PHYS 201 - GENERAL PHYSICS I Gyroscopic Motion and Stability Question Bank Set 4

PHYS 201 - GENERAL PHYSICS I -

Gyroscopic Motion and Stability

Question Bank - Set 4

Liberty University

Question 1

Question

A thin uniform rod of length Land mass Mis pivoted at one end and free

to rotate in a vertical plane without friction. The system is initially at rest

and then the rod is given a sharp impulse at the free end perpendicular to the

rod. Determine the angular velocity of the rod just after the impulse is given.

Consider the rotation axis perpendicular to the rod at the pivot point.

Solution

Step 1: We can begin by considering the conservation of angular momentum.

The angular momentum of the rod just after the impulse is given consists of

two parts: the angular momentum of the pivot point (which is zero) and the

angular momentum of the rod rotating about the pivot point.

Step 2: The angular momentum of the rod is given by the equation:

L=Iω

where Lis the angular momentum, Iis the moment of inertia of the rod about

the pivot point, and ωis the angular velocity of the rod.

Step 3: The moment of inertia of a thin rod rotating about an axis perpen-

dicular to the rod and passing through one end is given by:

I=1

3ML2

Step 4: After the impulse is given, the angular momentum of the rod is equal

to the impulse’s angular impulse (torque ∆L) applied to the rod. The angular

impulse is given by:

∆L=Iω

Step 5: Therefore, we have:

∆L=1

3ML2ω

Step 6: The impulse is perpendicular to the rod, so the direction of the

angular momentum is perpendicular to the rod’s length. By conservation of

angular momentum, the initial angular momentum is zero.

Step 7: Setting ∆Lequal to zero, we find:

3ML2ω= 0

Step 8: Solving for ω, we get:

ω= 0

Step 9: Thus, the angular velocity of the rod just after the impulse is given

is ω= 0.

Question 2

Question

A disk of radius Rand mass Mis spinning with an angular velocity ωabout an

axis perpendicular to its plane. The disk is mounted on a frictionless vertical

axle through its center, allowing it to spin freely. A slight disturbance causes

the disk to tilt so that the axis of rotation makes an angle θwith the vertical.

Determine the precession angular velocity of the disk.

Solution

Step 1: Identify the known quantities and define relevant variables: Given: -

Radius of the disk, R- Mass of the disk, M- Angular velocity of the spinning

disk, ω- Angle of tilt from the vertical, θ

We are asked to find the precession angular velocity of the disk.

Let: - I= moment of inertia of the disk about its central axis - α= angular

acceleration of precession - Ω= precession angular velocity

Step 2: Find the moment of inertia, I: The moment of inertia of a disk

rotating about its central axis is given by the equation: I=1

2MR2

Step 3: Apply Euler’s Equations for Rotation: Euler’s Equations for Rota-

tion can be applied in this problem. The relevant equation is: IdΩ

dt =ω·L·cos(θ)

where Lis the angular momentum of the spinning disk.

Step 4: Calculate the angular momentum, L: The angular momentum of

the spinning disk is given by: L=Iω

Step 5: Substitute the expression for Linto Euler’s equation: IdΩ

dt =ω·Iω ·

cos(θ)

Step 6: Solve for the precession angular velocity, Ω:dΩ

dt =ωcos(θ)

Therefore, the precession angular velocity of the disk is given by Ω = ωcos(θ)

Question 3

Question

A disk with radius rand mass mrotates about an axis perpendicular to the

disk with angular speed ω. The disk is placed on a horizontal surface and a

small nail is driven into the center of the disk to act as a pivot point. The disk

remains frictionless. What is the minimum angular speed at which the disk will

remain stable?

Solution

To find the minimum angular speed at which the disk will remain stable, we

need to consider the stability condition for a spinning disk on a pivot.

Step 1: Determine the condition for stability. For a spinning disk to remain

stable on a pivot, the center of mass of the disk must lie vertically below the

pivot point.

Step 2: Find the center of mass of the disk. The center of mass of the disk

lies at a distance r

2from the pivot point along the vertical axis.

Step 3: Set up the stability condition. The condition for stability can be

expressed as: mg(r

2)≥1

2mω2(r)2

Step 4: Solve for the minimum angular speed ω.

mg(r

2)≥1

2mω2(r)2

2mω2(r)2≤mg(r

ω2≥2g

ω≥√2g

Thus, the minimum angular speed at which the disk will remain stable is

ω≥√2g

Question 4

Question

A disc of mass Mand radius Rrolls along a horizontal surface without slipping.

The disc is then given a sharp impulse at the edge of its rim, causing it to

start spinning. The angular velocity of the spinning disc is ω, and the angular

momentum is L. Determine the expressions for the kinetic energy Kand the

total mechanical energy Eof the disc in terms of M,R,ω, and L.

Solution

Step 1: The kinetic energy Kof the disc consists of two parts: translational

kinetic energy and rotational kinetic energy. The translational kinetic energy

is given by 1

2Mv2, where vis the linear velocity of the disc. Since the disc is

rolling without slipping, v=Rω. The rotational kinetic energy is 1

2Iω2, where

Iis the moment of inertia. For a disc rotating about its center, I=1

2MR2.

Therefore, the total kinetic energy Kis

K=1

2M(Rω)2+1

2(1

2MR2)ω2=1

2MR2ω2+1

4MR2ω2=3

4MR2ω2

Step 2: The total mechanical energy Eof the disc is the sum of the kinetic

energy Kand the potential energy U. Since the disc is rolling on a horizontal

surface, the potential energy remains constant. Therefore, the total mechanical

energy Eis equal to the kinetic energy K, which can be written as

E=K=3

4MR2ω2

Question 5

Question

A thin circular hoop of mass Mand radius Ris rolling without slipping along a

horizontal surface with a linear velocity of v. The hoop is struck by an impulse

which is perpendicular to the plane of the hoop at its center. Find the angular

velocity of the hoop immediately after the impulse and the distance it travels

before coming to rest. Assume the impulse causes the hoop to have only a pure

rotational motion.

Given: M= 2 kg, R= 0.5m, v= 5 m/s

Solution

Step 1: Conservation of Angular Momentum

The impulse causes a torque with respect to the center of mass which results

only in a change in angular momentum, so we can use the law of conservation

of angular momentum to find the angular velocity after the impulse.

The initial angular momentum of the hoop is given by:

Linitial =Iω = (MR2)ω

Where Iis the moment of inertia of the hoop and ωis the initial angular

velocity.

The final angular momentum of the hoop is given by:

Lfinal =Iω′

Where ω′is the final angular velocity after the impulse is applied.

Since angular momentum is conserved, we have:

Linitial =Lfinal

MR2ω=MR2ω′

ω′=ω=v

R=5m/s

0.5m= 10 rad/s

Step 2: Distance Traveled Before Coming to Rest

The distance the hoop travels before coming to rest can be found using the

work-energy principle.

The work done by the impulse is equal to the change in kinetic energy of the

system. The initial kinetic energy of the hoop is due to both linear and rotational

motion, while the final kinetic energy is solely due to rotational motion.

The work done by the impulse is given by:

W= ∆KE =KEfinal −KEinitial

W=1

2Iω′2−(1

2Iω2+1

2Mv2)

Substitute the known values:

W=1

2MR2ω′2−(1

2MR2ω2+1

2Mv2)

Solve for the work Wand then use the work-energy principle to find the

distance traveled by the hoop before coming to rest.

Question 6

Question

A thin hoop with radius Rand mass Mis rotating about a vertical axis at an

angular speed ω. The hoop is released from rest in a horizontal position. What

is the angular speed of the hoop when it reaches the vertical position?

Solution

Step 1: Find the angular momentum of the hoop when it is released The initial

angular momentum of the hoop is given by:

Linitial =Iω

Where Iis the moment of inertia of the hoop and ωis the initial angular speed.

For a hoop rotating about an axis passing through its center (like in this

case):

I=MR2

Therefore, the initial angular momentum is:

Linitial =MR2ω

Step 2: Find the angular speed when the hoop reaches the vertical position

When the hoop reaches the vertical position, the radius is still Rbut the moment

of inertia changes since the axis of rotation is now at the edge of the hoop.

The final moment of inertia I′is given by:

I′=M(R2+R2) = 2MR2

The conservation of angular momentum states that:

Linitial =Lfinal

MR2ω= 2MR2ω′

ω′=ω

Therefore, the angular speed of the hoop when it reaches the vertical position

is ω

Question 7

Question

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

which makes an angle θwith the horizontal. The sphere starts from rest at the

top of the incline. What is the final linear velocity of the center of mass of the

sphere when it reaches the bottom of the incline?

Solution

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

The net torque about the center of mass of the sphere is due to the gravitational

force. The torque due to gravity is τ=m·g·R·sin(θ), where mis the mass of

the sphere, gis the acceleration due to gravity, and Ris the radius of the sphere.

The net torque is also equal to the moment of inertia of a solid sphere about

its center of mass times the angular acceleration: τ=I·α. Since the sphere is

rolling without slipping, we have a=R·α, where ais the linear acceleration

down the incline, and αis the angular acceleration.

Step 2: Solve for the acceleration down the incline.

Setting these two torque expressions equal, we have:

m·g·R·sin(θ) = 2

5mR2·α

⇒a=Rα =5

2gsin(θ)

Step 3: Determine the final velocity of the sphere at the bottom of the

incline.

Using the kinematic equation for motion along a straight line:

v2=u2+ 2as

Where: - vis the final velocity (which we want to find) - uis the initial velocity,

which is 0 since the sphere starts from rest - ais the acceleration down the

incline - sis the distance down the incline, which is the length of the incline L,

given by L=Rsin(θ)/cos(θ) = Rtan(θ)

Substitute u= 0,a= (5/2)gsin(θ), and s=Rtan(θ)into the kinematic

equation:

v2= 0 + 2 (5

2gsin(θ))(Rtan(θ))

v=√5gR sin(θ) tan(θ)

Question 8

Question

A car with a mass of 1500 kg is approaching a turn. As the car enters the

turn, it experiences a centripetal acceleration of 3 m/s2towards the center of

the turn, which has a radius of 50 meters. If the coeﬀicient of static friction

between the tires of the car and the road is 0.6, what is the maximum speed

the car can have to avoid sliding off the road as it turns?

Solution

Step 1: Determine the maximum frictional force that can act on the car. Given

that the car is experiencing a centripetal acceleration of 3 m/s2towards the

center of the turn, the maximum static frictional force that can act on the car

is:

ffriction =m·ac

ffriction = 1500 kg ·3m/s2

ffriction = 4500 N

Step 2: Calculate the maximum force of static friction that can act on the

car. The maximum force of static friction that can act on the car is determined

by:

ffriction,max =µs·N

where: - µs= 0.6is the coeﬀicient of static friction, - Nis the normal force

acting on the car.

Step 3: Determine the normal force acting on the car. The normal force

acting on the car is equal to the gravitational force acting on the car, since the

car is not moving vertically. Therefore,

N=mg

where: - g= 9.81 m/s2is the acceleration due to gravity.

Step 4: Substitute the known values to find the maximum speed the car can

have. Since the car is experiencing a centripetal force due to its velocity and

acceleration, the maximum speed the car can have without sliding off the road

is obtained from the equation:

ffriction,max =mv2

where: - vis the velocity of the car, - R= 50 m is the radius of the turn.

By setting the two expressions for the maximum frictional force equal to

each other, we have:

µs·mg =mv2

Solving for v:

v=√µs·g·R

v=√0.6×9.81 ×50

v=√294.3

v≈17.16 m/s

Therefore, the maximum speed the car can have to avoid sliding off the road

is approximately 17.16 m/s.

Question 9

Question

A disk of mass Mand radius Rrotates with an angular velocity ωabout a fixed

horizontal axis, passing through the center of the disk. The disk is supported

at one end of a light, horizontal rod of length L, the other end being attached

to a fixed point. Determine the angular velocity of precession Ω.

Solution

Let’s consider the forces acting on the disk. The only significant forces acting on

the disk are the gravitational force mg acting downward at the center of mass

and the normal force Nacting upward at the pivot point. The gravitational

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

mg cos θperpendicular to the rod.

Step 1: Set up equations of motion for the rotation and the pre-

cession

For rotation about the pivot point: Summing torques about the pivot point:

Rmg sin θ=Iα

Where I=1

2MR2is the moment of inertia of the disk and αis the angular

acceleration.

For precession: Summing forces in the xdirection:

N=mg cos θ

Summing torques about the pivot point:

mLg sin θ=IΩω

Where Ωis the angular velocity of precession.

Step 2: Relate the two equations

From the equation for rotation about the pivot point, we have:

Rmg sin θ=1

2MR2α

α=2Rmg sin θ

MR2=2gsin θ

Substitute αinto the equation for precession:

mLg sin θ=IΩ2gsin θ

mLg =1

2MR2Ω2gsin θ

mLg =MgΩ sin θ

Ω = mLg

Mg sin θ

Thus, the angular velocity of precession Ωis L

Rcot θ.

Question 10

Question

A thin horizontal circular disc of radius Rand mass Mis rotating about a

vertical axis passing through its center at an angular speed ω. A small piece of

the disc of size land mass mbreaks off tangentially and flies off horizontally.

Find the angular speed of the disc just after the mass mbreaks off.

Solution

Step 1: Let Idisc be the moment of inertia of the disc about the vertical axis

through its center. The moment of inertia of the thin disc about this axis is

given by Idisc =1

2MR2.

Step 2: Upon breaking off the mass mtangentially from the disc, the angular

momentum of the disc-mass system is conserved. The initial angular momentum

is Idiscω, where ωis the initial angular speed of the disc.

Step 3: The final angular momentum is (Idisc −mR2)ω′, where ω′is the

angular speed of the disc just after the mass mbreaks off.

Step 4: Using the conservation of angular momentum, we have Idiscω=

(Idisc −mR2)ω′.

Step 5: Substituting Idisc =1

2MR2into the equation gives 1

2MR2ω=

2MR2−mR2)ω′.

Step 6: Solving for ω′, we find ω′=M

M−2mω. Thus, the angular speed of the

disc just after the mass mbreaks off is M

M−2mω.

Question 11

Question

A thin uniform rod of length Land mass Mrotates in a horizontal plane about

one end with an angular velocity ω. The other end is pivoted to allow motion in

any direction. Show that the angular velocity of a rod when it makes an angle

θwith the vertical is given by ω′=ω

√1+ 3

2sin2θ.

Solution

Step 1: Calculate the angular velocity when the rod is horizontal (θ= 0). In

this case, the angular velocity ω0=ω.

Step 2: Apply conservation of angular momentum. The initial angular mo-

mentum is L0=Iω0, where Iis the moment of inertia of the rod rotating about

the end making an angle θ= 0.

Step 3: Calculate the moment of inertia for the rod about the pivoted end.

For a uniform rod rotating about an end, the moment of inertia is I=1

3ML2.

Step 4: Use conservation of angular momentum to find the final angular

velocity when the rod makes an angle θwith the vertical. The final angular

momentum is Lf=Iω′. Since angular momentum is conserved, L0=Lf.

Step 5: Substitute the values of L0,I,Lf, and ω0into the conservation of

angular momentum equation. This gives us Iω0=Iω′.

Step 6: Solve for ω′to find the angular velocity when the rod makes an angle

θwith the vertical. We have ω′=Iω0

3ML2ω

3ML2=ω.

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

vertical is given by ω′=ω

√1+ 3

2sin2θ.

Question 12

Question

A solid disk of radius Rand mass Mis rotating with an angular speed ωabout

its axis perpendicular to the plane of the disk. The disk is supported on a

frictionless axle, but a small piece of the disk suddenly breaks free and flies off

at a speed of vtangential to the edge of the disk. What is the resulting angular

speed of the disk after the piece breaks free?

Solution

1. We can start by considering the conservation of angular momentum. Initially,

the total angular momentum of the system (disk + piece) is given by the sum

of the angular momentum of the disk and the angular momentum of the piece

that breaks free. This is given by:

Linitial =Idiskω+mrv

where Idisk =1

2MR2is the moment of inertia of the disk, mis the mass of the

piece that breaks free, ris the radius of the disk, and vis the speed at which

the piece breaks free.

2. After the piece breaks free, the total angular momentum of the system is

only due to the disk, and it is given by:

Lfinal =Idiskωfinal

3. Since angular momentum is conserved, we have Linitial =Lfinal:

Idiskω+mrv =Idiskωfinal

4. Substituting the moment of inertia of the disk Idisk =1

2MR2and rear-

ranging the equation, we get:

2MR2ω+mrv =1

2MR2ωfinal

5. Solving for ωfinal, we find:

ωfinal =1

2(2MR2ω+ 2mrv

MR2)

6. Simplifying the expression further, we get:

ωfinal =Mω +mv

Therefore, the resulting angular speed of the disk after the piece breaks free

is Mω+mv

2M.

Question 13

Question

A thin uniform rod of length Land mass Mis pivoted at one end and allowed

to swing freely in a vertical plane. The rod is released from rest when it makes

an angle of θ0with the vertical. Determine the angular velocity of the rod just

before it becomes vertical. Given: L= 1.5m, M= 2 kg, θ0= 30◦

Solution

Step 1: The potential energy when the rod is released is given by U=−M gL cos θ,

where Mis the mass of the rod, gis the acceleration due to gravity, Lis the

length of the rod, and θis the angle of the rod with the vertical. Step 2: The ki-

netic energy at the lowest point is all rotational and is given by K=1

2Iω2, where

Iis the moment of inertia of the rod about the pivot point, and ωis the angular

velocity of the rod. Step 3: The only force doing work as the rod swings is grav-

ity, so the total mechanical energy E=K+Uis conserved. Step 4: At the start-

ing angle θ0= 30◦, we have U=−MgL cos(30◦)and K= 0. Step 5: At the low-

est point of the swing, the rod is vertical with θ= 90◦, so U=−MgL cos(90◦) =

0and K=1

2Iω2. Step 6: Setting the initial and final total mechanical energies

equal gives −M gL cos(30◦) = 1

2Iω2

f, where ωfis the angular velocity just before

the rod becomes vertical. Step 7: The moment of inertia of a thin rod rotating

about one end is I=1

3ML2. Step 8: Substituting the values into the energy

conservation equation, we get −Mg(1.5) cos(30◦) = 1

2(1

3M(1.5)2)ω2

f. Step 9:

Simplifying, we get −(2)(9.81)(1.5) (√3

2)=1

2(1

3(2)(1.5)2)ω2

f. Step 10: Solving

for ωf, we find ωf=√3(2)(9.81)(1.5)√3

(1/3)(2)(1.5)2. Step 11: Therefore, the angular velocity

of the rod just before it becomes vertical is ωf≈4.67 rad/s.

Question 14

Question

A spinning top has a moment of inertia of 0.02 kg m2and spins with an angular

velocity of 10 rad/s. The top begins to wobble due to a slight disturbance,

causing its axis of rotation to precess. Calculate the precession frequency of the

spinning top.

Solution

Step 1: The precession frequency of the spinning top can be calculated using

the formula:

Precession frequency =τ

where τis the torque applied to the spinning top and Lis the angular

momentum of the spinning top.

Step 2: To find the torque applied to the spinning top, we can use the

formula:

τ=I·α

where Iis the moment of inertia of the spinning top and αis the angular

acceleration of the spinning top.

Step 3: Since the spinning top wobbles and its axis of rotation precesses, we

can consider the torque as the torque due to the precession. By treating the

wobbling motion as a precession, we can relate the angular acceleration αto

the precession frequency Ωp:

α= Ωp×10

Step 4: Substituting α= Ωp×10 into the torque formula gives:

τ=I×Ωp×10

Step 5: The angular momentum Lof the spinning top is given by:

L=I×ω

where ωis the angular velocity of the spinning top.

Step 6: Substituting the values into the formulas, we get:

τ= 0.02 ×Ωp×10

L= 0.02 ×10

Step 7: Finally, substituting the torque and angular momentum into the

precession frequency formula gives:

Precession frequency =0.02 ×10

0.02 ×10 = 1 rad/s

Therefore, the precession frequency of the spinning top is 1rad/s.

Question 15

Question

A rigid body consists of a uniform disk of radius Rand mass Mattached to a

thin rod of length Land mass m. The disk and rod lie in the same plane and

rotate about a horizontal axis through one end of the rod. The body is rotating

with an angular velocity ω. Determine the angular momentum of the system

and the gyroscopic torque if the system tilts by an angle θfrom the vertical.

Solution

Step 1: The angular momentum of the system is given by the sum of the angular

momentum of the disk and the rod:

L=Idiskωdisk +Irodωrod

where

Idisk =1

2MR2, ωdisk =ω

and

Irod =1

3mL2, ωrod =ωcos θ

Substitute the values into the equation:

L=(1

2MR2)ω+(1

3mL2)ωcos θ

L=1

2MR2ω+1

3mL2ωcos θ

Step 2: The gyroscopic torque due to the tilting is given by:

τ=Itotalα

where

Itotal =Idisk +Irod

and

α= ˙ωrod =−ωsin θ

Substitute the expressions for Itotal and αinto the equation:

τ= (Idisk +Irod)(−ωsin θ)

τ=(1

2MR2+1

3mL2)(−ωsin θ)

τ=−(1

2MR2+1

3mL2)ωsin θ

Question 16

Question

A gyroscope consists of a disk mounted on an axle. The disk has a radius of 0.1

m and a mass of 2 kg. If the gyroscope is spinning at 200 rad/s, calculate the

magnitude of its angular momentum.

Solution

Step 1: Recall the formula for angular momentum of a rotating object:

L=Iω

where Lis the angular momentum, Iis the moment of inertia, and ωis the

angular velocity.

Step 2: Calculate the moment of inertia of the gyroscope. For a solid disk

rotating about its central axis, the moment of inertia is given by:

I=1

2mr2

where mis the mass of the disk and ris the radius.

Substitute the given values: m= 2 kg and r= 0.1m.

I=1

2×2×(0.1)2

I= 0.01 kg m2

Step 3: Now, substitute the moment of inertia and the angular velocity into

the angular momentum formula:

L= 0.01 ×200

L= 2 kg m2/s

Therefore, the magnitude of the angular momentum of the gyroscope is 2 kg

m²/s.

Question 17

Question

A thin uniform rod of length Land mass mrotates with constant angular speed

ωabout one end perpendicular to the rod. The rod starts to fall under the

influence of gravity. What is the angular speed of the rod when it is in the

vertical position?

Solution

Step 1: Let’s denote the angular speed of the rod when it is in the vertical

position as ωv.

Step 2: At the vertical position, the center of mass of the rod is at a distance

of L/2from the axis of rotation. Thus, the potential energy of the rod at this

position is given by:

U=mgh =mg (L

2)cos(90◦) = mgl

where h= (L/2) cos(90◦) = L/2is the vertical height.

Step 3: The total initial energy of the system is the sum of the kinetic energy

and potential energy. At the initial position (horizontal), the total energy is

given by:

Ei=KEi+P Ei=1

2Iω2+ 0,

where I=1

3mL2is the moment of inertia of the rod rotating about one end.

Step 4: At the vertical position, the total energy is given by:

Ev=KEv+P Ev=1

2Iω2

v+mgl

Step 5: Since the system is isolated, the total energy is conserved. Therefore,

we have Ei=Ev:1

2Iω2=1

2Iω2

v+mgl

Step 6: Substituting the expressions for the moment of inertia and the po-

tential energy into the conservation of energy equation, we get:

6mL2ω2=1

6mL2ω2

v+mgl

Step 7: Solving for ωv, we find:

ω2

v=ω2−3g

2L.

Step 8: Finally, the angular speed of the rod when it is in the vertical position

(ωv) is given by:

ωv=√ω2−3g

2L.

Question 18

Question

A solid sphere of mass mand radius ris initially rolling without slipping along a

horizontal surface at a speed of v0. The sphere then encounters a ramp inclined

at an angle θ. Find the minimum coeﬀicient of static friction between the sphere

and the ramp to prevent slipping as the sphere rolls up the ramp.

Solution

Step 1: The forces acting on the sphere when it is rolling up the ramp are

gravity (mg) acting downwards, the normal force (N) acting perpendicular to

the ramp, static friction (fs) acting along the ramp, and the force of tension

(T) along the direction of the ramp.

Step 2: The forces in the perpendicular direction cancel out, so N=mg cos θ.

The forces in the parallel direction are responsible for accelerating the sphere

up the ramp.

Step 3: The net force up the ramp is given by fs−mg sin θ=ma, where a

is the acceleration up the ramp.

Step 4: The acceleration can be expressed in terms of angular acceleration

by a=αr, where αis the angular acceleration of the sphere.

Step 5: The torque equation about the center of mass of the sphere is fsr=

Iα, where Iis the moment of inertia of the sphere about its center of mass.

Step 6: The moment of inertia of a solid sphere about its center of mass is

5mr2.

Step 7: Substituting I=2

5mr2and a=αr into the torque equation gives

fsr=2

5mr2α.

Step 8: Combining the net force equation and the torque equation, we get

fs−mg sin θ=2

5mrα.

Step 9: In order to find the minimum coeﬀicient of static friction, we must

consider the point where the static friction force is at its maximum, giving us

fs=µsN.

Step 10: Substituting fs=µsNinto the equation from step 8, we have

µsN−mg sin θ=2

5mrα.

Step 11: Substituting N=mg cos θinto the equation, we get µsmg cos θ−

mg sin θ=2

5mrα.

Step 12: We know that α=a

r=fs

mr , so we substitute α=fs

mr into the

equation.

Step 13: By solving the equation for µs, we find that the minimum coeﬀicient

of static friction needed to prevent slipping is µs=2

5(sin θ+µkcos θ

cos θ), where µkis

the coeﬀicient of kinetic friction.

Step 14: Therefore, the minimum coeﬀicient of static friction required is

µs=2

5(sin θ+µkcos θ

cos θ).

Question 19

Question

A thin circular hoop of mass mand radius Ris rotating about a vertical axis

through its center with an angular speed ω. A small mass mis placed at a

distance rfrom the axis of the hoop. Assuming the hoop is free to rotate about

the vertical axis, calculate the angular velocity of the hoop about the vertical

axis once the mass is allowed to fall.

Solution

Step 1: The hoop and the mass are initially rotating together about the vertical

axis with angular velocity ω. When the small mass falls, the hoop will start

rotating about the vertical axis with a new angular velocity Ω. We will apply

the conservation of angular momentum.

The initial angular momentum of the system (Isysω) is equal to the final

angular momentum of the system (IsysΩ).

Step 2: The moment of inertia of the hoop is Ihoop =mR2and the moment

of inertia of the mass about the vertical axis is Imass =mr2. The total moment

of inertia of the system is the sum of the moments of inertia of the hoop and

the mass.

So, Isys =Ihoop +Imass =mR2+mr2.

Step 3: Using the conservation of angular momentum, we can write:

Isysω=IsysΩ

(mR2+mr2)ω= (mR2+mr2)Ω

Step 4: Solving for Ω, we get:

Ω = mR2+mr2

mR2+mr2ω

Ω = ω

Therefore, the angular velocity of the hoop about the vertical axis once the

mass is allowed to fall is Ω = ω.

Question 20

Question

A disk of mass Mand radius Ris rotating with an angular velocity ωabout its

central axis. The disk is mounted on a shaft so that it is free to rotate about

a vertical axis through its center. At a certain instant, a force Fis applied

tangent to the rim of the disk. If the coeﬀicient of kinetic friction between the

axle and the disk is µk, determine the time it takes for the disk to stop rotating.

Assume the disk is initially rotating clockwise as seen from above.

Solution

Step 1: We begin by identifying the torque acting on the disk caused by the

force F. The torque from the force Fis given by τ=RF , where Ris the radius

of the disk.

Step 2: Next, we can write the equation of motion for the rotational motion

of the disk. The torque τgenerated by the force Fis equal to the moment of

inertia Iof the disk times the angular acceleration α:τ=Iα.

Step 3: The moment of inertia Ifor a thin disk rotating about an axis

through its center perpendicular to the disk is I=1

2MR2.

Step 4: From step 1, we have τ=RF , and from step 3, we have I=1

2MR2.

Substituting these into the equation from step 2 gives RF =1

2MR2α, which

simplifies to 2F=αR.

Step 5: The angular acceleration αis related to the angular velocity ωby

the equation α=dω

dt . Therefore, we have 2F=Rdω

dt .

Step 6: Separating variables and integrating both sides with respect to time,

we get ∫0

ωiωdω =∫t

02F/Rdt, where ωiis the initial angular velocity of the disk.

Step 7: Solving the integrals gives 1

2(0 −ωi) = 2F

Rt. Therefore, −1

2ωi=2F

Rt.

Step 8: Rearranging the equation gives t=−R

2Fωi. Since we want to find

the time it takes for the disk to stop rotating, we consider ωi=ω(the final

angular velocity is 0). Thus, the time it takes for the disk to stop rotating is

t=Rω

2F.

Question 21

Question

A rigid body with mass mis rotating about a fixed axis with an angular velocity

ω. At a certain instant, a torque τis applied to the body perpendicular to its

angular velocity vector. Determine the rate at which the kinetic energy of the

body changes in terms of I, the moment of inertia of the body with respect to

the rotation axis.

Solution

Step 1: The initial kinetic energy of the rotating body is given by the expression

2Iω2

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

Step 2: The final kinetic energy of the body can be written as

2I(ω+ ∆ω)2=1

2I(ω2+ 2ω∆ω+ (∆ω)2)

Step 3: The change in kinetic energy, ∆KE, is therefore

∆KE =1

2I(ω2+ 2ω∆ω+ (∆ω)2)−1

2Iω2

Step 4: Simplifying this expression, we get

∆KE =1

2I(2ω∆ω+ (∆ω)2)

Step 5: Since ∆ωis very small, we can neglect the term (∆ω)2in comparison

to 2ω∆ω. Therefore,

∆KE ≈1

2I(2ω∆ω)

Step 6: We also know that torque τis given by the expression

τ=Iα

where αis the angular acceleration.

Step 7: The rate at which work is done by the torque is given by

τ∆θ= ∆KE

where ∆θis the change in angle.

Step 8: Substituting τ=Iα into the above equation gives

Iα∆θ=1

2I(2ω∆ω)

Step 9: Simplifying this expression, we get

α∆θ=ω∆ω

Step 10: Finally, the rate at which the kinetic energy of the body changes,

dKE

dt , is given by

dKE

dt =αω =ω∆ω

∆θ

Step 11: Therefore, the rate at which the kinetic energy of the body changes

in terms of Iis

dKE

dt =ω∆ω

∆θ

Question 22

Question

A bicycle wheel with a radius of 0.3 meters and a mass of 1.5 kg rotates at a

constant angular speed of 4 radians per second. The wheel is set spinning about

a horizontal axle with the axis of rotation at one end of the axle. The moment

of inertia of the wheel about this axis is I= 0.09 kg ·m2.

Calculate the magnitude of the angular momentum of the wheel.

Solution

Step 1: Recall the definition of angular momentum, L, which is given by the

formula:

L=Iω,

where Iis the moment of inertia and ωis the angular speed.

Step 2: Substitute the given values into the formula. Here, I= 0.09 kg ·m2

and ω= 4 rad/s.

L= 0.09 ·4 = 0.36 kg ·m2/s.

Therefore, the magnitude of the angular momentum of the wheel is 0.36 kg ·

m2/s.

Question 23

Question

A gyroscope consists of a metal disk of radius 0.1 m and mass 1.5 kg rotating

at an angular speed of 300 rad/s. The disk is mounted on a frictionless axle,

which is supported by bearings on both ends. Suddenly, a torque is applied

perpendicular to the plane of the disk resulting in an angular impulse of 200

N*s. Determine the final angular speed of the gyroscope.

Solution

Step 1: Calculate the moment of inertia of the gyroscope disk.

Moment of inertia, I=1

2mr2

I=1

2×1.5kg ×(0.1m)2

I= 0.0075 kg m2

Step 2: Use the principle of conservation of angular momentum to find the

final angular speed.

Initial angular momentum, Li=I·ωi

Li= 0.0075 kg m2×300 rad/s

Li= 2.25 kg m2/s

Step 3: Calculate the final angular momentum using the angular impulse.

Angular impulse, τ∆t=I·(ωf−ωi)

200 N s = 0.0075 kg m2×(ωf−300)

Step 4: Solve for the final angular speed, ωf.

ωf=200 N s

0.0075 kg m2+ 300 rad/s

ωf= 2666.67 + 300

ωf= 2966.67 rad/s

Therefore, the final angular speed of the gyroscope is 2966.67 rad/s.

Question 24

Question

A gyroscopic rigid rotor of mass mand radius ris spun up to an angular velocity

ω0. It is then tilted at a small angle θ0from the vertical and released. Determine

the precessional period of the rotor.

Solution

Step 1: The precessional period Tpcan be calculated using the following formula:

Tp=4π2I

mgr

where Iis the moment of inertia of the rotor, mis the mass of the rotor, g

is the acceleration due to gravity, and ris the radius of the rotor. The moment

of inertia of a solid cylinder about its central axis is given by I=1

2mr2.

Step 2: Substituting I=1

2mr2into the equation for Tp, we get:

Tp=4π2(1

2mr2)

mgr

Tp=2π2mr

Step 3: Therefore, the precessional period of the rotor, Tp, is given by Tp=

2π2mr

Question 25

Question

A disk with radius 0.1 m is rotating about an axis perpendicular to the disk

and passing through its center at a rate of 30 rad/s. The disk is initially at

rest. A small marble is placed on the disk at a distance of 0.05 m from the axis

of rotation. What is the path of the marble relative to the disk if the disk’s

rotation suddenly stops? Assume no external forces are present and neglect any

friction between the marble and the disk.

Solution

Step 1: Find the initial velocity of the marble due to the rotation of the disk.

The initial velocity of the marble due to the rotation of the disk is given by

the equation:

v=r×ω

where ris the distance of the marble from the axis of rotation and ωis the

angular velocity of the disk.

Substitute r= 0.05 m and ω= 30 rad/s:

v= 0.05 ×30 = 1.5m/s

Step 2: Determine the initial direction of the velocity of the marble.

Since the marble is on the rotating disk, its initial velocity is tangential to

the disk’s surface. Therefore, the initial velocity of the marble is along the

tangent to the disk at the point where the marble is placed.

Step 3: Analyze the path of the marble relative to the disk when the disk

suddenly stops rotating.

When the disk’s rotation suddenly stops, the marble will continue to move in

a straight line in the direction of its initial velocity. This is because no external

forces are present to change the marble’s motion, and there is no friction acting

between the marble and the disk to decelerate it.

Therefore, the path of the marble relative to the disk will be a straight line

tangential to the point on the disk where the marble was initially placed.

PHYS 201 - GENERAL PHYSICS I Gyroscopic Motion and Stability Question Bank Set 4

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