PHYS 201 - GENERAL PHYSICS I -
Gyroscopic Motion and Stability
Question Bank - Set 6
Liberty University
Question 1
Question
A solid cylinder of radius Rand mass Mis rolling without slipping on a hor-
izontal surface. Initially, it has an angular velocity ω0and is rotating about
its symmetry axis. At a certain instant, a horizontal force Fis applied at the
top of the cylinder perpendicular to the axis of rotation. Determine the angular
acceleration of the cylinder immediately after the force is applied.
Solution
1. The rotational analog of Newton’s second law states:
τ=Iα
where τis the net torque acting on the cylinder, Iis the moment of inertia of
the cylinder, and αis the angular acceleration.
2. The torque applied by the force Fis τ=F r, where ris the radius of the
cylinder.
3. The moment of inertia of a solid cylinder rotating about its symmetry
axis is I=1
2MR2.
4. Substituting these expressions into the rotational analog of Newton’s
second law, we have:
F r =1
2MR2α
5. Solving for α, the angular acceleration, we get:
α=2F
MR
Thus, the angular acceleration of the cylinder immediately after the force is
applied is 2F
MR .
Question 2
Question
A circular disc of radius Rand mass Mis spinning about its axis at an angular
velocity ω0. The disc is dropped onto a horizontal surface where there is enough
friction to prevent slipping. As a result, the disc starts rolling without slipping.
Determine the angular velocity of the spinning disc after it starts rolling.
Solution
1. Conservation of Angular Momentum: When the disc is dropped onto
the surface, the friction generates a torque that causes the spinning disc to start
rolling. The net torque on the disc is zero, so angular momentum is conserved.
2. The initial angular momentum of the spinning disc is given by:
Li=Iω0
where I=1
2MR2is the moment of inertia of the disc about its center.
3. As the disc starts rolling, its angular momentum changes. After the disc
starts rolling without slipping, its angular momentum is the sum of the spinning
and rolling angular momenta:
Lf=Irollωroll
where Iroll =1
4MR2is the moment of inertia of the disc about its center when
rolling and ωroll is the angular velocity of the disc after it starts rolling.
4. Since angular momentum is conserved, we have:
Iω0=Irollωroll
5. Substitute the expressions for Iand Iroll:
1
2MR2ω0=1
4MR2ωroll
6. Solve for ωroll:
ωroll = 2ω0
7. Result: The angular velocity of the spinning disc after it starts rolling
without slipping is 2ω0.
Question 3
Question
A uniform disc of radius Rand mass Mis spinning with an angular velocity
ωabout an axis through its center. The disc is then placed on a horizontal
surface and released. Calculate the time it takes for the disc to come to rest due
to friction between the disc and the surface. Assume the coefficient of kinetic
friction between the disc and the surface is µk.
2
Solution
1. The initial angular momentum of the disc is given by Li=Iω, where Iis the
moment of inertia of the disc. The moment of inertia of a disc about its center
is I=1
2MR2; thus, Li=1
2MR2ω.
2. As the disc slows down, the frictional force fkacting on the disc will cause
a torque which will work against the initial angular momentum. The frictional
torque Tis given by T=r×fk, where ris the radius of the disc.
3. The frictional force fkcan be determined using fk=µkN, where Nis
the normal force. Since the disc is at rest, N=Mg, where gis the acceleration
due to gravity.
4. The frictional torque is then T=rfk=rµkMg. This torque will act in
the direction opposite the initial angular momentum, causing the disc to slow
down.
5. The net torque acting on the disc is given by τ=−T=−rµkMg.
6. The torque τcauses an angular acceleration αaccording to Newton’s
second law for rotation: τ=Iα. Therefore, −rµkM g =1
2MR2α.
7. The angular acceleration αis related to the angular velocity ωby the
equation α=dω
dt .
8. Rearranging the previous equation in terms of time, we get 1
2MR2dω
dt =
−rµkMg. This can be simplified to dω
dt =−2rµkg
R.
9. Integrating both sides with respect to time, we get ∫ω
0dω =∫t
0−2rµkg
Rdt.
This simplifies to ω−0 = −2rµkg
Rt−0.
10. Therefore, t=R
2rµkgω. Substituting ω= 0 (when the disc comes to
rest), we get t=R
2rµkg·0 = 0.
Hence, it takes the disc a time of 0 seconds to come to rest due to friction
between the disc and the surface.
Question 4
Question
A disk of mass mand radius Rrotates with an angular velocity ωabout its cen-
tral axis. The disk is mounted on a stationary axle through its center. A small
object of mass Mis placed on the edge of the disk. The coefficient of kinetic
friction between the object and the disk is µ. Find the angular acceleration of
the disk and the object right after the object is released and starts sliding on
the disk.
Solution
Step 1: To find the angular acceleration of the disk, we can consider the torques
acting on the system. The only external torque acting on the disk-object system
is due to friction. The frictional torque can be found using the equation τ=
µ·R·F, where Fis the frictional force.
3
Step 2: The frictional force Fcan be found by considering the forces acting
on the object. The forces acting on the object are the gravitational force and
the frictional force. The net force in the radial direction is given by Mg −F=
Maradial.
Step 3: The acceleration of the object in the radial direction is the same as
the tangential acceleration of the object on the disk. Therefore, aradial =Rα,
where αis the angular acceleration of the disk.
Step 4: We can now substitute the expression for Fin terms of αinto the
equation Mg −µ·R·F=Maradial and solve for α.
Step 5: After finding an expression for the angular acceleration α, we can
calculate its numerical value using the given parameters m,R,ω,M,g, and µ.
Question 5
Question
A uniform rectangular plate of mass Mand side lengths aand bis rotating
about an axis passing through its center making an angle θwith respect to its
length a. Determine the moment of inertia of the plate about an axis parallel
to the sides of length aand passing through one corner of the plate.
Solution
Step 1: The moment of inertia of the plate about its center and parallel to the
sides of length acan be calculated using the parallel-axis theorem. Let’s denote
this moment of inertia as Ic.
Step 2: The moment of inertia about an axis passing through one corner
of the plate and parallel to the sides of length acan be found by adding the
moment of inertia about the center to the product of its mass and the square
of the distance between the two axes, a.
Step 3: The moment of inertia about one corner of the plate is given by
I=Ic+Ma2.
Step 4: Given that the moment of inertia about its center and parallel to
the sides of length ais Ic=1
12 M(a2+b2), we substitute this into the equation
for the moment of inertia about one corner.
Step 5: Thus, the moment of inertia of the plate about the axis passing
through one corner and parallel to the sides of length ais I=1
12 M(a2+b2) +
Ma2. Simplifying this expression gives the final answer.
Question 6
Question
A uniform wheel of mass Mand radius R, initially at rest, is free to rotate
about its axis passing through its center. A small mass mis slowly dropped on
4
the rim from a height habove the center of the wheel. Assuming the mass m
sticks to the rim, find the angular velocity of the wheel and the linear velocity
of the small mass just before it makes contact with the rim.
Solution
Step 1: The total angular momentum of the system (wheel +mass) is conserved
since no external torques act on it. Step 2: The initial angular momentum of
the system is zero since the wheel is at rest initially. Step 3: When the mass
mis dropped, it acquires angular momentum equal to mgh, where gis the
acceleration due to gravity. Step 4: The final angular momentum of the system
is mgh +Iω, where Iis the moment of inertia of the wheel and ωis the final
angular speed of the wheel. Step 5: Setting the initial angular momentum equal
to the final angular momentum, we have mgh =Iω. Step 6: The moment of
inertia of the wheel about its center is I=1
2MR2. Step 7: Substituting this
into the equation mgh =Iω, we get mgh =1
2MR2ω. Step 8: Solving for ω, we
find ω=2mgh
MR2. Step 9: The linear velocity of the mass just before it reaches
the rim is v=Rω = 2gh√m
M.
Therefore, the angular velocity of the wheel is 2mgh
MR2and the linear velocity
of the small mass just before it makes contact with the rim is 2gh√m
M.
Question 7
Question
A solid circular disk of radius Rand mass Mis rotating with an angular velocity
ωabout a perpendicular axis through its center. The disk is dropped onto a
horizontal surface with a coefficient of kinetic friction µkand comes to rest after
sliding a distance d.
Calculate the distance dthe disk slides before coming to rest.
Solution
Step 1: The frictional force acting on the disk while it is sliding is given by
fk=µk·mg, where mis the mass of the disk. This frictional force will provide
a torque that opposes the rotational motion of the disk.
Step 2: The torque due to friction is τ=fk·R. This torque will work to
decrease the angular velocity of the disk until it comes to rest.
Step 3: The work done by this torque is equal to the change in kinetic
energy of the disk. The initial kinetic energy is due to its rotation and is given
by KEi=1
2Iω2, where Iis the moment of inertia of the disk. The final kinetic
energy is zero since the disk comes to rest.
Step 4: Equating the work done by torque to the change in kinetic energy,
we have τ·d=1
2Iω2, where dis the distance the disk slides before coming to
rest.
5
Step 5: Substituting τ=fk·R=µk·mg ·Rand I=1
2MR2into the above
equation, we get µk·mg ·R·d=1
2
1
2MR2ω2.
Step 6: Solving for d, we find d=MRω2
4µkgas the distance the disk slides before
coming to rest.
Therefore, the distance dthe disk slides before coming to rest is M Rω2
4µkg.
Question 8
Question
A gyroscope consists of a thin uniform disk of radius Rand mass M, mounted
at the end of a thin, massless rod of length L. The gyroscope is spinning with
an angular velocity ω. Calculate the angular momentum of the gyroscope about
an axis through the center of the disk perpendicular to the plane of the disk.
Solution
Step 1: The angular momentum of a rotating object is given by the formula
L=Iω, where Iis the moment of inertia and ωis the angular velocity.
Step 2: The moment of inertia of a disk rotating about an axis through its
center perpendicular to the plane of the disk is given by I=1
2MR2.
Step 3: Substituting the moment of inertia into the formula for angular
momentum, we get
L=1
2MR2ω.
Therefore, the angular momentum of the gyroscope about an axis through
the center of the disk perpendicular to the plane of the disk is 1
2MR2ω.
Question 9
Question
A gyroscope consists of a disk of mass mand radius Rmounted on a frictionless
axle. The disk is rotating with an angular speed ωabout the axle. In order
to change the orientation of the gyroscope, a torque is applied at right angles
to the axle. If the magnitude of the torque applied is T, determine the rate at
which the gyroscope’s angular momentum is precessing.
Solution
1. The angular momentum of the gyroscope is given by the equation L=Iω,
where Iis the moment of inertia of the gyroscope.
2. The moment of inertia for a disk rotating about an axis passing through
its center is I=1
2mR2.
3. The initial angular momentum of the gyroscope is L0=Iω.
6
4. When a torque is applied perpendicular to the axle, the gyroscope starts
precessing. The torque causes a change in angular momentum perpendicular to
both the torque and the angular momentum, resulting in precession.
5. As the torque is perpendicular to the axis of rotation, it does not change
the magnitude of angular momentum, but changes its direction.
6. The rate of change of angular momentum, dL
dt , is equal to the torque
applied, T.
7. Since the torque is perpendicular to the axis of rotation (angular velocity
ω), the gyroscope precesses about the vertical z-axis.
8. The rate at which the gyroscope’s angular momentum is precessing can
be determined by: dLprecession
dt =T
Therefore, the rate at which the gyroscope’s angular momentum is precessing
is equal to the magnitude of the torque applied.
Question 10
Question
A uniform rod of length Land mass Mis pivoted at one end and set into
oscillation in a vertical plane. The rod is released from rest at an angle of θ0
with the vertical. Determine the period of the small oscillations.
Solution
Step 1: First, let’s find the moment of inertia of the rod about the pivot point.
The moment of inertia of a uniform rod rotating about an axis perpendicular
to the rod and passing through one end is I=1
3ML2.
Step 2: Next, we will find the gravitational torque acting on the rod when
it is at an angle θwith the vertical. The torque is τ=Mg L
2sin(θ).
Step 3: Using the negative sign convention for restoring torques, we find the
equation of motion: I¨
θ=−τ=−Mg L
2sin(θ).
Step 4: To simplify this, we will use the small-angle approximation sin(θ)≈
θfor small θ.
Step 5: Substituting into the equation of motion, we get I¨
θ=−Mg L
2θ.
Step 6: Rearranging, we have ¨
θ+3g
2Lθ= 0.
Step 7: This is a second order differential equation with solutions of the form
θ=Asin(ωt) + Bcos(ωt), with ω=√3g
2L.
Step 8: The period of oscillation is T=2π
ω= 2π√2L
3g. Thus, the period of
small oscillations of the rod is T= 2π√2L
3g.
7
Question 11
Question
A gyroscope consists of a 0.5 kg disk with a radius of 0.1 m rotating at 2000 RPM
(revolutions per minute) about its axis. The gyroscope is spinning clockwise
when viewed from the top. Determine the angular momentum of the gyroscope
in both magnitude and direction.
Solution
Step 1: Calculate the angular velocity of the gyroscope in rad/s. Given that
the gyroscope is rotating at 2000 RPM, we first convert this into radians per
second:
Angular velocity ω= 2000 RPM×2πrad
60 s=2000 ×2π
60 rad/s =2000π
60 rad/s =100π
3rad/s
Step 2: Calculate the angular momentum of the gyroscope. The angular
momentum of an object is given by the formula: L=Iω, where Lis the
angular momentum, Iis the moment of inertia, and ωis the angular velocity.
The moment of inertia of a disk rotating about its central axis is given by
I=1
2MR2, where Mis the mass of the disk and Ris its radius.
Substitute the values: M= 0.5kg, R= 0.1m, ω=100π
3rad/s.
I=1
2×0.5×(0.1)2= 0.025 kg m2
Calculating the angular momentum:
L=Iω = 0.025 ×100π
3
L=100π
3×0.025
L=100π×0.025
3=100π×0.025
3
L=100π×0.025
3=100π×0.025
3=100π
120 kg m2/s
The magnitude of angular momentum is 100π
120 kg m²/s. Since the gyroscope
is spinning clockwise, the direction of the angular momentum is in the direction
of the rotation, which is clockwise when viewed from the top.
Question 12
Question
A uniform solid cylinder of radius Rand mass Mis rolling without slipping on
a horizontal surface with angular speed ω. At time t= 0, a constant horizontal
8
force Fis applied at a point on the rim of the cylinder perpendicular to the
plane of the cylinder. Find the angular acceleration of the cylinder as a function
of time.
Solution
Step 1: We will start by finding the torque acting on the cylinder due to the
applied force F. The torque τis given by:
τ=r×F=rF sin θ
where ris the radius of the cylinder and θis the angle between the force and
the radius vector.
Step 2: The torque causes the angular acceleration αof the cylinder. Using
Newton’s second law for rotation, we have:
τ=Iα
where Iis the moment of inertia of the cylinder. For a solid cylinder rotating
about an axis passing through its center and perpendicular to its symmetry
axis, I=1
2MR2.
Step 3: Substituting the expression for torque and moment of inertia into
the equation τ=Iα, we obtain:
rF sin θ=1
2MR2α
Step 4: We know that the linear acceleration of a point on the rim of the
cylinder is a=Rα. Since the cylinder is rolling without slipping, a=Rα =αR,
we can write:
F=Ma
Step 5: Differentiating both sides of this equation with respect to time, we
get:
dF
dt =Md
dt(Rα)
Step 6: Substituting F=Ma and Rα =ainto the above equation and using
chain rule, we obtain:
Mdv
dt =Mα +Ma
Step 7: Rearranging the terms in the equation, we find:
α=dv
dt −a
Therefore, the angular acceleration of the cylinder as a function of time is
α=dv
dt −a.
9
Question 13
Question
A gyroscope wheel of radius Rand mass Mis spinning at an angular velocity ω
about a horizontal axis through its center of mass. The gyroscope is mounted in
a frame that allows it to rotate freely in the vertical plane. Initially, the frame
is vertical, and the gyroscope spins horizontally about the vertical axis of the
frame. Calculate the precession frequency of the gyroscope.
Solution
Step 1: The precession frequency of the gyroscope can be calculated using the
formula:
Precession frequency =mgR
Iω
where m= mass of the gyroscope, g= acceleration due to gravity, R= radius
of the gyroscope wheel, I= moment of inertia of the gyroscope wheel, and ω=
angular velocity of the gyroscope.
Step 2: The moment of inertia of the gyroscope wheel can be calculated
using the formula:
I=1
2MR2
Step 3: Substitute the given values into the expressions for moment of inertia
and precession frequency:
I=1
2MR2
Precession frequency =mgR
1
2MR2ω
Step 4: Simplify the expression for the precession frequency:
Precession frequency =2mg
Rω
Therefore, the precession frequency of the gyroscope is 2mg
Rω .
Question 14
Question
A thin uniform rod of length Land mass Mis rotating at a constant angular
velocity ωabout a fixed vertical axis passing through one end of the rod. The
other end of the rod is held horizontally by a pivot (like a rotating baton).
Determine the angular momentum of the rod about the pivot point.
10
Solution
Let’s denote the pivot point as point Oat the end of the rod where the axis
of rotation passes through. We need to find the angular momentum of the rod
about this point.
Step 1: Determine the moment of inertia of the rod about point O. The
moment of inertia of a thin rod rotating about one end is I=1
3ML2
Step 2: Calculate the angular momentum. The angular momentum of the
rod about the pivot point Ois given by:
L=Iω =1
3ML2ω
Therefore, the angular momentum of the rod about the pivot point is 1
3ML2ω.
Question 15
Question
A thin hoop of mass Mand radius Rinitially at rest is allowed to roll without
slipping on a horizontal surface. A small mass mis attached to a string wound
around the hoop’s circumference. The mass is released from rest at a point that
is level with the center of the hoop. Determine the speed of the mass mwhen
it reaches the ground.
Solution
Step 1: The potential energy of mass mis converted to its kinetic energy when
it reaches the ground. At the starting point, the potential energy of mass mis
given by:
Ui=mghinitial
where hinitial is the initial height of mass m.
Step 2: At the bottommost point, the potential energy of mass mis entirely
converted to its kinetic energy, given by:
Kf=1
2mv2
Step 3: The initial potential energy of mass mcan be calculated as:
Ui=mghinitial =mg (R
2+R)=mg (3R
2
Step 4: By the conservation of mechanical energy, the initial potential energy
is converted to the final kinetic energy. Thus, we have:
mghinitial =1
2mv2
11
Step 5: Substituting the values, we get:
mg (3R
2)=1
2mv2
Step 6: Solving for v, we find:
v=√3gR
Therefore, the speed of the mass mwhen it reaches the ground is √3gR.
Question 16
Question
A cylindrical gyroscope has a radius of 0.1m and a mass of 0.5kg. It is spinning
at an angular velocity of 100 rad/s. The gyroscope is initially horizontal and
supported only at one end. Calculate the minimum force required to prevent
the gyroscope from falling when the other end is released.
Solution
Step 1: The angular momentum of the gyroscope is given by the formula L=Iω,
where Iis the moment of inertia and ωis the angular velocity. Given: Radius,
r= 0.1m Mass, m= 0.5kg Angular velocity, ω= 100 rad/s
The moment of inertia of a cylindrical gyroscope is I=1
2mr2. Substituting
the given values:
I=1
2×0.5×(0.1)2= 0.0025 kg m2
Angular momentum, L=Iω = 0.0025 ×100 = 0.25 kg m2/s
Step 2: When the other end is released, the gyroscope would start to fall
under the influence of gravity. In order to prevent the gyroscope from falling,
a minimum force equal and opposite to the weight of the gyroscope must be
applied. The weight of the gyroscope is given by W=mg.
Given that g= 9.8m/s2, the weight is:
W= 0.5×9.8 = 4.9N
Therefore, to prevent the gyroscope from falling, the minimum force required
is F=W= 4.9N.
Question 17
Question
A uniform rod of mass mand length lis pivoted at one end and set into rotation
about a vertical axis. The other end of the rod is attached to a small object
12
of mass m. The system initially rotates with angular velocity ω0. Determine
the angular velocity of the system when the rod is in the horizontal position (i)
before and (ii) after the object is moved towards the center of the rod a distance
l
2. Assume that the object slides along the rod without friction.
Solution
(i) Before the object is moved: Step 1: We begin by finding the moment of
inertia of the system about the pivot point. The moment of inertia of the rod
about the pivot point is Irod =1
3ml2. The moment of inertia of the object about
the pivot point is Iobj =m(l
2)2=1
4ml2. The total moment of inertia of the
system is I=Irod +Iobj =1
3ml2+1
4ml2=7
12 ml2.
Step 2: Conservation of angular momentum: Initial angular momentum
L0=Iω0=(7
12 ml2)ω0. Final angular momentum Lf=Iωf, where ωfis the
angular velocity when the rod is in the horizontal position.
Since there are no external torques, angular momentum is conserved: L0=
Lf. Thus, (7
12 ml2)ω0=(7
12 ml2)ωf. Solving for ωf, we get ωf=ω0.
Therefore, the angular velocity of the system before the object is moved
towards the center of the rod is ω0.
(ii) After the object is moved: Step 1: The new total moment of inertia of
the system when the object is moved towards the center of the rod a distance
l
2is: I′=1
3ml2+1
4m(l
2)2=7
12 ml2+1
8ml2=11
24 ml2.
Step 2: Conservation of angular momentum: Initial angular momentum is
the same as in part (i), L0=(7
12 ml2)ω0. Final angular momentum L′
f=I′ω′
f,
where ω′
fis the angular velocity after moving the object.
Conservation of angular momentum gives (7
12 ml2)ω0=(11
24 ml2)ω′
f. Solving
for ω′
f, we find ω′
f=2
3ω0.
Therefore, the angular velocity of the system after the object is moved to-
wards the center of the rod a distance l
2is 2
3ω0.
Question 18
Question
A thin circular hoop of mass Mand radius Ris initially at rest on a frictionless
horizontal table. A small object of mass mis placed on the hoop and given an
initial velocity vin a tangential direction. The object starts to slide along the
hoop. Find the angular velocity of the hoop when the object reaches the top of
the hoop.
Solution
Step 1: Identify the conservation of energy principle.
The system has gravitational potential energy initially (when the object is
at the bottom of the hoop) and finally (when the object is at the top of the
13
hoop) along with the kinetic energy at both points. Assuming no energy is lost
due to friction or air resistance, we can apply the principle of conservation of
energy:
Initial energy =Final energy
Step 2: Write the expression for the initial energy.
The initial energy consists of the kinetic energy of the object and the grav-
itational potential energy of the hoop and object. The kinetic energy of the
object is 1
2mv2and the potential energy at the bottom is −(M+m)gR (taking
the height −Ras the reference level).
Therefore, the initial energy is:
Ei=1
2mv2+0+(−M−m)gR
Step 3: Write the expression for the final energy.
At the top of the hoop, the object has no kinetic energy and the hoop has
rotational kinetic energy. The potential energy is −(M+m)g(R+R).
Therefore, the final energy is:
Ef= 0 + 1
2Iω2−(M+m)g2R
where I=MR2is the moment of inertia of the hoop and ωis the angular
velocity of the hoop.
Step 4: Apply the conservation of energy principle.
Setting the initial energy equal to the final energy:
1
2mv2−(M+m)gR =1
2MR2ω2−2(M+m)gR
Step 5: Solve for the angular velocity ω.
Solving for ω:
ω=√2
MR2(1
2mv2−(M+m)gR + 2(M+m)gR)
Thus, the angular velocity of the hoop when the object reaches the top is
given by the above expression.
Question 19
Question
A solid sphere of mass mand radius ris set to roll without slipping along a
horizontal surface with angular speed ω. The sphere then rolls up a frictionless
incline with an angle of elevation θ.
14
If the sphere reaches a maximum height hon the incline before reversing
direction, determine the coefficient of kinetic friction between the sphere and
the incline.
(Note: Assume the sphere rolls perfectly without slipping and air resistance
is negligible.)
Solution
Step 1: Write down the equations for conservation of energy. The total me-
chanical energy of the system at the bottom of the incline is equal to the total
mechanical energy at the maximum height, neglecting any energy losses due to
friction:
K0+U0=Kmax +Umax
where K0=1
2Iω2is the kinetic energy of rotation at the bottom of the incline,
U0=mgh is the gravitational potential energy at the bottom of the incline,
Kmax =1
2Iω2
max is the maximum kinetic energy of rotation at the maximum
height, and Umax =mghmax is the maximum gravitational potential energy at
the maximum height.
Step 2: Express the moment of inertia in terms of the radius of the sphere.
For a solid sphere, I=2
5mr2.
Step 3: Calculate the angular speed at the bottom of the incline. Using
the fact that the sphere rolls without slipping, we have ω=v
r, where vis the
linear speed at the bottom of the incline. Since the sphere is rolling without
slipping, the linear speed is related to the angular speed by v=rω. Thus,
ω=v
r=v
v
r=v
v/r =v
v/r = 1.
Step 4: Find the angular speed at the maximum height. Since energy is
conserved, we have ωmax =√5gh
7r.
Step 5: Write down the relation between the linear speed and the angular
speed. At the maximum height, the linear speed is given by vmax =rωmax.
Step 6: Use the work-energy principle to determine the coefficient of kinetic
friction. The work done by the friction force as the sphere rolls up the incline
is equal to the change in mechanical energy:
Wfriction = ∆K+ ∆U=Kmax −K0+Umax −U0
Step 7: Determine the work done by the friction force. The work done by
the friction force is given by Wfriction =fk·d, where fkis the force of kinetic
friction and dis the distance along the incline.
Step 8: Express the distance along the incline in terms of the height and
angle. The distance along the incline is d=h/ sin(θ).
Step 9: Determine the force of kinetic friction in terms of the coefficient
of kinetic friction. The force of kinetic friction is given by fk=µkN, where
N=mg cos(θ)is the normal force.
15
Step 10: Substitute the expressions for work done by friction and force of
kinetic friction into the work-energy equation.
µkmg cos(θ)h
sin(θ)=5
7mgh −1
2(2
5mr2)(5gh
7r2)
Step 11: Solve for the coefficient of kinetic friction µk.
µk=5
7(1−1
5)cot(θ)
Therefore, the coefficient of kinetic friction between the sphere and the in-
cline is µk=20
35 cot(θ).
Question 20
Question
A gyroscope consists of a disk of radius Rand mass Mspinning at an angular
velocity ωabout its central axis. The gyroscope is mounted on a frictionless
pivot at the center which allows it to rotate freely. Initially, the gyroscope is
spinning with the axis of rotation vertical. At a certain instant, a torque τis
applied to the gyroscope about its center which causes the gyroscope to precess
with a constant angular velocity Ωabout a vertical axis.
If the gyroscopic precession causes the gyroscope’s axis to make an angle θ
with the vertical, determine the angular momentum of the gyroscope about the
pivot point in terms of R,M,ω,θ, and g, the acceleration due to gravity.
Solution
Step 1: The angular momentum of the gyroscope about the pivot point can be
written as the sum of the angular momentum due to the spinning of the disk
and the angular momentum due to the precession:
L=Lspin +Lprecession
Step 2: The angular momentum due to the spinning of the disk can be calculated
as Lspin =Iω, where Iis the moment of inertia of the disk. For a disk rotating
about its central axis, the moment of inertia is given by I=1
2MR2:
Lspin =1
2MR2ω
Step 3: The angular momentum due to the precession can be calculated as
Lprecession =IΩ, where Iis the moment of inertia of the disk. The moment of
inertia for precession about the pivot point can be approximated as the moment
of inertia about the center times the cosine of the angle θ:
Lprecession =(1
2MR2)Ω cos(θ)
16
Step 4: Summing the two angular momenta, we get:
L=1
2MR2ω+(1
2MR2)Ω cos(θ)
Therefore, the angular momentum of the gyroscope about the pivot point is
1
2MR2ω+(1
2MR2)Ω cos(θ).
Question 21
Question
A disk of mass mand radius ris rotating with an angular velocity ωabout a
vertical axis passing through its center. The disk is mounted on a frictionless
horizontal axle at one edge of the disk. The disk is brought to a stop in tseconds
by absorbing the kinetic energy of the rotating disk. Find the magnitude and
direction of the torque applied to the disk. Given: m= 2 kg, r= 0.5m,
ω= 10 rad/s, t= 5 s.
Solution
Step 1: Find the initial kinetic energy of the rotating disk. The rotational
kinetic energy of the disk is given by:
K=1
2Iω2
Where Iis the moment of inertia of the disk. For a disk rotating about an
axis perpendicular to its plane at one edge, the moment of inertia is I=1
2mr2.
Substitute m= 2 kg, r= 0.5m, and ω= 10 rad/s into the equation to find K.
Step 2: Find the final kinetic energy of the disk. When the disk is brought
to a stop, the final kinetic energy is zero.
Step 3: Calculate the work done on the disk. The work done on the disk is
equal to the change in kinetic energy:
W=Kfinal −Kinitial =−Kinitial
The negative sign indicates that work is done on the system to stop the rotation.
Step 4: Calculate the torque applied to the disk. The work done on the
system is equal to the torque applied multiplied by the angle through which it
acts:
W=τθ
Since the disk rotates through 2πradians (one full rotation) during the stopping
process, θ= 2π. Thus, τ=−Kinitial
θ. Substitute the calculated value of Kinitial
and θ= 2πinto the equation to find the torque applied to the disk.
Step 5: Determine the direction of the torque. Since the disk is rotating
about a vertical axis, the torque required to stop it must be in the opposite
direction to the rotation.
17
Question 22
Question
A thin circular disk of radius Rand mass Mis rotating about an axis through
its center with an angular velocity ω. The disk is placed on a horizontal table.
Suddenly, a small force Fis applied perpendicular to the plane of the disk at
its rim. Determine the subsequent motion of the disk.
Solution
Step 1: Determine the direction of precession.
The force Fapplied perpendicularly at the rim of the disk will create a
torque about the center of the disk. This torque will cause the disk to precess.
The direction of the precession can be determined using the right-hand rule.
Step 2: Calculate the torque.
The torque acting on the disk is given by τ=F r, where ris the radius of
the disk.
Step 3: Calculate the angular acceleration.
The torque τwill cause an angular acceleration of the disk. The angular
acceleration αcan be calculated using the equation τ=Iα, where Iis the
moment of inertia of the disk.
Step 4: Determine the precession frequency.
The precession frequency Ωis related to the angular acceleration αby the
equation Ω = α
ω. This gives the rate at which the disk precesses.
Step 5: Describe the subsequent motion.
The disk will start precessing about the vertical axis passing through its
center. The rate of precession depends on the angular acceleration and the
initial angular velocity of the disk.
Question 23
Question
A solid cylindrical disk of mass Mand radius Ris rotating about its central
axis with an angular velocity ω. The disk is mounted on a pivot at one end of a
massless rod of length L, with the other end free. The system is released from
rest in a horizontal position. Calculate the angular velocity of the disk just as
the rod becomes vertical. Assume the moment of inertia of the disk about its
central axis is 1
2MR2and neglect any friction.
Solution
Step 1: When the rod becomes vertical, the center of mass of the disk will
follow a circular path around the pivot point. The forces acting on the system
18
are gravity mg acting at the center of mass of the disk downward, the normal
force provided by the pivot point, and the tension Tin the rod.
Step 2: The net force acting on the system in the vertical direction is equal
to the centripetal force required to keep the disk moving in a circular path at
that moment:
T−mg =Mv2
L
where vis the speed of the center of mass of the disk at that moment.
Step 3: The centripetal acceleration of the center of mass of the disk at that
moment is given by:
a=v2
L
Step 4: The torque about the pivot point due to the tension Tis equal to
the net external torque acting on the system, causing the disk to rotate:
T·R=I·α
where Iis the moment of inertia of the disk and αis the angular acceleration
of the disk.
Step 5: The angular acceleration αis related to the linear acceleration aby:
α=a
R
Step 6: Substitute the expressions for Tand αinto the torque equation:
(mg +Mv2
L)·R=1
2MR2·v2
L·R
Step 7: By rearranging and solving the equation, we can find the angular
velocity ωjust as the rod becomes vertical:
ω=√3gL
4R
Question 24
Question
A uniform square plate of side length aand mass mis spinning about an axis
through its center perpendicular to the plate at an angular speed ω. The plate
is placed on a frictionless table. Determine the ratio of the kinetic energy of the
plate to the kinetic energy of a particle of mass mmoving in a circle with the
same angular speed ωand radius a/2.
19
Solution
Step 1: Let’s first calculate the moment of inertia of the square plate spinning
about its axis passing through its center. The moment of inertia of a square
plate rotating about an axis passing through its center and perpendicular to the
plate is given by I=1
6ma2.
Step 2: The kinetic energy of the plate is given by
KEplate =1
2Iω2=1
2(1
6ma2)ω2=1
12ma2ω2
Step 3: Now, let’s calculate the kinetic energy of a particle of mass mmoving
in a circle with radius a/2. The kinetic energy of a particle moving in a circle
is given by KE =1
2mv2, where vis the speed of the particle.
Step 4: The speed of the particle moving in a circle with radius a/2and
angular velocity ωis given by v=a
2ω.
Step 5: Substituting v=a
2ωinto the equation for kinetic energy, we get
KEparticle =1
2m(a
2ω)2
=1
2m(a2ω2
4)=1
8ma2ω2
Step 6: Finally, we find the ratio of the kinetic energy of the plate to the
kinetic energy of the particle:
KEplate
KEparticle
=
1
12 ma2ω2
1
8ma2ω2=1
12 ×8
1=2
3
Therefore, the ratio of the kinetic energy of the plate to the kinetic energy
of the particle is 2
3.
Question 25
Question
A bicycle wheel has a radius of 0.5 m and a mass of 2 kg. The wheel is spinning
with an angular speed of 10 rad/s around its axis of rotation. If the wheel is
laying on its side with the axis horizontal, what torque would need to be applied
to keep it from falling over due to gravity?
Solution
Step 1: The moment of inertia for a solid disk rotating about an axis through its
center is given by the formula I=1
2MR2, where Mis the mass of the disk and
Ris the radius of the disk. Given that R= 0.5m and M= 2 kg, we calculate
the moment of inertia I:
I=1
2(2 kg)(0.5m)2= 0.5kg ·m2
20
Step 2: The gravitational force acting on the wheel can be calculated as
mg, where mis the mass of the wheel and gis the acceleration due to gravity
(9.81 m/s2). The lever arm of this force is equal to the radius of the wheel
(0.5 m). The torque due to gravity is given by the formula τgravity =mgR.
Substitute m= 2 kg, g= 9.81 m/s2, and R= 0.5m:
τgravity = (2 kg)(9.81 m/s2)(0.5m) = 9.81 Nm
Step 3: The torque needed to keep the wheel from falling over must be equal
in magnitude but opposite in direction to the torque due to gravity. Thus, the
torque needed to prevent the wheel from falling over is 9.81 Nm in the opposite
direction.
Therefore, a torque of 9.81 Nm would need to be applied to the bicycle wheel
to keep it from falling over due to gravity.
21
