Since the
sphere is rolling without slipping, the linear acceleration of the center of mass
is related to the angular acceleration:
acm =Rα
acm =5gsin(30◦)
2
Step 3: Find the final velocity of the sphere at the bottom of the incline.
Using energy conservation, the initial mechanical energy of the sphere is equal
to its final mechanical energy:
1
2Iω2=1
2Mv2+1
2Iv
R2
Substitute ω=v
Rand I=2
5MR2:
1
22
5MR2v
R2
=1
2Mv2+1
22
5MR2v
R2
Solve for vto get the final velocity of the sphere.
Question 5
Question
A disc of radius 0.5 m is rotating with an angular acceleration of −2 rad/s2. At
t= 0, the angular velocity of the disc is 3 rad/s. What is the angular velocity
of the disc after it has rotated through an angle of 30 radians?
4
Solution
Step 1: Find the angular velocity of the disc using the equation ω=ω0+αt
Given: Initial angular velocity, ω0= 3 rad/s
Angular acceleration, α=−2 rad/s2
Time, t=? (to be determined)
Using the given equation, we can solve for t:
ω=ω0+αt
3 = 3 + (−2)t
−2t= 0
t= 0 s
Step 2: Find the final angular velocity using the equation ω2=ω2
0+ 2αθ
Given: Initial angular velocity, ω0= 3 rad/s
Angular acceleration, α=−2 rad/s2
Angular displacement, θ= 30 rad
Using the given equation, we can solve for the final angular velocity ω:
ω2=ω2
0+ 2αθ
ω2= 32+ 2(−2)(30)
ω2= 9 −120
ω2=−111
ω=√−111
ω= 0 rad/s
Therefore, the angular velocity of the disc after it has rotated through an
angle of 30 radians is 0 rad/s.
Question 6
Question
A thin rod of mass mand length Lis rotating about one end with an angular
speed ω. Calculate the angular momentum of the rod about the rotation axis.
5
Solution
Step 1: The angular momentum of an object rotating about an axis is given by
the formula L=Iω, where Lis the angular momentum, Iis the moment of
inertia, and ωis the angular speed.
Step 2: To find the moment of inertia of the rod rotating about one end, we
need to use the parallel axis theorem. The moment of inertia of a rod rotating
about its center is 1
12 mL2.
Step 3: The distance between the axis of rotation and the center of mass
of the rod is L
2. Therefore, the moment of inertia about one end can be found
using the parallel axis theorem: I=1
12 mL2+mL
22.
Step 4: Simplifying the equation gives I=1
3mL2.
Step 5: Now, we can calculate the angular momentum by substituting the
moment of inertia into the formula: L=1
3mL2ω.
Step 6: Therefore, the angular momentum of the rod about the rotation axis
is 1
3mL2ω.
Question 7
Question
A solid sphere of radius Rstarts from rest and rolls without slipping down a
30◦incline. What is the acceleration of the center of mass of the sphere along
the incline?
Solution
Step 1: The acceleration of the center of mass of the sphere along the incline
can be found using the rotational kinematics equations. The equation relating
linear acceleration aand angular acceleration αfor a rolling object is a=Rα,
where Ris the radius of the sphere.
Step 2: The acceleration of the center of mass can also be expressed using
the linear acceleration aand angular velocity ωas a=Rα =Rdω
dt .
Step 3: The kinematic constraint for rolling without slipping is a=Rα =
Rdω
dt . For the sphere rolling without slipping down an incline, the linear accel-
eration acan be related to the angular velocity ωas a=Rdω
dt =Rα.
Step 4: Considering that the sphere is rolling without slipping, its linear
acceleration adown the incline can be related to the angular acceleration α
through the equation a=Rα =Rdω
dt =Rd
dt v
R, where vis the linear velocity
of the sphere.
Step 5: In the rotational kinematics equation a=Rα =Rd
dt v
R, we can
substitute v=Rω to find a=Rd
dt Rω
R.
Step 6: Simplifying the expression a=Rd
dt Rω
R, we get a=Rdω
dt , which
means that the linear acceleration of the center of mass of the sphere along the
incline is equal to Rdω
dt .
6
Step 7: Therefore, the acceleration of the center of mass of the sphere along
the incline is Rdω
dt .
Question 8
Question
A disk of radius 0.3 m starts from rest and accelerates with a constant angular
acceleration of 0.5 rad/s2. Find the angular velocity of the disk when it has
completed 3 full revolutions.
Solution
Step 1: Calculate the total angular displacement when the disk completes 3 full
revolutions.
Total angular displacement = 2π×number of revolutions
= 2π×3
= 6πrad
Step 2: Use the kinematic equation for rotational motion to find the final
angular velocity.
ω2=ω2
0+ 2αθ
Where: ω= final angular velocity, ω0= initial angular velocity (which is 0
rad/s), α= angular acceleration, θ= angular displacement.
Step 3: Substitute the known values into the kinematic equation.
ω2= 0 + 2(0.5)(6π)
ω2= 6π
Step 4: Solve for the final angular velocity.
ω=√6π≈4.91 rad/s
Therefore, the angular velocity of the disk when it completes 3 full revolu-
tions is approximately 4.91 rad/s.
Question 9
Question
A disk of radius 0.2 m starts from rest and rotates with a constant angular
acceleration of 2.0 rad/s2.
7
Part (a)
Determine the angular velocity of the disk after 3.0 seconds.
Part (b)
Find the number of revolutions the disk has made after 3.0 seconds.
Solution
Part (a)
Step 1: The angular velocity of the disk is given by the equation ω=ω0+αt,
where ωis the final angular velocity, ω0is the initial angular velocity, αis the
angular acceleration, and tis the time.
Step 2: Substituting the known values ω0= 0, α= 2.0 rad/s2, and t= 3.0 s
into the equation, we have ω= 0 + 2.0×3.0=6.0 rad/s.
Therefore, the angular velocity of the disk after 3.0 seconds is 6.0 rad/s .
Part (b)
Step 1: The number of revolutions made by the disk can be found using the
formula θ=θ0+ω0t+1
2αt2, where θis the total angular displacement, θ0is
the initial angular displacement (usually 0), ω0is the initial angular velocity, α
is the angular acceleration, and tis the time.
Step 2: Since the disk starts from rest, ω0= 0, and calculating θwe have
θ=0+0+1
2×2.0×(3.0)2= 9.0 rad.
Step 3: Converting the total angular displacement to the number of revolu-
tions, we recall that 2πrad = 1 revolution. Therefore, the number of revolutions
is 9.0 rad
2π=9.0
2π≈1.43 revolutions.
Hence, the number of revolutions the disk has made after 3.0 seconds is
1.43 revolutions .
Question 10
Question
A disk of radius Ris initially at rest. A constant force is applied tangentially
to the disk at a distance rfrom the center. The force is applied for a time T,
after which it is removed. Determine the angular velocity of the disk after the
force is removed.
Solution
Step 1: Calculate the torque exerted on the disk by the applied force. The
torque τis given by τ=rF sin θ, where Fis the force applied and θis the angle
8
between the force and the lever arm. Since the force is tangential to the disk,
θ= 90◦and sin θ= 1. Thus, τ=rF .
Step 2: Use the rotational analog of Newton’s second law: τ=Iα, where
Iis the moment of inertia of the disk and αis the angular acceleration. For a
disk, I=1
2mR2, where mis the mass of the disk.
Step 3: Substituting the expressions for torque and moment of inertia into
the rotational Newton’s second law equation gives rF =1
2mR2α. Rearranging
for angular acceleration gives α=2rF
mR2.
Step 4: The final angular velocity ωfof the disk can be calculated using the
kinematic equation for rotational motion, ωf=ωi+αt, where ωiis the initial
angular velocity (which is zero in this case). Thus, ωf=αT .
Step 5: Substituting the expression for angular acceleration into the equation
for final angular velocity gives ωf=2rF
mR2T. So, the angular velocity of the disk
after the force is removed is 2rF
mR2T.
Question 11
Question
A wheel with a radius of 0.5 m starts from rest and accelerates with a constant
angular acceleration of 2 rad/s2. How long will it take for the wheel to reach
an angular speed of 10 rad/s?
Solution
Step 1: Let’s first identify the given values: Initial angular speed, ω0= 0 rad/s
Final angular speed, ω= 10 rad/s
Angular acceleration, α= 2 rad/s2
Radius of the wheel, r= 0.5 m
Step 2: Next, let’s use the rotational kinematic equation relating initial
angular speed, final angular speed, angular acceleration, and time:
ω=ω0+αt
Substitute in the given values:
10 = 0 + 2t
Step 3: Solve for the time t:
2t= 10 ⇒t=10
2= 5 s
Step 4: Therefore, it will take 5 seconds for the wheel to reach an angular
speed of 10 rad/s.
9
Question 12
Question
A solid sphere of radius Rand mass Mis released from the top of a rough
inclined plane that makes an angle θwith the horizontal. The sphere rolls
without slipping down the incline. Calculate the acceleration of the center of
mass of the sphere.
Solution
Step 1: The acceleration of the center of mass of the sphere can be found
by considering the forces acting on the sphere. The forces involved are the
gravitational force mg acting vertically downward, the normal force Nacting
perpendicular to the incline, and the friction force facting parallel to the incline
in the direction opposite to the motion.
Step 2: The gravitational force can be resolved into components parallel and
perpendicular to the incline. The component of the gravitational force parallel
to the incline is mg sin(θ) and the component perpendicular to the incline is
mg cos(θ).
Step 3: The net force causing the acceleration down the incline is given by
f=macm, where acm is the acceleration of the center of mass.
Step 4: The friction force fcan be calculated using the relationship f=µN ,
where µis the coefficient of friction between the sphere and the incline. The
normal force Ncan be calculated using the relationship N=mg cos(θ).
Step 5: Substituting the expression for the friction force f, we have µmg cos(θ) =
macm.
Step 6: Now, we can solve for the acceleration of the center of mass acm
and obtain acm =µg cos(θ). Thus, the acceleration of the center of mass of the
sphere rolling down the incline is µg cos(θ).
Question 13
Question
A thin rod of length Land mass Mis rotating about an axis passing through
one of its ends with an angular velocity ω. If the rod is suddenly brought to rest
without any external torque, what is the tension in the rod just after it stops
rotating?
Solution
Let’s denote the position of the center of mass of the rod by x. The moment
of inertia of a rod rotating about an axis passing through one of its ends is
I=1
3ML2. The kinetic energy of the rod while rotating is given by KE =1
2Iω2,
and the potential energy of the rod due to its center of mass is P E =−Mgx.
10
When the rod stops rotating, all of its initial kinetic energy is converted into
potential energy. This means that KEinitial =KEfinal +P Efinal, which gives us
KEinitial =KEfinal +P Efinal
1
2Iω2=P Efinal
Step 1: Substitute the expressions for the moment of inertia Iand potential
energy P Efinal.
1
21
3ML2ω2=−Mgx
Step 2: Solve for x.1
6ML2ω2=Mgx
x=1
6Lω2
Step 3: The tension in the rod just after stopping is equal to the force it
exerts on its center of mass. This force is a combination of the force due to
gravity and the tension force.
T=Mg +Mω2L
6
Therefore, the tension in the rod just after it stops rotating is T=Mg +
Mω2L
6.
Question 14
Question
A solid sphere of radius Rand mass Mrolls without slipping down an inclined
plane that makes an angle θwith the horizontal. If the sphere starts from rest at
the top of the incline, determine its speed when it reaches the bottom. Assume
the incline is frictionless.
Solution
Step 1: Find the acceleration of the sphere down the incline. The net force
along the incline is due to the component of the gravitational force along the
incline:
Fnet =ma =Mg sin θ
where ais the acceleration, Mis the mass of the sphere, gis the acceleration
due to gravity, and θis the angle of the incline.
11
Step 2: Find the angular acceleration of the sphere. Since the sphere rolls
without slipping, the acceleration of the center of mass ais related to the angular
acceleration αby:
a=Rα
where αis the angular acceleration and Ris the radius of the sphere.
Step 3: Find the speed of the sphere at the bottom of the incline. Since the
sphere starts from rest, we can use the following kinematic equation:
v2=u2+ 2as
where vis the final velocity, uis the initial velocity (which is 0 in this case), a
is the acceleration, and sis the displacement down the incline.
Step 4: Substitute the values into the equations and solve for the final ve-
locity. From step 1: a=gsin θFrom step 2: α=a
R=gsin θ
RFrom step 3:
v2= 0 + 2(gsin θ)(h)
where his the height of the incline.
Step 5: Now, we can substitute the expression for hto get the final velocity
in terms of R,M,g, and θ.
h=R(1 −cos θ)
Thus, the final velocity is:
v=p2gR sin θ(1 −cos θ)
Question 15
Question
A thin rod of length Lis pivoted about one end and is initially at rest. A small
bug is placed on the rod a distance xfrom the pivot point. At time t= 0, the
bug starts crawling along the rod with a constant velocity vrelative to the rod.
What is the angular velocity of the rod as a function of time?
Solution
Step 1: Let rbe the distance from the pivot to the bug at time t. Then, the
bug’s distance from the pivot at time tis given by r=x+vt.
Step 2: The bug’s velocity relative to the pivot point is given by vbug =v.
Step 3: The bug’s velocity can also be expressed in terms of the radial
velocity of the bug along the rod and the angular velocity of the rod, using
vbug =rω, where ωis the angular velocity of the rod.
Step 4: Substituting r=x+vt into vbug =rω, we get v= (x+vt)ω.
Step 5: Simplifying the equation, we find ω=v
x+vt .
Thus, the angular velocity of the rod as a function of time is ω(t) = v
x+vt .
12
Question 16
Question
A solid sphere of radius Rand mass Mis rotating about an axis passing through
its center with an angular speed ω. If the radius of gyration of the sphere is k,
determine the angular momentum of the sphere.
Solution
Step 1: The angular momentum Lof an object rotating about an axis is given
by the equation L=Iω, where Iis the moment of inertia and ωis the angular
speed.
Step 2: The moment of inertia Iof a solid sphere rotating about an axis
passing through its center is given by I=2
5MR2.
Step 3: Given that the radius of gyration kis defined as k=qI
M, we can
express the moment of inertia Iin terms of kas I=Mk2.
Step 4: Substituting the expression for Iinto the equation for angular mo-
mentum, we have L=Mk2ω.
Step 5: The angular momentum of the sphere rotating about the given axis
is given by L=Mk2ω.
Question 17
Question
A flywheel with a radius of 0.5 m starts from rest and accelerates uniformly to
an angular velocity of 10 rad/s in 5 seconds. What is the angular acceleration
of the flywheel?
Solution
Step 1: First, we need to find the angular acceleration of the flywheel using the
formula for angular acceleration, which is given by
α=ωf−ω0
t,
where - αis the angular acceleration, - ωfis the final angular velocity, and - ω0
is the initial angular velocity.
Step 2: Substituting the given values into the formula, we have
α=10 rad/s −0
5 s = 2 rad/s2.
Step 3: Therefore, the angular acceleration of the flywheel is 2 rad/s2.
13
Question 18
Question
A uniform disc of radius Rand mass Mis initially at rest. A constant force F
is applied tangentially to the edge of the disc. What is the angular acceleration
of the disc?
Solution
1. We can find the torque τexerted on the disc by the force Fat the edge of
the disc using the equation τ=F r, where r=Ris the radius of the disc. Thus,
τ=F R.
2. The moment of inertia Iof a uniform disc rotating about an axis per-
pendicular to the plane of the disc and passing through its center is given by
I=1
2MR2.
3. Using Newton’s second law for rotational motion, τ=Iα, where αis the
angular acceleration of the disc.
4. Substituting the expressions for torque τ, moment of inertia I, and solving
for angular acceleration α, we get
F R =1
2MR2α.
5. Solving for α, we find that the angular acceleration of the disc is
α=2F
M.
Therefore, the angular acceleration of the disc is 2F
M.
Question 19
Question
A disk with radius Ris spinning at an angular velocity ω0. A bug lands on
the edge of the disk and starts crawling towards the center at a constant speed
of vbug. At the same time, the disk is slowing down with a constant angular
acceleration α. Given that the bug reaches the center of the disk after making
exactly one full revolution, determine the bug’s position as a function of time.
Solution
Step 1: Find the angular position of the bug as a function of time.
Let θbe the angle the bug has traveled in radians, and tbe the time since
the bug landed on the disk. The bug’s angular velocity can be expressed as:
ωbug(t) = vbug
r
14
where ris the distance from the bug to the center at time t. Since the bug is
moving towards the center, rcan be expressed as R−vbugt. Thus:
ωbug(t) = vbug
R−vbugt
Integrating ωbug(t) with respect to time gives the angular position θas a
function of time:
θ(t) = Zωbug(t)dt =Zvbug
R−vbugtdt
Step 2: Use the given information that the bug makes exactly one full revo-
lution to solve for an expression of θ(t).
Since the bug travels 2πradians when it reaches the center:
Z2π
0
dθ =ZT
0
vbug
R−vbugtdt
where Tis the time it takes for the bug to reach the center.
Solving the integration on the right side and using the condition for one full
revolution, we can then determine the position of the bug as a function of time.
Question 20
Question
A solid cylinder of mass Mand radius Ris initially at rest. A constant force F
is applied tangentially to the edge of the cylinder, causing it to start rotating
about its central axis. The cylinder acquires an angular velocity ωafter a certain
time. What is the angular acceleration of the cylinder in terms of F,M,R, and
ω?
Solution
Step 1: The torque τacting on the cylinder can be calculated from the equation
τ=Iα, where Iis the moment of inertia of the cylinder about its central axis
and αis the angular acceleration. The torque can also be written as τ=F R,
where Fis the applied force.
Step 2: The moment of inertia of a solid cylinder rotating about its central
axis is I=1
2MR2.
Step 3: Setting the two expressions for torque equal to each other, we have
F R =1
2MR2α.
Step 4: Solving for α, we get α=2F
MR .
Step 5: Therefore, the angular acceleration of the cylinder in terms of F,
M,R, and ωis α=2F
MR .
15
Question 21
Question
A solid sphere of radius Rstarts from rest and rolls without slipping down an
inclined plane that makes an angle θwith the horizontal. If the sphere’s moment
of inertia about its center is I=2
5MR2, where Mis the mass of the sphere,
determine its linear acceleration down the incline in terms of gand θ.
Solution
Step 1: The acceleration of the center of mass of the sphere down the incline
can be determined by considering the forces acting on it along the incline. The
forces include the gravitational force mg sin(θ) down the incline, the normal
force Nperpendicular to the incline, and the frictional force fopposing the
motion. The net force down the incline is mg sin(θ)−f. The frictional force
can be determined using the condition for no slipping, f=µN =µmg cos(θ),
where µis the coefficient of static friction.
Step 2: The torque about the center of the sphere due to the frictional force
is fR, causing a clockwise angular acceleration. The torque due to the gravita-
tional force about the center is mgR sin(θ), causing a counterclockwise angular
acceleration. Since the sphere rolls without slipping, the linear acceleration a
of the center of mass is related to the angular acceleration αby a=Rα.
Step 3: Equating the torques gives us fR =Iα. Substituting fand Iin
terms of Mand Rgives µmg cos(θ)R=2
5MR2α. Substituting α=a
Ryields
µg cos(θ) = 2
5Ma.
Step 4: Since the linear acceleration down the incline is given as a, the final
expression for the acceleration in terms of gand θis a=5
2µg cos(θ) .
Question 22
Question
A wheel of radius 0.5 m starts rotating from rest with a constant angular accel-
eration of 2 rad/s2. Find the time it takes for the wheel to make 10 complete
revolutions.
Solution
Step 1: Determine the angular velocity of the wheel after a certain time. The
angular acceleration of the wheel is α= 2 rad/s2. We can use the following
kinematic equation to find the angular velocity of the wheel after a certain
time:
ω=ω0+αt
16
where ωis the final angular velocity, ω0is the initial angular velocity (which is
0 since the wheel starts from rest), αis the angular acceleration, and tis the
time.
Substitute the given values into the equation:
ω= 0 + 2t
ω= 2t
Step 2: Determine the time it takes for the wheel to make 10 complete
revolutions. The angular displacement for one complete revolution is 2πradians.
Therefore, the angular displacement for 10 complete revolutions is 10×2π= 20π
radians.
We can use the following kinematic equation to find the time it takes for the
wheel to make 10 complete revolutions:
θ=θ0+ω0t+1
2αt2
where θis the angular displacement, θ0is the initial angular displacement, ω0
is the initial angular velocity, αis the angular acceleration, and tis the time.
Substitute the given values into the equation:
20π=0+0+1
2×2t2
20π=t2
t=√20π≈7.98 s
Therefore, it takes approximately 7.98 seconds for the wheel to make 10
complete revolutions.
Question 23
Question
A thin hoop of radius Rand mass Mrolls without slipping down a ramp inclined
at an angle θ. If the hoop’s initial velocity at the top of the ramp is v0, determine
its angular velocity ωwhen it reaches the bottom of the ramp.
Solution
Step 1: We will start by finding the speed of the hoop at the bottom of the
ramp using conservation of energy. At the top of the ramp, the hoop has
gravitational potential energy which is all converted to kinetic energy at the
bottom (rotational and translational). Thus, we have
mgR sin θ=1
2Mv2
0+1
2Iω2
17
where Iis the moment of inertia of the hoop and is equal to MR2and ω=v
R.
Step 2: Simplifying the equation, we have
mgR sin θ=1
2Mv2
0+1
2MR2v0
R22
mgR sin θ=1
2Mv2
0+1
2Mv2
0
mgR sin θ=Mv2
0
Step 3: Solving for the speed vat the bottom of the ramp,
v=pgR sin θ
Step 4: Finally, we can find the angular velocity at the bottom of the ramp
by
ω=v
R=√gR sin θ
R=rgsin θ
R
Question 24
Question
A thin, uniform rod of length Land mass Mis rotated about an axis perpen-
dicular to the rod and passing through one end of the rod. What is the moment
of inertia of this rotating rod about this axis?
Solution
Step 1: Let’s consider the moment of inertia of a thin rod of length Land mass
Mabout an axis passing through its end and perpendicular to the rod. This
can be expressed using the formula for the moment of inertia of a rod rotating
about an axis perpendicular to the rod and passing through one end:
I=1
3ML2
Step 2: Hence, the moment of inertia of a thin, uniform rod of length Land
mass Mabout an axis perpendicular to the rod and passing through one end is
given by the expression I=1
3ML2.
Question 25
Question
A solid sphere of radius Rand mass Mrolls without slipping down a 30-degree
incline. Starting from rest at the top, what is its speed when it reaches the
bottom of the incline? Assume the sphere rolls without slipping and the moment
of inertia of a solid sphere about its center is 2
5MR2.
18
Solution
Step 1: Calculate the acceleration of the sphere down the incline due to gravity.
The acceleration of the sphere down the incline can be calculated using the
component of gravity that acts parallel to the incline.
a=gsin θ
where θ= 30◦is the angle of the incline.
Step 2: Calculate the angular acceleration of the sphere. Since the sphere is
rolling without slipping, the linear acceleration of the center of mass ais related
to the angular acceleration αthrough:
a=Rα
and we know:
a=gsin θ
Therefore, we have:
Rα =gsin θ
α=gsin θ
R
Step 3: Calculate the final angular velocity of the sphere. The final angular
velocity of the sphere can be calculated using the rotational kinematic equation:
v2=u2+ 2as
where: - uis the initial angular velocity (which is zero in this case) - sis the
distance traveled down the incline - vis the final angular velocity
Step 4: Calculate the final linear velocity of the sphere. The final linear
velocity of the sphere can be calculated using the relation between linear and
angular velocity of a rolling object:
v=Rω
where ωis the final angular velocity.
Step 5: Substitute the given values and solve for v. The final linear velocity
vof the sphere when it reaches the bottom of the incline is:
v=Rp2gsin θ
Therefore, the speed of the sphere when it reaches the bottom of the incline
is v=R√2gsin 30◦.
Question 26
Question
A disk of radius Rstarts rotating with an angular acceleration αat time t= 0. It
reaches an angular velocity ωafter a time T. What is the angular displacement
θof a point on the rim of the disk at time t=T?
19
Solution
Step 1: Find the angular velocity ωof the disk at time Tusing the formula for
angular kinematics:
ω=αT
Step 2: The angular displacement θof a point on the rim of the disk at time
t=Tcan be calculated using the equation for angular displacement in terms
of initial angular velocity, initial angular acceleration, and time:
θ=1
2αT 2
Step 3: Substituting the expression for αfrom Step 1 into the equation for
θgives:
θ=1
2ω
TT2
θ=1
2ωT
Therefore, the angular displacement θof a point on the rim of the disk at
time t=Tis 1
2ωT .
Question 27
Question
A disk of radius 0.3 m starts from rest and accelerates with a constant angular
acceleration of 2.5 rad/s2.
1. Find the angular velocity of the disk after 4 seconds.
2. Determine the angular displacement of the disk during this time interval.
Solution
1. To find the angular velocity of the disk after 4 seconds, we can use the
equation for rotational kinematics:
ωf=ωi+αt
where
ωfis the final angular velocity,
ωiis the initial angular velocity (which is 0 since the disk starts from rest),
αis the angular acceleration, and
tis the time.
20
Step 1: Calculate the final angular velocity:
ωf= 0 + 2.5×4 = 10 rad/s
Therefore, the angular velocity of the disk after 4 seconds is 10 rad/s.
2. To determine the angular displacement of the disk during this time inter-
val, we can use the equation:
θ=ωit+1
2αt2
where
θis the angular displacement,
ωiis the initial angular velocity,
αis the angular acceleration, and
tis the time.
Since the initial angular velocity is 0, the equation simplifies to:
θ=1
2αt2
Step 2: Calculate the angular displacement:
θ=1
2×2.5×42= 20 rad
Therefore, the angular displacement of the disk during this time interval is
20 radians.
Question 28
Question
A wheel of radius 0.5 m starts from rest and accelerates uniformly at 2 rad/s2
for 10 seconds. What is the angular velocity of the wheel after 10 seconds?
Solution
Step 1: We can use the equation for rotational kinematics to find the final
angular velocity of the wheel:
ωf=ωi+αt
where: - ωfis the final angular velocity, - ωiis the initial angular velocity
(which is 0 because the wheel starts from rest), - αis the angular acceleration
(2 rad/s2), and - tis the time (10 s).
Step 2: Substitute the known values into the equation:
ωf= 0 + 2 ×10 = 20 rad/s
Therefore, the angular velocity of the wheel after 10 seconds is 20 rad/s.
21
Question 29
Question
A disk of radius Rand mass Mis rotating about a fixed axis with an angular
velocity ω. A small piece of clay of mass mfalls onto the disk at a radial distance
rfrom the center. Assuming the clay sticks to the disk after collision, find the
new angular velocity of the system.
Solution
Step 1: Conservation of Angular Momentum
The initial angular momentum of the system is given by:
Li=Idiskω
where Idisk is the moment of inertia of the disk.
Let’s express the moment of inertia of the disk in terms of its mass and
radius:
Idisk =1
2MR2
Therefore, the initial angular momentum is:
Li=1
2MR2ω
Step 2: Angular Momentum of Clay-Disk System after Collision
After the clay falls onto the disk and sticks to it, the new moment of inertia
of the system is given by:
Ifinal =1
2(M+m)R2
The final angular momentum of the system is:
Lf=Ifinalωf
where ωfis the final angular velocity.
Step 3: Conservation of Angular Momentum
Since angular momentum is conserved, we have:
Li=Lf
Substitute in the expressions for Liand Lf:
1
2MR2ω=1
2(M+m)R2ωf
Step 4: Solving for Final Angular Velocity
Solving for ωf:
ωf=M
M+mω
Therefore, the new angular velocity of the system after the collision is M
M+m
times the initial angular velocity ω.
22
Question 30
Question
A disk with radius Rstarts from rest and rotates with a constant angular
acceleration α. Find an expression for the magnitude of the acceleration of a
point on its rim as a function of time t.
Solution
Step 1: Find the angular velocity ω(t) of the disk as a function of time t. The
angular acceleration αis the rate of change of angular velocity with respect to
time, so we have α=dω
dt . Integrating both sides with respect to time gives:
α=Zdω
dt dt →αdt =dω
Integrating over the initial conditions t= 0 and ω= 0 gives:
Zω
0
dω =Zt
0
αdt →ω=αt
Step 2: Find the linear velocity v(t) of a point on the rim of the disk as a
function of time t. The linear velocity of a point on the rim of the disk is given
by v=Rω. Substitute ω=αt into this expression to get:
v(t) = Rαt
Step 3: Find the acceleration a(t) of a point on the rim of the disk as a
function of time t. The acceleration of a point on the rim of the disk is the rate
of change of linear velocity, so we have:
a(t) = dv
dt
From the expression for v(t), we have:
a(t) = d
dt(Rαt) = Rα
Therefore, the magnitude of the acceleration of a point on the rim of the
disk as a function of time tis constant and equal to Rα.
Question 31
Question
A thin rod of length Land mass Mis rotating about one end with an angular
speed ω. The rod is then stopped in time t0. What is the angular acceleration
of the rod during this time?
23
Solution
Step 1: The moment of inertia of the rod about the end where it is rotating is
I=1
3ML2. Step 2: Since the rod is rotating and then comes to a stop, the
final angular speed ωfis 0. Step 3: Using the kinematic equation for rotational
motion, ωf=ω+αt, we can solve for the angular acceleration α. Step 4:
Substituting the given values into the equation, we have 0 = ω+αt0. Step 5:
Solving for the angular acceleration α, we get α=−ω
t0. Step 6: Therefore, the
angular acceleration of the rod during the stopping time t0is −ω
t0
.
Question 32
Question
A thin rod of length Land mass Mis free to rotate about one end. Initially,
the rod is at rest in a vertical position with the top end pointed downward. At
t= 0, the rod is released from rest.
What is the speed of the top end of the rod at the moment the rod makes
an angle θwith the vertical?
Solution
Step 1: Find the angular velocity ωof the rod when it makes an angle θwith
the vertical.
The energy of the system is conserved, so we can equate the initial potential
energy to the sum of final potential and kinetic energy. Initially, the rod is in a
purely vertical position with potential energy Ui= 0.
When the rod is at an angle θ, the potential energy Uf=−MgL(1 −cos(θ))
and the kinetic energy is due to the rotation: K=1
2Iω2, where I=1
3ML2is
the moment of inertia of a rod rotating about one end.
Setting Ui=K+Uf:
0 = 1
2Iω2−MgL(1 −cos(θ))
1
21
3ML2ω2=MgL(1 −cos(θ))
1
6Mω2L2=MgL(1 −cos(θ))
ω2= 6g(1 −cos(θ))
Step 2: Find the velocity vtop of the top end of the rod.
vtop =rω, where ris the length of the rod.
Substitute ω2= 6g(1 −cos(θ)) and r=L:
vtop =p6gL(1 −cos(θ))
24
Therefore, the speed of the top end of the rod when it makes an angle θwith
the vertical is p6gL(1 −cos(θ)) .
Question 33
Question
A disk starts from rest and accelerates uniformly for 3.00 s. During this time, it
rotates through 47.0 revolutions. Calculate the angular acceleration of the disk.
Solution
Step 1: Find the initial and final angular velocity of the disk. The angular
displacement of the disk can be calculated using the formula:
θ= Number of revolutions ×2π
Given that the disk rotates through 47.0 revolutions, we have:
θ= 47.0×2π= 94πrad
The initial angular velocity is zero, and the final angular velocity can be calcu-
lated using the formula:
ωf=ωi+αt
where ωi= 0 (initial angular velocity), t= 3.00 s (time), and αis the angular
acceleration.
Step 2: Calculate the final angular velocity. Substitute the values into the
formula to get:
ωf= 0 + α·3.00
ωf= 3.00α
Step 3: Calculate the angular velocity in terms of revolutions per minute
(RPM). To convert the angular velocity from rad/s to RPM, we use the conver-
sion factor 1 rad/s = 60
2πRPM.
ωf= 3.00α·60
2π=90
παRPM
Step 4: Find the angular acceleration. Since the disk starts from rest, the
initial angular velocity is zero and thus
α=ωf−ωi
t=ωf−0
3.00
Substitute the expression for ωf:
α=90
πα·1
3.00
25
1 = 90
π·1
3=30
π
α=30
πrad/s2
Therefore, the angular acceleration of the disk is 30
πrad/s2.
Question 34
Question
A wheel starts from rest and rotates with a constant angular acceleration of
2.0 rad/s2. If the wheel makes 10 full rotations, what is its final angular speed?
Solution
Step 1: First, we need to find the final angular speed of the wheel after making
10 full rotations. We can use the rotational kinematic equation:
ω2
f=ω2
i+ 2αθ
where - ωfis the final angular speed, - ωiis the initial angular speed (which
is 0 since the wheel starts from rest), - αis the angular acceleration, and - θis
the angle rotated (which is 10 ×2πsince the wheel makes 10 full rotations).
Step 2: Plugging in the given values:
ω2
f= 0 + 2 ×2.0×10 ×2π
ω2
f= 80π
ωf=√80π
ωf≈28.28 rad/s
Therefore, the final angular speed of the wheel after making 10 full rotations
is approximately 28.28 rad/s.
Question 35
Question
A thin rod of length 1.5 m rotates about an axis perpendicular to the rod and
passing through one end with an angular speed of 2 rad/s. A small object with
mass 0.2 kg is attached to the other end of the rod. What is the magnitude of
the angular momentum of the system about the axis of rotation?
26
Solution
Step 1: Calculate the moment of inertia of the rod. The moment of inertia of
the rod rotating about an axis perpendicular to the rod and passing through
one end is given by Irod =1
3mL2, where mis the mass of the rod and Lis the
length of the rod. Since the rod is thin, its mass can be approximated by λL,
where λis the linear mass density. Given that the length of the rod is 1.5 m
and mass density is λ=m
L, we have:
m=λL =1
1.5×1.5 = 1 kg
The moment of inertia of the rod is:
Irod =1
3×1×(1.5)2= 0.5 kg m2
Step 2: Calculate the moment of inertia of the object. The moment of inertia
of the object with mass mobject = 0.2 kg about an axis perpendicular to the rod
and passing through the object is given by Iobject =mobjectL2. The moment of
inertia of the object is:
Iobject = 0.2×(1.5)2= 0.45 kg m2
Step 3: Calculate the total moment of inertia of the system. The total
moment of inertia of the system is the sum of the moment of inertia of the rod
and the moment of inertia of the object:
Itotal =Irod +Iobject = 0.5+0.45 = 0.95 kg m2
Step 4: Calculate the angular momentum of the system. The angular mo-
mentum of the system is given by L=Itotalω, where ω= 2 rad/s is the angular
speed. Substitute the values:
L= 0.95 ×2=1.9 kg m2/s
Therefore, the magnitude of the angular momentum of the system about the
axis of rotation is 1.9 kg m2/s.
27
Question 2
Question
A wheel starts at 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 variables and the unknown we are looking for. Let
ω0= 0 rad/s be the initial angular velocity of the wheel, α= 2 rad/s2be the
angular acceleration of the wheel, and t= 3 s be the time at which we want to
find the angular velocity of the wheel.
Step 2: Use the equation for angular motion under constant acceleration.
The equation for angular motion under constant acceleration is:
ω=ω0+αt
where: ωis the final angular velocity, ω0is the initial angular velocity, αis the
angular acceleration, and tis the time.
Step 3: Substitute the values into the equation.
ω= 0 + 2 ×3
Step 4: Calculate the final angular velocity.
ω= 6 rad/s
Therefore, after 3 seconds, the angular velocity of the wheel is 6 rad/s.
Question 3
Question
A uniform circular disc of radius Rand mass Mrotates about its axis passing
through its center with an angular velocity ω. A small ball is placed on top of
the disc at a distance rfrom the center (as shown in the figure below).
r
R
Center
Ball
ω
If the disc suddenly comes to a stop, find the initial angular speed of the
ball relative to the the disc just before the disc stops rotating.
2
Solution
Step 1: Let vbe the velocity of the ball relative to the disc just before the disc
stops. This velocity can be found using the relative circular motion equation:
v=ωR
Step 2: The initial angular momentum of the ball about the center of the
disc is given by:
Li=m·(v·r)
Step 3: For the ball-disc system, angular momentum is conserved. The final
angular momentum of the ball is zero since it is no longer rotating. Therefore,
the change in angular momentum is given by:
∆L=Li
Step 4: The change in angular momentum is also given by:
∆L=Idisc ·ω
Step 5: Substituting the expressions for Li,ω, and Idisc =1
2MR2into the
above equation, we have:
m·(v·r) = 1
2MR2·ω
Step 6: Substituting v=ωR into the above equation:
m·ωR ·r=1
2MR2·ω
Step 7: Solving for ωgives:
ω=2m
M·r
R
Hence, the initial angular speed of the ball relative to the disc just before
the disc stops rotating is 2m
M·r
R.
Question 4
Question
A solid sphere with radius Rand moment of inertia Iabout its center of mass
starts from rest and rolls without slipping down a 30-degree incline. The sphere’s
mass is M. Find the final velocity of the sphere when it reaches the bottom of
the incline.
3
Solution
Step 1: Determine the acceleration of the sphere down the incline. The net
torque about the center of mass of the sphere is equal to the moment of inertia
times the angular acceleration. The torque is due to the gravitational force
component parallel to the incline, so we have:
τ=2
5MR2α=MgR sin(30◦)R
2
5R2α=gR sin(30◦)
α=5gsin(30◦)
2R
acm =Rα =5gsin(30◦)
2
Step 2: Determine the linear acceleration of the center of mass. Since the
sphere is rolling without slipping, the linear acceleration of the center of mass
is related to the angular acceleration:
acm =Rα
acm =5gsin(30◦)
2
Step 3: Find the final velocity of the sphere at the bottom of the incline.
Using energy conservation, the initial mechanical energy of the sphere is equal
to its final mechanical energy:
1
2Iω2=1
2Mv2+1
2Iv
R2
Substitute ω=v
Rand I=2
5MR2:
1
22
5MR2v
R2
=1
2Mv2+1
22
5MR2v
R2
Solve for vto get the final velocity of the sphere.
Question 5
Question
A disc of radius 0.5 m is rotating with an angular acceleration of −2 rad/s2. At
t= 0, the angular velocity of the disc is 3 rad/s. What is the angular velocity
of the disc after it has rotated through an angle of 30 radians?
4
Solution
Step 1: Find the angular velocity of the disc using the equation ω=ω0+αt
Given: Initial angular velocity, ω0= 3 rad/s
Angular acceleration, α=−2 rad/s2
Time, t=? (to be determined)
Using the given equation, we can solve for t:
ω=ω0+αt
3 = 3 + (−2)t
−2t= 0
t= 0 s
Step 2: Find the final angular velocity using the equation ω2=ω2
0+ 2αθ
Given: Initial angular velocity, ω0= 3 rad/s
Angular acceleration, α=−2 rad/s2
Angular displacement, θ= 30 rad
Using the given equation, we can solve for the final angular velocity ω:
ω2=ω2
0+ 2αθ
ω2= 32+ 2(−2)(30)
ω2= 9 −120
ω2=−111
ω=√−111
ω= 0 rad/s
Therefore, the angular velocity of the disc after it has rotated through an
angle of 30 radians is 0 rad/s.
Question 6
Question
A thin rod of mass mand length Lis rotating about one end with an angular
speed ω. Calculate the angular momentum of the rod about the rotation axis.
5
Solution
Step 1: The angular momentum of an object rotating about an axis is given by
the formula L=Iω, where Lis the angular momentum, Iis the moment of
inertia, and ωis the angular speed.
Step 2: To find the moment of inertia of the rod rotating about one end, we
need to use the parallel axis theorem. The moment of inertia of a rod rotating
about its center is 1
12 mL2.
Step 3: The distance between the axis of rotation and the center of mass
of the rod is L
2. Therefore, the moment of inertia about one end can be found
using the parallel axis theorem: I=1
12 mL2+mL
22.
Step 4: Simplifying the equation gives I=1
3mL2.
Step 5: Now, we can calculate the angular momentum by substituting the
moment of inertia into the formula: L=1
3mL2ω.
Step 6: Therefore, the angular momentum of the rod about the rotation axis
is 1
3mL2ω.
Question 7
Question
A solid sphere of radius Rstarts from rest and rolls without slipping down a
30◦incline. What is the acceleration of the center of mass of the sphere along
the incline?
Solution
Step 1: The acceleration of the center of mass of the sphere along the incline
can be found using the rotational kinematics equations. The equation relating
linear acceleration aand angular acceleration αfor a rolling object is a=Rα,
where Ris the radius of the sphere.
Step 2: The acceleration of the center of mass can also be expressed using
the linear acceleration aand angular velocity ωas a=Rα =Rdω
dt .
Step 3: The kinematic constraint for rolling without slipping is a=Rα =
Rdω
dt . For the sphere rolling without slipping down an incline, the linear accel-
eration acan be related to the angular velocity ωas a=Rdω
dt =Rα.
Step 4: Considering that the sphere is rolling without slipping, its linear
acceleration adown the incline can be related to the angular acceleration α
through the equation a=Rα =Rdω
dt =Rd
dt v
R, where vis the linear velocity
of the sphere.
Step 5: In the rotational kinematics equation a=Rα =Rd
dt v
R, we can
substitute v=Rω to find a=Rd
dt Rω
R.
Step 6: Simplifying the expression a=Rd
dt Rω
R, we get a=Rdω
dt , which
means that the linear acceleration of the center of mass of the sphere along the
incline is equal to Rdω
dt .
6
Step 7: Therefore, the acceleration of the center of mass of the sphere along
the incline is Rdω
dt .
Question 8
Question
A disk of radius 0.3 m starts from rest and accelerates with a constant angular
acceleration of 0.5 rad/s2. Find the angular velocity of the disk when it has
completed 3 full revolutions.
Solution
Step 1: Calculate the total angular displacement when the disk completes 3 full
revolutions.
Total angular displacement = 2π×number of revolutions
= 2π×3
= 6πrad
Step 2: Use the kinematic equation for rotational motion to find the final
angular velocity.
ω2=ω2
0+ 2αθ
Where: ω= final angular velocity, ω0= initial angular velocity (which is 0
rad/s), α= angular acceleration, θ= angular displacement.
Step 3: Substitute the known values into the kinematic equation.
ω2= 0 + 2(0.5)(6π)
ω2= 6π
Step 4: Solve for the final angular velocity.
ω=√6π≈4.91 rad/s
Therefore, the angular velocity of the disk when it completes 3 full revolu-
tions is approximately 4.91 rad/s.
Question 9
Question
A disk of radius 0.2 m starts from rest and rotates with a constant angular
acceleration of 2.0 rad/s2.
7
Part (a)
Determine the angular velocity of the disk after 3.0 seconds.
Part (b)
Find the number of revolutions the disk has made after 3.0 seconds.
Solution
Part (a)
Step 1: The angular velocity of the disk is given by the equation ω=ω0+αt,
where ωis the final angular velocity, ω0is the initial angular velocity, αis the
angular acceleration, and tis the time.
Step 2: Substituting the known values ω0= 0, α= 2.0 rad/s2, and t= 3.0 s
into the equation, we have ω= 0 + 2.0×3.0=6.0 rad/s.
Therefore, the angular velocity of the disk after 3.0 seconds is 6.0 rad/s .
Part (b)
Step 1: The number of revolutions made by the disk can be found using the
formula θ=θ0+ω0t+1
2αt2, where θis the total angular displacement, θ0is
the initial angular displacement (usually 0), ω0is the initial angular velocity, α
is the angular acceleration, and tis the time.
Step 2: Since the disk starts from rest, ω0= 0, and calculating θwe have
θ=0+0+1
2×2.0×(3.0)2= 9.0 rad.
Step 3: Converting the total angular displacement to the number of revolu-
tions, we recall that 2πrad = 1 revolution. Therefore, the number of revolutions
is 9.0 rad
2π=9.0
2π≈1.43 revolutions.
Hence, the number of revolutions the disk has made after 3.0 seconds is
1.43 revolutions .
Question 10
Question
A disk of radius Ris initially at rest. A constant force is applied tangentially
to the disk at a distance rfrom the center. The force is applied for a time T,
after which it is removed. Determine the angular velocity of the disk after the
force is removed.
Solution
Step 1: Calculate the torque exerted on the disk by the applied force. The
torque τis given by τ=rF sin θ, where Fis the force applied and θis the angle
8
between the force and the lever arm. Since the force is tangential to the disk,
θ= 90◦and sin θ= 1. Thus, τ=rF .
Step 2: Use the rotational analog of Newton’s second law: τ=Iα, where
Iis the moment of inertia of the disk and αis the angular acceleration. For a
disk, I=1
2mR2, where mis the mass of the disk.
Step 3: Substituting the expressions for torque and moment of inertia into
the rotational Newton’s second law equation gives rF =1
2mR2α. Rearranging
for angular acceleration gives α=2rF
mR2.
Step 4: The final angular velocity ωfof the disk can be calculated using the
kinematic equation for rotational motion, ωf=ωi+αt, where ωiis the initial
angular velocity (which is zero in this case). Thus, ωf=αT .
Step 5: Substituting the expression for angular acceleration into the equation
for final angular velocity gives ωf=2rF
mR2T. So, the angular velocity of the disk
after the force is removed is 2rF
mR2T.
Question 11
Question
A wheel with a radius of 0.5 m starts from rest and accelerates with a constant
angular acceleration of 2 rad/s2. How long will it take for the wheel to reach
an angular speed of 10 rad/s?
Solution
Step 1: Let’s first identify the given values: Initial angular speed, ω0= 0 rad/s
Final angular speed, ω= 10 rad/s
Angular acceleration, α= 2 rad/s2
Radius of the wheel, r= 0.5 m
Step 2: Next, let’s use the rotational kinematic equation relating initial
angular speed, final angular speed, angular acceleration, and time:
ω=ω0+αt
Substitute in the given values:
10 = 0 + 2t
Step 3: Solve for the time t:
2t= 10 ⇒t=10
2= 5 s
Step 4: Therefore, it will take 5 seconds for the wheel to reach an angular
speed of 10 rad/s.
9
Question 12
Question
A solid sphere of radius Rand mass Mis released from the top of a rough
inclined plane that makes an angle θwith the horizontal. The sphere rolls
without slipping down the incline. Calculate the acceleration of the center of
mass of the sphere.
Solution
Step 1: The acceleration of the center of mass of the sphere can be found
by considering the forces acting on the sphere. The forces involved are the
gravitational force mg acting vertically downward, the normal force Nacting
perpendicular to the incline, and the friction force facting parallel to the incline
in the direction opposite to the motion.
Step 2: The gravitational force can be resolved into components parallel and
perpendicular to the incline. The component of the gravitational force parallel
to the incline is mg sin(θ) and the component perpendicular to the incline is
mg cos(θ).
Step 3: The net force causing the acceleration down the incline is given by
f=macm, where acm is the acceleration of the center of mass.
Step 4: The friction force fcan be calculated using the relationship f=µN ,
where µis the coefficient of friction between the sphere and the incline. The
normal force Ncan be calculated using the relationship N=mg cos(θ).
Step 5: Substituting the expression for the friction force f, we have µmg cos(θ) =
macm.
Step 6: Now, we can solve for the acceleration of the center of mass acm
and obtain acm =µg cos(θ). Thus, the acceleration of the center of mass of the
sphere rolling down the incline is µg cos(θ).
Question 13
Question
A thin rod of length Land mass Mis rotating about an axis passing through
one of its ends with an angular velocity ω. If the rod is suddenly brought to rest
without any external torque, what is the tension in the rod just after it stops
rotating?
Solution
Let’s denote the position of the center of mass of the rod by x. The moment
of inertia of a rod rotating about an axis passing through one of its ends is
I=1
3ML2. The kinetic energy of the rod while rotating is given by KE =1
2Iω2,
and the potential energy of the rod due to its center of mass is P E =−Mgx.
10
When the rod stops rotating, all of its initial kinetic energy is converted into
potential energy. This means that KEinitial =KEfinal +P Efinal, which gives us
KEinitial =KEfinal +P Efinal
1
2Iω2=P Efinal
Step 1: Substitute the expressions for the moment of inertia Iand potential
energy P Efinal.
1
21
3ML2ω2=−Mgx
Step 2: Solve for x.1
6ML2ω2=Mgx
x=1
6Lω2
Step 3: The tension in the rod just after stopping is equal to the force it
exerts on its center of mass. This force is a combination of the force due to
gravity and the tension force.
T=Mg +Mω2L
6
Therefore, the tension in the rod just after it stops rotating is T=Mg +
Mω2L
6.
Question 14
Question
A solid sphere of radius Rand mass Mrolls without slipping down an inclined
plane that makes an angle θwith the horizontal. If the sphere starts from rest at
the top of the incline, determine its speed when it reaches the bottom. Assume
the incline is frictionless.
Solution
Step 1: Find the acceleration of the sphere down the incline. The net force
along the incline is due to the component of the gravitational force along the
incline:
Fnet =ma =Mg sin θ
where ais the acceleration, Mis the mass of the sphere, gis the acceleration
due to gravity, and θis the angle of the incline.
11
Step 2: Find the angular acceleration of the sphere. Since the sphere rolls
without slipping, the acceleration of the center of mass ais related to the angular
acceleration αby:
a=Rα
where αis the angular acceleration and Ris the radius of the sphere.
Step 3: Find the speed of the sphere at the bottom of the incline. Since the
sphere starts from rest, we can use the following kinematic equation:
v2=u2+ 2as
where vis the final velocity, uis the initial velocity (which is 0 in this case), a
is the acceleration, and sis the displacement down the incline.
Step 4: Substitute the values into the equations and solve for the final ve-
locity. From step 1: a=gsin θFrom step 2: α=a
R=gsin θ
RFrom step 3:
v2= 0 + 2(gsin θ)(h)
where his the height of the incline.
Step 5: Now, we can substitute the expression for hto get the final velocity
in terms of R,M,g, and θ.
h=R(1 −cos θ)
Thus, the final velocity is:
v=p2gR sin θ(1 −cos θ)
Question 15
Question
A thin rod of length Lis pivoted about one end and is initially at rest. A small
bug is placed on the rod a distance xfrom the pivot point. At time t= 0, the
bug starts crawling along the rod with a constant velocity vrelative to the rod.
What is the angular velocity of the rod as a function of time?
Solution
Step 1: Let rbe the distance from the pivot to the bug at time t. Then, the
bug’s distance from the pivot at time tis given by r=x+vt.
Step 2: The bug’s velocity relative to the pivot point is given by vbug =v.
Step 3: The bug’s velocity can also be expressed in terms of the radial
velocity of the bug along the rod and the angular velocity of the rod, using
vbug =rω, where ωis the angular velocity of the rod.
Step 4: Substituting r=x+vt into vbug =rω, we get v= (x+vt)ω.
Step 5: Simplifying the equation, we find ω=v
x+vt .
Thus, the angular velocity of the rod as a function of time is ω(t) = v
x+vt .
12
Question 16
Question
A solid sphere of radius Rand mass Mis rotating about an axis passing through
its center with an angular speed ω. If the radius of gyration of the sphere is k,
determine the angular momentum of the sphere.
Solution
Step 1: The angular momentum Lof an object rotating about an axis is given
by the equation L=Iω, where Iis the moment of inertia and ωis the angular
speed.
Step 2: The moment of inertia Iof a solid sphere rotating about an axis
passing through its center is given by I=2
5MR2.
Step 3: Given that the radius of gyration kis defined as k=qI
M, we can
express the moment of inertia Iin terms of kas I=Mk2.
Step 4: Substituting the expression for Iinto the equation for angular mo-
mentum, we have L=Mk2ω.
Step 5: The angular momentum of the sphere rotating about the given axis
is given by L=Mk2ω.
Question 17
Question
A flywheel with a radius of 0.5 m starts from rest and accelerates uniformly to
an angular velocity of 10 rad/s in 5 seconds. What is the angular acceleration
of the flywheel?
Solution
Step 1: First, we need to find the angular acceleration of the flywheel using the
formula for angular acceleration, which is given by
α=ωf−ω0
t,
where - αis the angular acceleration, - ωfis the final angular velocity, and - ω0
is the initial angular velocity.
Step 2: Substituting the given values into the formula, we have
α=10 rad/s −0
5 s = 2 rad/s2.
Step 3: Therefore, the angular acceleration of the flywheel is 2 rad/s2.
13
Question 18
Question
A uniform disc of radius Rand mass Mis initially at rest. A constant force F
is applied tangentially to the edge of the disc. What is the angular acceleration
of the disc?
Solution
1. We can find the torque τexerted on the disc by the force Fat the edge of
the disc using the equation τ=F r, where r=Ris the radius of the disc. Thus,
τ=F R.
2. The moment of inertia Iof a uniform disc rotating about an axis per-
pendicular to the plane of the disc and passing through its center is given by
I=1
2MR2.
3. Using Newton’s second law for rotational motion, τ=Iα, where αis the
angular acceleration of the disc.
4. Substituting the expressions for torque τ, moment of inertia I, and solving
for angular acceleration α, we get
F R =1
2MR2α.
5. Solving for α, we find that the angular acceleration of the disc is
α=2F
M.
Therefore, the angular acceleration of the disc is 2F
M.
Question 19
Question
A disk with radius Ris spinning at an angular velocity ω0. A bug lands on
the edge of the disk and starts crawling towards the center at a constant speed
of vbug. At the same time, the disk is slowing down with a constant angular
acceleration α. Given that the bug reaches the center of the disk after making
exactly one full revolution, determine the bug’s position as a function of time.
Solution
Step 1: Find the angular position of the bug as a function of time.
Let θbe the angle the bug has traveled in radians, and tbe the time since
the bug landed on the disk. The bug’s angular velocity can be expressed as:
ωbug(t) = vbug
r
14
where ris the distance from the bug to the center at time t. Since the bug is
moving towards the center, rcan be expressed as R−vbugt. Thus:
ωbug(t) = vbug
R−vbugt
Integrating ωbug(t) with respect to time gives the angular position θas a
function of time:
θ(t) = Zωbug(t)dt =Zvbug
R−vbugtdt
Step 2: Use the given information that the bug makes exactly one full revo-
lution to solve for an expression of θ(t).
Since the bug travels 2πradians when it reaches the center:
Z2π
0
dθ =ZT
0
vbug
R−vbugtdt
where Tis the time it takes for the bug to reach the center.
Solving the integration on the right side and using the condition for one full
revolution, we can then determine the position of the bug as a function of time.
Question 20
Question
A solid cylinder of mass Mand radius Ris initially at rest. A constant force F
is applied tangentially to the edge of the cylinder, causing it to start rotating
about its central axis. The cylinder acquires an angular velocity ωafter a certain
time. What is the angular acceleration of the cylinder in terms of F,M,R, and
ω?
Solution
Step 1: The torque τacting on the cylinder can be calculated from the equation
τ=Iα, where Iis the moment of inertia of the cylinder about its central axis
and αis the angular acceleration. The torque can also be written as τ=F R,
where Fis the applied force.
Step 2: The moment of inertia of a solid cylinder rotating about its central
axis is I=1
2MR2.
Step 3: Setting the two expressions for torque equal to each other, we have
F R =1
2MR2α.
Step 4: Solving for α, we get α=2F
MR .
Step 5: Therefore, the angular acceleration of the cylinder in terms of F,
M,R, and ωis α=2F
MR .
15
Question 21
Question
A solid sphere of radius Rstarts from rest and rolls without slipping down an
inclined plane that makes an angle θwith the horizontal. If the sphere’s moment
of inertia about its center is I=2
5MR2, where Mis the mass of the sphere,
determine its linear acceleration down the incline in terms of gand θ.
Solution
Step 1: The acceleration of the center of mass of the sphere down the incline
can be determined by considering the forces acting on it along the incline. The
forces include the gravitational force mg sin(θ) down the incline, the normal
force Nperpendicular to the incline, and the frictional force fopposing the
motion. The net force down the incline is mg sin(θ)−f. The frictional force
can be determined using the condition for no slipping, f=µN =µmg cos(θ),
where µis the coefficient of static friction.
Step 2: The torque about the center of the sphere due to the frictional force
is fR, causing a clockwise angular acceleration. The torque due to the gravita-
tional force about the center is mgR sin(θ), causing a counterclockwise angular
acceleration. Since the sphere rolls without slipping, the linear acceleration a
of the center of mass is related to the angular acceleration αby a=Rα.
Step 3: Equating the torques gives us fR =Iα. Substituting fand Iin
terms of Mand Rgives µmg cos(θ)R=2
5MR2α. Substituting α=a
Ryields
µg cos(θ) = 2
5Ma.
Step 4: Since the linear acceleration down the incline is given as a, the final
expression for the acceleration in terms of gand θis a=5
2µg cos(θ) .
Question 22
Question
A wheel of radius 0.5 m starts rotating from rest with a constant angular accel-
eration of 2 rad/s2. Find the time it takes for the wheel to make 10 complete
revolutions.
Solution
Step 1: Determine the angular velocity of the wheel after a certain time. The
angular acceleration of the wheel is α= 2 rad/s2. We can use the following
kinematic equation to find the angular velocity of the wheel after a certain
time:
ω=ω0+αt
16
where ωis the final angular velocity, ω0is the initial angular velocity (which is
0 since the wheel starts from rest), αis the angular acceleration, and tis the
time.
Substitute the given values into the equation:
ω= 0 + 2t
ω= 2t
Step 2: Determine the time it takes for the wheel to make 10 complete
revolutions. The angular displacement for one complete revolution is 2πradians.
Therefore, the angular displacement for 10 complete revolutions is 10×2π= 20π
radians.
We can use the following kinematic equation to find the time it takes for the
wheel to make 10 complete revolutions:
θ=θ0+ω0t+1
2αt2
where θis the angular displacement, θ0is the initial angular displacement, ω0
is the initial angular velocity, αis the angular acceleration, and tis the time.
Substitute the given values into the equation:
20π=0+0+1
2×2t2
20π=t2
t=√20π≈7.98 s
Therefore, it takes approximately 7.98 seconds for the wheel to make 10
complete revolutions.
Question 23
Question
A thin hoop of radius Rand mass Mrolls without slipping down a ramp inclined
at an angle θ. If the hoop’s initial velocity at the top of the ramp is v0, determine
its angular velocity ωwhen it reaches the bottom of the ramp.
Solution
Step 1: We will start by finding the speed of the hoop at the bottom of the
ramp using conservation of energy. At the top of the ramp, the hoop has
gravitational potential energy which is all converted to kinetic energy at the
bottom (rotational and translational). Thus, we have
mgR sin θ=1
2Mv2
0+1
2Iω2
17
where Iis the moment of inertia of the hoop and is equal to MR2and ω=v
R.
Step 2: Simplifying the equation, we have
mgR sin θ=1
2Mv2
0+1
2MR2v0
R22
mgR sin θ=1
2Mv2
0+1
2Mv2
0
mgR sin θ=Mv2
0
Step 3: Solving for the speed vat the bottom of the ramp,
v=pgR sin θ
Step 4: Finally, we can find the angular velocity at the bottom of the ramp
by
ω=v
R=√gR sin θ
R=rgsin θ
R
Question 24
Question
A thin, uniform rod of length Land mass Mis rotated about an axis perpen-
dicular to the rod and passing through one end of the rod. What is the moment
of inertia of this rotating rod about this axis?
Solution
Step 1: Let’s consider the moment of inertia of a thin rod of length Land mass
Mabout an axis passing through its end and perpendicular to the rod. This
can be expressed using the formula for the moment of inertia of a rod rotating
about an axis perpendicular to the rod and passing through one end:
I=1
3ML2
Step 2: Hence, the moment of inertia of a thin, uniform rod of length Land
mass Mabout an axis perpendicular to the rod and passing through one end is
given by the expression I=1
3ML2.
Question 25
Question
A solid sphere of radius Rand mass Mrolls without slipping down a 30-degree
incline. Starting from rest at the top, what is its speed when it reaches the
bottom of the incline? Assume the sphere rolls without slipping and the moment
of inertia of a solid sphere about its center is 2
5MR2.
18
Solution
Step 1: Calculate the acceleration of the sphere down the incline due to gravity.
The acceleration of the sphere down the incline can be calculated using the
component of gravity that acts parallel to the incline.
a=gsin θ
where θ= 30◦is the angle of the incline.
Step 2: Calculate the angular acceleration of the sphere. Since the sphere is
rolling without slipping, the linear acceleration of the center of mass ais related
to the angular acceleration αthrough:
a=Rα
and we know:
a=gsin θ
Therefore, we have:
Rα =gsin θ
α=gsin θ
R
Step 3: Calculate the final angular velocity of the sphere. The final angular
velocity of the sphere can be calculated using the rotational kinematic equation:
v2=u2+ 2as
where: - uis the initial angular velocity (which is zero in this case) - sis the
distance traveled down the incline - vis the final angular velocity
Step 4: Calculate the final linear velocity of the sphere. The final linear
velocity of the sphere can be calculated using the relation between linear and
angular velocity of a rolling object:
v=Rω
where ωis the final angular velocity.
Step 5: Substitute the given values and solve for v. The final linear velocity
vof the sphere when it reaches the bottom of the incline is:
v=Rp2gsin θ
Therefore, the speed of the sphere when it reaches the bottom of the incline
is v=R√2gsin 30◦.
Question 26
Question
A disk of radius Rstarts rotating with an angular acceleration αat time t= 0. It
reaches an angular velocity ωafter a time T. What is the angular displacement
θof a point on the rim of the disk at time t=T?
19
Solution
Step 1: Find the angular velocity ωof the disk at time Tusing the formula for
angular kinematics:
ω=αT
Step 2: The angular displacement θof a point on the rim of the disk at time
t=Tcan be calculated using the equation for angular displacement in terms
of initial angular velocity, initial angular acceleration, and time:
θ=1
2αT 2
Step 3: Substituting the expression for αfrom Step 1 into the equation for
θgives:
θ=1
2ω
TT2
θ=1
2ωT
Therefore, the angular displacement θof a point on the rim of the disk at
time t=Tis 1
2ωT .
Question 27
Question
A disk of radius 0.3 m starts from rest and accelerates with a constant angular
acceleration of 2.5 rad/s2.
1. Find the angular velocity of the disk after 4 seconds.
2. Determine the angular displacement of the disk during this time interval.
Solution
1. To find the angular velocity of the disk after 4 seconds, we can use the
equation for rotational kinematics:
ωf=ωi+αt
where
ωfis the final angular velocity,
ωiis the initial angular velocity (which is 0 since the disk starts from rest),
αis the angular acceleration, and
tis the time.
20
Step 1: Calculate the final angular velocity:
ωf= 0 + 2.5×4 = 10 rad/s
Therefore, the angular velocity of the disk after 4 seconds is 10 rad/s.
2. To determine the angular displacement of the disk during this time inter-
val, we can use the equation:
θ=ωit+1
2αt2
where
θis the angular displacement,
ωiis the initial angular velocity,
αis the angular acceleration, and
tis the time.
Since the initial angular velocity is 0, the equation simplifies to:
θ=1
2αt2
Step 2: Calculate the angular displacement:
θ=1
2×2.5×42= 20 rad
Therefore, the angular displacement of the disk during this time interval is
20 radians.
Question 28
Question
A wheel of radius 0.5 m starts from rest and accelerates uniformly at 2 rad/s2
for 10 seconds. What is the angular velocity of the wheel after 10 seconds?
Solution
Step 1: We can use the equation for rotational kinematics to find the final
angular velocity of the wheel:
ωf=ωi+αt
where: - ωfis the final angular velocity, - ωiis the initial angular velocity
(which is 0 because the wheel starts from rest), - αis the angular acceleration
(2 rad/s2), and - tis the time (10 s).
Step 2: Substitute the known values into the equation:
ωf= 0 + 2 ×10 = 20 rad/s
Therefore, the angular velocity of the wheel after 10 seconds is 20 rad/s.
21
Question 29
Question
A disk of radius Rand mass Mis rotating about a fixed axis with an angular
velocity ω. A small piece of clay of mass mfalls onto the disk at a radial distance
rfrom the center. Assuming the clay sticks to the disk after collision, find the
new angular velocity of the system.
Solution
Step 1: Conservation of Angular Momentum
The initial angular momentum of the system is given by:
Li=Idiskω
where Idisk is the moment of inertia of the disk.
Let’s express the moment of inertia of the disk in terms of its mass and
radius:
Idisk =1
2MR2
Therefore, the initial angular momentum is:
Li=1
2MR2ω
Step 2: Angular Momentum of Clay-Disk System after Collision
After the clay falls onto the disk and sticks to it, the new moment of inertia
of the system is given by:
Ifinal =1
2(M+m)R2
The final angular momentum of the system is:
Lf=Ifinalωf
where ωfis the final angular velocity.
Step 3: Conservation of Angular Momentum
Since angular momentum is conserved, we have:
Li=Lf
Substitute in the expressions for Liand Lf:
1
2MR2ω=1
2(M+m)R2ωf
Step 4: Solving for Final Angular Velocity
Solving for ωf:
ωf=M
M+mω
Therefore, the new angular velocity of the system after the collision is M
M+m
times the initial angular velocity ω.
22
Question 30
Question
A disk with radius Rstarts from rest and rotates with a constant angular
acceleration α. Find an expression for the magnitude of the acceleration of a
point on its rim as a function of time t.
Solution
Step 1: Find the angular velocity ω(t) of the disk as a function of time t. The
angular acceleration αis the rate of change of angular velocity with respect to
time, so we have α=dω
dt . Integrating both sides with respect to time gives:
α=Zdω
dt dt →αdt =dω
Integrating over the initial conditions t= 0 and ω= 0 gives:
Zω
0
dω =Zt
0
αdt →ω=αt
Step 2: Find the linear velocity v(t) of a point on the rim of the disk as a
function of time t. The linear velocity of a point on the rim of the disk is given
by v=Rω. Substitute ω=αt into this expression to get:
v(t) = Rαt
Step 3: Find the acceleration a(t) of a point on the rim of the disk as a
function of time t. The acceleration of a point on the rim of the disk is the rate
of change of linear velocity, so we have:
a(t) = dv
dt
From the expression for v(t), we have:
a(t) = d
dt(Rαt) = Rα
Therefore, the magnitude of the acceleration of a point on the rim of the
disk as a function of time tis constant and equal to Rα.
Question 31
Question
A thin rod of length Land mass Mis rotating about one end with an angular
speed ω. The rod is then stopped in time t0. What is the angular acceleration
of the rod during this time?
23
Solution
Step 1: The moment of inertia of the rod about the end where it is rotating is
I=1
3ML2. Step 2: Since the rod is rotating and then comes to a stop, the
final angular speed ωfis 0. Step 3: Using the kinematic equation for rotational
motion, ωf=ω+αt, we can solve for the angular acceleration α. Step 4:
Substituting the given values into the equation, we have 0 = ω+αt0. Step 5:
Solving for the angular acceleration α, we get α=−ω
t0. Step 6: Therefore, the
angular acceleration of the rod during the stopping time t0is −ω
t0
.
Question 32
Question
A thin rod of length Land mass Mis free to rotate about one end. Initially,
the rod is at rest in a vertical position with the top end pointed downward. At
t= 0, the rod is released from rest.
What is the speed of the top end of the rod at the moment the rod makes
an angle θwith the vertical?
Solution
Step 1: Find the angular velocity ωof the rod when it makes an angle θwith
the vertical.
The energy of the system is conserved, so we can equate the initial potential
energy to the sum of final potential and kinetic energy. Initially, the rod is in a
purely vertical position with potential energy Ui= 0.
When the rod is at an angle θ, the potential energy Uf=−MgL(1 −cos(θ))
and the kinetic energy is due to the rotation: K=1
2Iω2, where I=1
3ML2is
the moment of inertia of a rod rotating about one end.
Setting Ui=K+Uf:
0 = 1
2Iω2−MgL(1 −cos(θ))
1
21
3ML2ω2=MgL(1 −cos(θ))
1
6Mω2L2=MgL(1 −cos(θ))
ω2= 6g(1 −cos(θ))
Step 2: Find the velocity vtop of the top end of the rod.
vtop =rω, where ris the length of the rod.
Substitute ω2= 6g(1 −cos(θ)) and r=L:
vtop =p6gL(1 −cos(θ))
24
Therefore, the speed of the top end of the rod when it makes an angle θwith
the vertical is p6gL(1 −cos(θ)) .
Question 33
Question
A disk starts from rest and accelerates uniformly for 3.00 s. During this time, it
rotates through 47.0 revolutions. Calculate the angular acceleration of the disk.
Solution
Step 1: Find the initial and final angular velocity of the disk. The angular
displacement of the disk can be calculated using the formula:
θ= Number of revolutions ×2π
Given that the disk rotates through 47.0 revolutions, we have:
θ= 47.0×2π= 94πrad
The initial angular velocity is zero, and the final angular velocity can be calcu-
lated using the formula:
ωf=ωi+αt
where ωi= 0 (initial angular velocity), t= 3.00 s (time), and αis the angular
acceleration.
Step 2: Calculate the final angular velocity. Substitute the values into the
formula to get:
ωf= 0 + α·3.00
ωf= 3.00α
Step 3: Calculate the angular velocity in terms of revolutions per minute
(RPM). To convert the angular velocity from rad/s to RPM, we use the conver-
sion factor 1 rad/s = 60
2πRPM.
ωf= 3.00α·60
2π=90
παRPM
Step 4: Find the angular acceleration. Since the disk starts from rest, the
initial angular velocity is zero and thus
α=ωf−ωi
t=ωf−0
3.00
Substitute the expression for ωf:
α=90
πα·1
3.00
25
1 = 90
π·1
3=30
π
α=30
πrad/s2
Therefore, the angular acceleration of the disk is 30
πrad/s2.
Question 34
Question
A wheel starts from rest and rotates with a constant angular acceleration of
2.0 rad/s2. If the wheel makes 10 full rotations, what is its final angular speed?
Solution
Step 1: First, we need to find the final angular speed of the wheel after making
10 full rotations. We can use the rotational kinematic equation:
ω2
f=ω2
i+ 2αθ
where - ωfis the final angular speed, - ωiis the initial angular speed (which
is 0 since the wheel starts from rest), - αis the angular acceleration, and - θis
the angle rotated (which is 10 ×2πsince the wheel makes 10 full rotations).
Step 2: Plugging in the given values:
ω2
f= 0 + 2 ×2.0×10 ×2π
ω2
f= 80π
ωf=√80π
ωf≈28.28 rad/s
Therefore, the final angular speed of the wheel after making 10 full rotations
is approximately 28.28 rad/s.
Question 35
Question
A thin rod of length 1.5 m rotates about an axis perpendicular to the rod and
passing through one end with an angular speed of 2 rad/s. A small object with
mass 0.2 kg is attached to the other end of the rod. What is the magnitude of
the angular momentum of the system about the axis of rotation?
26
Solution
Step 1: Calculate the moment of inertia of the rod. The moment of inertia of
the rod rotating about an axis perpendicular to the rod and passing through
one end is given by Irod =1
3mL2, where mis the mass of the rod and Lis the
length of the rod. Since the rod is thin, its mass can be approximated by λL,
where λis the linear mass density. Given that the length of the rod is 1.5 m
and mass density is λ=m
L, we have:
m=λL =1
1.5×1.5 = 1 kg
The moment of inertia of the rod is:
Irod =1
3×1×(1.5)2= 0.5 kg m2
Step 2: Calculate the moment of inertia of the object. The moment of inertia
of the object with mass mobject = 0.2 kg about an axis perpendicular to the rod
and passing through the object is given by Iobject =mobjectL2. The moment of
inertia of the object is:
Iobject = 0.2×(1.5)2= 0.45 kg m2
Step 3: Calculate the total moment of inertia of the system. The total
moment of inertia of the system is the sum of the moment of inertia of the rod
and the moment of inertia of the object:
Itotal =Irod +Iobject = 0.5+0.45 = 0.95 kg m2
Step 4: Calculate the angular momentum of the system. The angular mo-
mentum of the system is given by L=Itotalω, where ω= 2 rad/s is the angular
speed. Substitute the values:
L= 0.95 ×2=1.9 kg m2/s
Therefore, the magnitude of the angular momentum of the system about the
axis of rotation is 1.9 kg m2/s.
27
