PHYS 201 - GENERAL PHYSICS I -
Gyroscopic Motion and Stability
Question Bank - Set 10
Liberty University
Question 1
Question
A thin, uniform rod of length Land mass Mis pivoted at one end. The rod is
set into motion with an angular velocity ω0about a vertical axis through the
pivot. Determine the angular velocity of the rod as it makes an angle θwith
the vertical.
Solution
Step 1: The rod has both translational kinetic energy and rotational kinetic
energy. The total kinetic energy can be expressed as the sum of the translational
and rotational kinetic energies:
K=1
2Iω2+1
2Mv2
CM
where Iis the moment of inertia of the rod about the pivot point, ωis the
angular velocity of the rod, Mis the mass of the rod, and vCM is the speed of
the center of mass of the rod.
Step 2: The moment of inertia of the rod about the pivot point is I=1
3ML2.
The speed of the center of mass of the rod can be related to the angular velocity
by vCM =Lω.
Step 3: Substituting the expressions for Iand vCM into the total kinetic
energy equation, we have:
K=1
2(1
3ML2)ω2+1
2M(Lω)2
Step 4: At an angle θwith the vertical, the potential energy of the rod due
to gravity is U=−M gL(1 −cos θ).
Step 5: Since there are no external torques acting on the system, the total
mechanical energy is conserved:
E=K+U=constant
Step 6: Therefore, we can write the conservation of energy equation as:
1
2(1
3ML2)ω2+1
2M(Lω)2−MgL(1 −cos θ) = constant
Step 7: Initially, when the rod is at its maximum angle θ, its angular velocity
is ω0. At this point, the total mechanical energy is:
E0=1
2(1
3ML2)ω2
0+1
2M(Lω0)2−MgL(1 −cos θ)
Question 2
Question
A wheel of mass 4 kg and radius 0.3 m rotating at a constant speed of 120 rad/s
about a fixed axis. Calculate the angular momentum of the wheel.
Solution
Step 1: Identify the given values. The mass of the wheel, m= 4 kg. The radius
of the wheel, r= 0.3m. The angular speed of the wheel, ω= 120 rad/s.
Step 2: Find the moment of inertia of the wheel. The moment of inertia of
a solid disk rotating about its central axis is given by the formula:
I=1
2mr2
Substitute the given values:
I=1
2×4×(0.3)2
I= 0.54 kg m2
Step 3: Calculate the angular momentum. The angular momentum of the
wheel is given by the formula:
L=Iω
Substitute the values of Iand ω:
L= 0.54 ×120
L= 64.8kg m2/s
Therefore, the angular momentum of the wheel is 64.8kg m2/s.
2
Question 3
Question
A thin circular hoop with radius Rand mass mis rotating about its vertical di-
ameter with an angular velocity ω. The hoop is released from rest in the vertical
position. Find the angular velocity of the hoop when it becomes horizontal.
Solution
Step 1: Identify the initial and final energies of the hoop.
The initial energy of the system is purely gravitational potential energy,
since the hoop is initially at rest:
Ui=mgh
where his the height of the center of mass of the hoop above its final lowest
point.
The final energy of the system is purely kinetic energy, since all potential
energy will have been converted to kinetic energy when the hoop becomes hor-
izontal:
Kf=1
2Iω2
f
where I=mR2is the moment of inertia of the hoop about its center of mass.
Step 2: Apply the conservation of mechanical energy.
Since there are no non-conservative forces doing work on the system and
there is no change in mechanical energy, we have:
Ui=Kf
mgh =1
2mR2ω2
f
Step 3: Solve for the final angular velocity ωf.
Solving for ωf, we have:
2mgh
mR2=ω2
f
ωf=√2gh
R
Thus, the angular velocity of the hoop when it becomes horizontal is √2gh
R.
Question 4
Question
A thin circular hoop with radius Rand mass Mis spinning about a vertical
axis through its center with an angular velocity ω. A small particle of mass m
3
is placed at a distance rfrom the center of the hoop along its axis of rotation.
Determine the required angular velocity ω′of the hoop for the particle to remain
in a vertical position, assuming there is no slipping or friction.
Solution
To determine the required angular velocity ω′of the hoop, we need to consider
the forces acting on the particle at distance rfrom the center of the hoop when
it is in a vertical position.
Step 1: Free Body Diagram The forces acting on the particle at distance
rare: - The gravitational force Fg=m·g, where gis the acceleration due to
gravity. - The normal force Nexerted by the hoop on the particle, which keeps
it in a vertical position.
Step 2: Acceleration in Vertical Direction The acceleration in the
vertical direction is given by:
∑Fy=m·ay
N−m·g=m·ay
N=m·(g+ay)
Step 3: Acceleration in Horizontal Direction The angular acceleration
of the hoop is given by:
α=a
R
where ais the linear acceleration of the particle.
Step 4: Equating Accelerations The linear acceleration of the particle
is the same as the linear acceleration of a point on the hoop:
a=r·α
Substituting the expression for angular acceleration:
a=r·a
R
a=r
R·a
a=r
R·g
Step 5: Equating Accelerations Equating the accelerations in the verti-
cal direction:
N=m·(g+r
R·g)
N=m·g(1 + r
R)
4
Step 6: Equation for Angular Velocity When the particle is in a ver-
tical position, the normal force Nprovides the centripetal force needed for the
particle’s circular motion:
N=m·ω′2·r
Step 7: Solving for ω′Equating the expressions for Nfrom Steps 5 and
6:
m·g(1 + r
R)=m·ω′2·r
ω′2=g·(1 + r
R)
r
ω′=√g·(1 + r
R)
r
Therefore, the required angular velocity ω′of the hoop for the particle to
remain in a vertical position is √g·(1+ r
R)
r.
Question 5
Question
A thin circular hoop of radius Rand mass Mis rotating about its symmetry
axis with an angular velocity ω. A small piece of putty with mass mis dropped
vertically onto the hoop. The putty sticks to the hoop and is rotating with the
hoop afterwards. What is the angular velocity of the hoop and the putty system
after the collision? Assume the hoop is initially at rest.
Solution
Step 1: Before collision, the hoop had angular momentum due to its rotation:
The initial angular momentum of the hoop is Li=Iω, where Iis the moment of
inertia of the hoop and ωis the initial angular velocity. For a hoop, I=M R2.
Step 2: When the putty is dropped onto the hoop, the system has angular
momentum conservation: The angular momentum of the system after the col-
lision is the sum of the initial angular momentum of the hoop and the angular
momentum of the putty: Lf= (I+mr2)ωf+mr(vf+Rωf), where ris the
radius at which the putty sticks to the hoop, vfis the final linear velocity of the
putty, and ωfis the final angular velocity of the hoop and the putty system.
Step 3: Consider the conservation of energy for the system: The initial
kinetic energy of the putty is all converted into rotational kinetic energy of the
hoop-putty system after the collision, neglecting any energy loss due to friction.
Write the expression for the total initial kinetic energy and the final kinetic
energy of the system.
Step 4: Apply the conservation of angular momentum and energy to solve
for the final angular velocity of the hoop and the putty system. Simplify the
equations to find ωf. This will give you the final angular velocity of the hoop
and the putty system.
5
Question 6
Question
A gyroscope initially at rest and free to pivot in a vertical plane is given an
angular velocity ωo. As it spins, the gyroscope begins to tilt with respect to the
vertical. Determine the angle θthrough which it tilts at a later time tgiven
that the moment of inertia about the pivot point is I, the gyroscope’s mass is
m, and the acceleration due to gravity is g.
Solution
Step 1: Draw a free-body diagram of the gyroscope at an angle θfrom the
vertical where the only force acting is the gravitational force.
Step 2: Apply Newton’s second law for rotational motion to the gyroscope.
The torque due to gravity about the pivot point is equal to the rate of change of
angular momentum. The torque due to gravity is given by τ=mgl sin θ, where
mis the mass of the gyroscope, gis the acceleration due to gravity, and lis the
distance from the pivot point to the center of mass of the gyroscope. The rate
of change of angular momentum is Idω
dt .
Step 3: Since the gyroscope initially has an angular velocity of ωoat t= 0,
we can find the angular velocity at a later time t. Integrate the rate of change
of angular velocity to find ω(t). This will give you the formula to calculate the
angle θas a function of time.
Step 4: Integrate the angular velocity with respect to time to find the angle θ
as a function of time. Apply initial conditions to find the integration constants.
Step 5: Evaluate the final expression for θat the time t.
Therefore, by following these steps, we can determine the angle θthrough
which the gyroscope tilts at a later time t.
Question 7
Question
A solid sphere of mass mand radius ris rotating about a horizontal axis passing
through its center with an angular velocity ω. The sphere is released from this
rotating state and allowed to fall vertically. Determine the minimum coefficient
of static friction needed between the sphere and the surface it rolls on so that
the sphere does not slip while it falls.
Solution
Step 1: The condition for no slipping during the fall is that the linear speed of
the sphere at the point of contact with the surface is equal to the product of
the radius and the angular velocity. This gives:
v=rω
6
Step 2: The linear speed at the point of contact, when the sphere falls
without slipping, is given by adding the translational and rotational kinetic
energy. For the translational kinetic energy, we have:
Ttrans =1
2mv2
Step 3: For the rotational kinetic energy, we have:
Trot =1
2Iω2
Where I=2
5mr2is the moment of inertia of a solid sphere.
Step 4: Equating the initial potential energy due to gravity to the sum of
translational and rotational kinetic energies gives:
mgh =1
2mv2+1
2(2
5mr2)ω2
Step 5: Substitute the relationship between vand ωin terms of r:
mgh =1
2m(rω)2+1
2(2
5mr2)ω2
Step 6: Now, solve for the minimum coefficient of static friction µsusing
the above equation and the condition for no slipping. Finally, the minimum
coefficient of static friction needed between the sphere and the surface it rolls
on so that the sphere does not slip while it falls is:
µs=5
7
Question 8
Question
A thin hoop of radius Rand mass Mis rolling without sliding on a horizontal
surface with a linear speed v. The hoop rotates about its own axis with an
angular speed ω. Calculate the total kinetic energy of the hoop.
Solution
Let’s first break down the kinetic energy into the translational and rotational
components.
Step 1: Find the translational kinetic energy The translational kinetic
energy of a rolling object is given by 1
2Mv2where vis the velocity of the center
of mass. For the hoop, the velocity of the center of mass vis equal to the linear
speed of the hoop, so the translational kinetic energy is:
KEtrans =1
2Mv2
7
Step 2: Find the rotational kinetic energy The rotational kinetic en-
ergy of a rotating object is given by 1
2Iω2where Iis the moment of inertia and
ωis the angular speed. For a hoop rotating about its own axis, the moment of
inertia is I=MR2. Thus, the rotational kinetic energy is:
KErot =1
2MR2ω2
Step 3: Calculate the total kinetic energy The total kinetic energy is
the sum of the translational and rotational kinetic energies:
KEtotal =KEtrans +KErot =1
2Mv2+1
2MR2ω2
Therefore, the total kinetic energy of the hoop rolling without sliding on a
horizontal surface is 1
2Mv2+1
2MR2ω2.
Question 9
Question
A thin circular loop of radius Rand mass Mis rotating about its vertical
diameter with an angular velocity ω. Determine the angular momentum of the
loop about the vertical axis.
Solution
Step 1: The angular momentum vector
Lof the loop is given by the formula:
L=Iω
where Iis the moment of inertia of the loop about the vertical axis and ω is the
angular velocity vector.
Step 2: The moment of inertia of a thin circular loop about its diameter is
I=1
2MR2. Thus, the angular momentum vector becomes:
L=1
2MR2ω
Step 3: Since the rotation is about the vertical diameter, the direction of
angular velocity is perpendicular to the vertical axis. Hence, the magnitude of
angular velocity |ω|is ω.
Step 4: Therefore, the angular momentum of the loop about the vertical
axis is:
L=1
2MR2ω
8
Question 10
Question
A uniform solid sphere of radius Rand mass Mis set rolling without slipping
on a horizontal surface with an initial angular speed ω0. Suddenly, a light rope
is wound around the sphere with one end of the rope attached to the surface.
Find the angular speed of the sphere when it has descended a distance h.
Solution
Step 1: We will begin by finding the total energy of the sphere when it has
descended a height h. The initial total energy of the system is given by
Ei=1
2Iω2
0+1
2Mv2
cm +Mgh,
where I=2
5MR2is the moment of inertia of a solid sphere, vcm =Rω0is the
velocity of the center of mass, and his the distance the sphere has descended.
Step 2: As the sphere descends, some of its gravitational potential energy is
converted to kinetic energy. At the bottom, when the sphere is in pure rolling
motion, the total energy is given by
Ef=1
2Iω2+1
2Mv2
cm.
Step 3: Since energy is conserved, we have Ei=Ef. Substituting the
expressions for Eiand Efand simplifying, we get
1
2(2
5MR2)ω2
0+1
2M(Rω0)2+Mgh =1
2(2
5MR2)ω2+1
2M(Rω)2.
Step 4: Solving for ωand simplifying, we find
ω=√5gh
7R.
Therefore, the angular speed of the sphere when it has descended a distance
his √5gh
7R.
Question 11
Question
A solid cylindrical gyroscope with a mass of 2 kg and a radius of 0.1 m is
initially spinning at 600 rpm. The gyroscope is mounted in a frame that allows
it to rotate about an axis through its center of mass. If a torque of 4 N·m is
applied to the gyroscope in a direction perpendicular to its spin axis, calculate
the angular acceleration of the gyroscope.
9
Solution
Step 1: Convert the initial angular velocity from rpm to rad/s. Given that the
initial angular velocity ωiof the gyroscope is 600 rpm and there are 60 seconds
in a minute, we can convert this to rad/s:
ωi=600 rpm
60 s/min ×2πrad
1rev = 20πrad/s
Step 2: Calculate the moment of inertia of the gyroscope. The moment of
inertia Iof a solid cylinder rotating about its central axis is given by:
I=1
2mr2
where mis the mass of the gyroscope and ris its radius. Substitute m= 2 kg
and r= 0.1m into the equation:
I=1
2(2 kg)(0.1m)2= 0.01 kg ·m2
Step 3: Calculate the angular acceleration using the torque equation. The
torque τapplied to the gyroscope is 4 N·m. The net torque on a spinning body
is related to its angular acceleration αby the equation:
τ=Iα
Solve for α:
α=τ
I=4N·m
0.01 kg ·m2= 400 rad/s2
Therefore, the angular acceleration of the gyroscope is 400 rad/s2.
Question 12
Question
A uniform thin rod of length Land mass Mis pivoted at one end such that
it can rotate freely in a vertical plane. A small bead of mass mslides without
friction along the rod. Initially, the rod is held horizontally and the bead is a
distance dfrom the pivot point. When the rod is released from rest, find the
position of the bead when the rod makes an angle θwith the vertical.
Solution
Step 1: We will start by analyzing the forces acting on the bead at an angle θ.
Step 2: The forces acting on the bead are the gravitational force mg acting
vertically downward and the tension in the rod Tacting along the rod towards
the pivot point.
10
Step 3: Resolving the forces along and perpendicular to the rod, we have:
Forces along the rod (x-direction): −Tsin θ=max(axis the acceleration of
the bead along the rod)
Forces perpendicular to the rod (y-direction): Tcos θ−mg =may= 0
(ay= 0 since the bead is constrained to move along the rod)
Step 4: Solving the y-direction equation for T yields: T=mg/ cos θ
Step 5: Substituting Tinto the equation of motion in the x-direction gives:
−mg tan θ=max
Step 6: The acceleration of the bead along the rod is related to the angular
acceleration of the rod by: ax=Rα
Step 7: where Ris the distance of the bead from the pivot (R=L−d) and
αis the angular acceleration of the rod.
Step 8: Substituting ax=Rα into the equation of motion gives: −mg tan θ=
mRα
Step 9: The torque about the pivot point is given by: τ=Iα
Step 10: For the rod, the moment of inertia is I=1
3M(L)2
Step 11: The torque applied by the gravitational force about the pivot point
is MgL sin θ
Step 12: Setting up the torque equation gives: MgL sin θ=1
3M(L)2α
Step 13: Solving for the angular acceleration αgives: α=3gL sin θ
L2
Step 14: Substituting αback into our previous equation gives: −mg tan θ=
mR ·3gL sin θ
L2
Step 15: Now, simplify to solve for sin θ:−gtan θ= 3gsin θ/L
Step 16: Rearranging the above equation leads to the final answer: sin θ=
3 tan θ
L
Question 13
Question
A bicycle wheel with a radius of 0.35 m and a mass of 1.5 kg rotates at a constant
angular speed of 3.0 rad/s. Suddenly, the wheel comes to a stop. What average
torque was required to stop the wheel in 4.0 s?
Solution
Step 1: Identify the given quantities. The radius of the bicycle wheel, r= 0.35
m. The mass of the bicycle wheel, m= 1.5kg. The initial angular speed of the
wheel, ω0= 3.0rad/s. The final angular speed of the wheel, ωf= 0 rad/s. The
time taken to stop the wheel, t= 4.0s.
Step 2: Calculate the moment of inertia of the bicycle wheel. The moment
of inertia of a solid cylinder rotating about its symmetry axis is I=1
2mr2.
Substitute the known values to find the moment of inertia:
I=1
2(1.5)(0.35)2= 0.0919 kg ·m2
11
Step 3: Calculate the initial angular momentum of the bicycle wheel. The
initial angular momentum is given by L0=Iω0. Substitute the values to find
the initial angular momentum:
L0= (0.0919)(3.0) = 0.2757 kg ·m2/s
Step 4: Calculate the final angular momentum of the bicycle wheel. Since
the wheel comes to a stop, the final angular momentum is Lf= 0.
Step 5: Use the torque-angular momentum theorem to find the average
torque. The torque-angular momentum theorem states that τavg =Lf−L0
t.
Substitute the values to find the average torque:
τavg =0−0.2757
4.0=−0.0689 N·m
Therefore, the average torque required to stop the bicycle wheel in 4.0 s is
−0.0689 Nm.
Question 14
Question
A thin rod of length Lwith a mass mis pivoted at one end and subjected to a
constant torque Tabout the pivot, perpendicular to the rod. Initially, the rod
is at rest in a horizontal position. Determine the angular acceleration of the rod
as a function of time.
Solution
Step 1: The torque about the pivot is given by τ=Iα, where Iis the moment
of inertia of the rod about the pivot and αis the angular acceleration.
Step 2: The moment of inertia of the rod about the pivot is I=1
3mL2.
Step 3: The torque τis given as τ=T.
Step 4: Therefore, T=1
3mL2α.
Step 5: Solving for α, we get α=3T
mL2.
Step 6: The angular acceleration of the rod as a function of time is α=3T
mL2,
which is constant.
Question 15
Question
A thin circular hoop of radius Rand mass Mis rotating about a vertical axis
passing through its center with an angular velocity ω. A small bead of mass m
is threaded onto the hoop and is free to slide along the hoop without friction.
Determine the critical angular velocity ωcat which the bead will just lift off the
hoop.
12
Solution
Step 1: At the critical angular velocity ωc, the normal force acting on the bead
is zero, which means the tension in the hoop at the top of the bead is 0.
Step 2: The forces acting on the bead are the gravitational force (mg) and
the centrifugal force (−mRω2
c) caused by the rotation.
Step 3: At the top of the hoop (θ=π), we can write the equation of motion
for the bead in the radial direction as:
−mRω2
c−mg =−mRω2
csin θ
Step 4: Substituting θ=πinto the equation, we get:
−mRω2
c−mg =−mRω2
csin π
−mRω2
c−mg = 0
Step 5: Solving for ωc, we have:
ωc=√g
R
Therefore, the critical angular velocity ωcat which the bead will just lift off
the hoop is √g
R.
Question 16
Question
A bicycle wheel has a radius of 0.3 meters and a mass of 2 kg. The wheel is
rotating at a constant angular speed of 10 rad/s. If a force of 20 N is applied
tangentially to the edge of the wheel, what is the resulting angular acceleration
of the wheel? Assume the wheel is a solid disk.
Solution
Step 1: Calculate the moment of inertia of the wheel using the formula for a
solid disk:
I=1
2·m·r2
where mis the mass of the wheel and ris the radius of the wheel.
I=1
2·2kg ·(0.3m)2= 0.09 kg ·m2
Step 2: Calculate the torque applied to the wheel using the formula:
τ=r·F
13
where τis the torque, ris the radius of the wheel, and Fis the force applied.
τ= 0.3m·20 N= 6 N·m
Step 3: Use the equation for torque and moment of inertia to find the angular
acceleration:
τ=I·α
where αis the angular acceleration.
6N·m= 0.09 kg ·m2·α
Step 4: Solve for α:
α=6N·m
0.09 kg ·m2= 66.67 rad/s2
Therefore, the resulting angular acceleration of the wheel is 66.67 rad/s2.
Question 17
Question
A uniformly solid disk of mass Mand radius Ris initially spinning about a
vertical axis through its center with an angular speed ω0. The disk is now
placed on a horizontal frictionless surface. Using the conservation of angular
momentum and energy, determine the angular speed ωof the disk just after it
has been released from rest in an inclined position at an angle θwith respect
to the vertical. The disk is free to rotate about a horizontal axis perpendicular
to the inclined plane and passing through its center, and it is released from rest
with the center of mass at a height habove the horizontal surface. Assume the
disk rolls without slipping.
Solution
Step 1: The angular momentum of the disk is conserved about the vertical axis.
Initially, the angular momentum is L0=I0ω0, where I0=1
2MR2is the moment
of inertia of the disk about the vertical axis. When the disk is released, the axis
of rotation changes to the contact point with the surface, thus the moment of
inertia of the disk about this axis is I=1
2MR2+Mh2.
Step 2: By conservation of angular momentum,
L0=Iω
1
2MR2ω0=(1
2MR2+Mh2)ω
ω=MR2ω0
MR2+ 2Mh2
14
Step 3: Next, we can use conservation of energy. The initial kinetic energy
is KE0=1
2I0ω2
0and the final kinetic energy is KE =1
2Iω2. The potential
energy at height his P E =Mgh.
Step 4: By conservation of energy,
KE0+P E =KE
1
2(1
2MR2)ω2
0+Mgh =1
2(1
2MR2+Mh2)ω2
Step 5: Substituting the expression for ωderived in Step 2 into the energy
conservation equation, we get
1
4MR2ω2
0+Mgh =1
4MR2(MR2ω0
MR2+ 2Mh2)2
+Mh2(MR2ω0
MR2+ 2Mh2)2
Step 6: After simplifying the equation and solving for ω, we find
ω=ω0
√1+2(h
R)2
Question 18
Question
A solid cylinder of mass Mand radius Rrolls without slipping down an inclined
plane making an angle θwith the horizontal. The cylinder is released from rest
at the top of the incline. What is the angular speed of the cylinder when it
reaches the bottom of the incline? Assume the incline is frictionless.
Solution
Step 1: Identify the relevant principles and formulas.
The conservation of energy principle can be used in this problem. The initial
potential energy of the cylinder at the top of the incline will be converted into
kinetic energy (translational and rotational) at the bottom of the incline.
Step 2: Calculate the initial potential energy. The initial potential energy
of the cylinder is given by:
P Einitial =Mgh
where his the height of the incline and gis the acceleration due to gravity.
Since the cylinder is released from rest, the initial potential energy is equal to
the initial kinetic energy plus the initial rotational kinetic energy:
P Einitial =KEtrans, initial +KErot, initial
Step 3: Calculate the final kinetic energy. The final kinetic energy of the
cylinder at the bottom of the incline is given by:
KEfinal =KEtrans, final +KErot, final
15
Since the cylinder is rolling without slipping, the final translational and rota-
tional kinetic energies are related by:
KEtrans, final =1
2Mv2
KErot, final =1
2Iω2
where vis the linear velocity of the cylinder and ωis the angular velocity.
Step 4: Equate the initial potential energy to the final total kinetic energy.
Since energy is conserved, we can set the initial potential energy equal to the
final total kinetic energy:
Mgh =1
2Mv2+1
2Iω2
Step 5: Substitute values and solve for the angular speed. The moment of
inertia of a solid cylinder about its center is I=1
2MR2. Substituting this into
the energy equation gives:
Mgh =1
2Mv2+1
4MR2ω2
gh =1
2v2+1
4R2ω2
Step 6: Relationship between linear and angular speed. Since the cylinder is
rolling without slipping, the linear speed vand the angular speed ωare related
by v=Rω. Substituting this relationship into the energy equation gives:
gh =1
2(Rω)2+1
4R2ω2
gh =3
4R2ω2
ω=√4gh
3R2
Thus, the angular speed of the cylinder when it reaches the bottom of the incline
is ω=√4gh
3R2.
Question 19
Question
A thin, uniform rod of length Land mass mis pivoted at one end and set into
circular motion about the pivot point at a constant angular speed ω. Calculate
the gyroscopic moment of inertia of the rod about the pivot point.
16
Solution
To find the gyroscopic moment of inertia of the rod about the pivot point,
we need to consider the rotational inertia of the rod with respect to an axis
perpendicular to the rod at its end and passing through the pivot point.
Step 1: The rotational inertia of a thin, uniform rod about an axis per-
pendicular to the rod and passing through one end is given by the formula
I=1
3mL2, where mis the mass of the rod and Lis its length.
Step 2: Since we are rotating the rod about the same pivot point, the
gyroscopic moment of inertia of the rod about the pivot point is equal to its
rotational inertia about an axis perpendicular to the rod at its end. Therefore,
the gyroscopic moment of inertia of the rod is also 1
3mL2.
Thus, the gyroscopic moment of inertia of the rod about the pivot point is
1
3mL2.
Question 20
Question
A uniform solid sphere of radius Rand mass Mis placed on a rough horizontal
surface. The sphere is given a quick spin about a vertical axis through its center
at an angular speed of ω0. The sphere then begins to translate in the x-direction
and rotate in a counterclockwise direction as seen from above. Show that for
sufficiently large ω0, the sphere will roll without slipping, and determine the
critical value of ω0at which rolling begins. Assume the coefficient of friction
between the sphere and the surface is µ.
Solution
Step 1: Start by drawing a free-body diagram for the sphere. The forces acting
on the sphere are the normal force Nand the frictional force f. There is also
a torque Nfriction due to friction. Step 2: Resolve the gravitational force mg
into components normal and tangential to the surface. The normal force N
balances the component of mg perpendicular to the surface, and the frictional
force fbalances the component of mg parallel to the surface. Step 3: Write
the torque equation for the sphere about its center of mass. The torque due
to the normal force and the frictional force must equal the rate of change of
angular momentum. Step 4: Use the equations of motion to express the normal
force and frictional force in terms of the angular speed ω. The normal force
will be N=mg and the frictional force f=µmg. Step 5: Substitute the
expressions for Nand finto the torque equation and solve for ω. Determine
for what values of ωrolling without slipping occurs. Step 6: Finally, compute
the critical value of ω0at which rolling begins by analyzing the frictional force.
This occurs when the frictional force reaches its maximum value, which happens
when rolling starts.
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Question 21
Question
A solid sphere of mass mand radius rrolls without slipping down an inclined
plane of height h. If the sphere starts from rest at the top of the incline,
determine the speed of the center of mass of the sphere when it reaches the
bottom.
Solution
Step 1: First, we need to analyze the forces acting on the sphere as it rolls down
the incline. Since the sphere is rolling without slipping, the frictional force is
static (i.e., fs=µsN, where fsis the static frictional force, µsis the coefficient
of static friction, and Nis the normal force). The forces acting on the sphere
are its weight mg acting downward, the normal force Nacting perpendicular to
the incline, and the static frictional force fsacting parallel to the incline in the
opposite direction of motion.
Step 2: The net force accelerating the sphere down the incline is the com-
ponent of the weight parallel to the incline minus the frictional force. The
component of the weight along the incline is mg sin θ, where θis the angle of
the incline. Therefore, the net force accelerating the sphere is (mg sin θ−fs).
Step 3: The acceleration of the sphere down the incline can be found by
using Newton’s second law:
ma =mg sin θ−µsN
where N=mg cos θis the normal force. Substitute Ninto the equation, we
have:
ma =mg sin θ−µsmg cos θ
a=g(sin θ−µscos θ)
Step 4: Next, we can determine the final speed of the sphere when it reaches
the bottom of the incline. Since the sphere starts from rest at the top, its initial
speed vi= 0. Using the kinematic equation v2
f=v2
i+ 2a∆x, where ∆x=h
(height of the incline), we can solve for the final speed vf:
v2
f= 0 + 2gh(sin θ−µscos θ)
vf=√2gh(sin θ−µscos θ)
Therefore, the speed of the center of mass of the sphere when it reaches the
bottom of the incline is √2gh(sin θ−µscos θ).
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Question 22
Question
A solid cylindrical object of mass mand radius ris rolling without slipping on a
horizontal surface with a constant speed v. The object encounters a small bump
which causes it to gain some height. Determine the magnitude and direction of
the precession of the object’s angular momentum vector about the vertical axis
through its center due to the bump.
Solution
Let’s denote the angular velocity of the object as ω, the angular velocity of
precession as Ω, and the angular momentum of the object as L. We will consider
the angular momentum of the object before and after encountering the bump.
Step 1: Determine the angular momentum of the object before encountering
the bump.
The angular momentum of the object before encountering the bump is given
by:
Lbefore =Iω
where I=1
2mr2is the moment of inertia of a solid cylinder. Since the object
is rolling without slipping, the angular velocity is related to the linear speed v
by ω=v
r. Thus, the angular momentum before encountering the bump is:
Lbefore =(1
2mr2)(v
r)=1
2mvr
Step 2: Determine the angular momentum of the object after encountering
the bump.
As the object gains height, the angular momentum of the object after en-
countering the bump will also have a component along the vertical axis:
Lafter =Iω+mgh
where his the height gained due to the bump. Since there is a precession about
the vertical axis, the angular momentum due to precession is Lprecession =IΩ.
Thus, the total angular momentum after encountering the bump is:
Lafter =Iω+mgh +IΩ
Since the object is still rotating about its geometric axis, the angular velocity can
be expressed in terms of the precession angular velocity as ω=−Ω. Substitute
in the expressions for I,ω, and Lbefore:
Lafter =(1
2mr2)(−Ω) + mgh +(1
2mr2)Ω
Step 3: Compute the precession angular velocity and direction.
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Equating the two expressions for angular momentum before and after en-
countering the bump, we have:
1
2mvr =(1
2mr2)(−Ω) + mgh +(1
2mr2)Ω
Solving for Ω:
Ω=v
r−(v
r+2gh
r)=−2gh
r
Therefore, the magnitude of the precession angular velocity is 2gh
rin the
clockwise direction when viewed from the top.
Question 23
Question
A solid cylinder of mass Mand radius Ris rolling without slipping along a
horizontal surface with a velocity v0when it encounters a step of height h.
What is the minimum initial velocity v0needed for the cylinder to successfully
navigate the step without tipping over?
Solution
Step 1: To determine the minimum initial velocity needed for the cylinder to
navigate the step without tipping over, we need to consider the conservation of
energy. At the top of the step, the kinetic energy of the cylinder will be fully
converted to potential energy at the maximum height that the cylinder reaches.
Step 2: The initial kinetic energy of the cylinder is due to both its transla-
tional and rotational motion. The kinetic energy can be written as:
KE =1
2Mv2
0+1
2Iω2
where Iis the moment of inertia of a solid cylinder (I=1
2MR2) and ωis the
angular velocity.
Step 3: Since the cylinder is rolling without slipping, the linear velocity v
and angular velocity ωare related by v=Rω. Substituting this relationship
into the kinetic energy equation gives:
KE =1
2Mv2
0+1
2(1
2MR2)(v0
R)2
Step 4: When the cylinder reaches the top of the step, all of the kinetic
energy is converted to potential energy. The potential energy at the top of the
step is Mgh, where gis the acceleration due to gravity.
20
Step 5: Setting the initial kinetic energy equal to the potential energy at the
top of the step gives:
1
2Mv2
0+1
2(1
2MR2)(v0
R)2=Mgh
Step 6: Simplifying and solving for v0, we find:
v0=√2gh
Therefore, the minimum initial velocity v0needed for the cylinder to suc-
cessfully navigate the step without tipping over is √2gh.
Question 24
Question
A solid disk of mass Mand radius Ris rotating about a vertical axis through
its center with an angular velocity ω. The disk is mounted on a frictionless
horizontal axle. A small mass mis dropped vertically onto the disk from a
height habove the center of the disk and sticks to the disk without rebounding.
Find the angular velocity of the disk and the common angular velocity of the
disk and mass immediately after the mass sticks.
Solution
Step 1: Conservation of angular momentum for the disk-mass system
The initial angular momentum of the system is given by:
Li=Idiskω
where the moment of inertia of a solid disk is Idisk =1
2MR2.
When the small mass mis added, the total moment of inertia becomes
Itotal =Idisk +mR2.
The final angular momentum of the system is:
Lf=Itotalωf
where ωfis the final angular velocity of the disk and mass.
Since angular momentum is conserved, we have:
Li=Lf
Idiskω=Itotalωf
1
2MR2ω=(1
2MR2+mR2)ωf
Step 2: Finding the final angular velocity of the system
21
Solving for ωf, we get:
ωf=
1
2MR2ω
1
2MR2+mR2=M ω
M+ 2m
Therefore, the final angular velocity of the disk and mass is Mω
M+2m.
Question 25
Question
A non-uniform disk of radius Rand mass Mrotates at an angular velocity ω
about a vertical axis through its center. The disk is placed on a horizontal table
and a small horizontal force Fis applied to the edge of the disk in the direction
perpendicular to the radius at a distance R
2from the center. Find the angular
acceleration of the disk immediately after the force is applied.
Solution
1. The torque τexerted by the force Fis given by τ=F r, where r=R
2is
the lever arm. 2. The moment of inertia Iof the disk is needed to calculate
the angular acceleration. For a non-uniform disk of mass Mand radius R, the
moment of inertia is I=∫r2dm. 3. Expressing dm in terms of the linear mass
density λ=M
πR2:dm =λdA =λ2πrdr. 4. The moment of inertia becomes:
I=∫R
0r2λ2πrdr = 2λπ ∫R
0r3dr = 2λπ [r4
4]R
0. 5. The moment of inertia
simplifies to: I=MR2
2. 6. The torque τleads to an angular acceleration α
through τ=Iα. Substituting the values, we get F R/2 = MR2
2α. 7. Solving for
α, we find: α=F
M. 8. Thus, the angular acceleration of the disk immediately
after the force is applied is F
M.
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