PHYS 201 - GENERAL PHYSICS I Gyroscopic Motion and Stability Question Bank Set 3

PHYS 201 - GENERAL PHYSICS I -

Gyroscopic Motion and Stability

Question Bank - Set 3

Liberty University

Question 1

Question

A disk of mass mand radius ris rotating at an angular speed of ω. The disk

is placed on a horizontal surface and released from rest. Determine the angular

speed of the disk when it comes to rest after rolling a distance don the surface.

Assume the surface is frictionless.

Solution

1. First, let’s consider the initial kinetic energy of the system when the disk

is released from rest. The initial kinetic energy KEiis solely attributed to the

rotational motion and is given by:

KEi=1

2Iω2

where Iis the moment of inertia of the disk.

2. The moment of inertia of a disk rotating about its center is I=1

2mr2.

3. Substituting the moment of inertia into the equation for initial kinetic

energy gives:

KEi=1

2(1

2mr2)ω2=1

4mr2ω2

4. As the disk rolls on a frictionless surface, the total mechanical energy of

the system remains constant. Therefore, the initial kinetic energy KEiwill be

equal to the final potential energy P Efwhen the disk comes to rest. The final

potential energy P Efcan be expressed as:

P Ef=mgd

where gis the acceleration due to gravity and dis the distance rolled by the

disk.

5. Setting the initial kinetic energy equal to the final potential energy:

4mr2ω2=mgd

6. Solving for the angular speed ωgives:

ω=√4gd

7. Therefore, the angular speed of the disk when it comes to rest after rolling

a distance don the surface is √4gd

r2.

Question 2

Question

A spinning disk has a radius of 0.2m and a mass of 2kg. If the disk is rotating

at 500 rpm about its central axis, determine the angular momentum of the disk

in kg m2/s.

Solution

Step 1: Convert the rotational speed from rpm to rad/s. Recall that 1rpm is

equal to π

30 rad/s.

Angular velocity ω= (500 rpm)×(π

rad

rpm)=25

3πrad/s

Step 2: Calculate the moment of inertia for a solid disk using the formula:

I=1

2MR2where Mis the mass and Ris the radius.

I=1

2(2 kg)(0.2m)2= 0.04 kg m2

Step 3: Use the formula for angular momentum L=Iω to find the angular

momentum of the disk.

L= (0.04 kg m2)( 25

3πrad/s) = 25

75πkg m2/s =1

3πkg m2/s

Therefore, the angular momentum of the spinning disk is 1

3πkg m2/s.

Question 3

Question

A solid cylinder of mass mand radius Ris floating in space with its axis of

rotation horizontal. Initially, the cylinder is not rotating. A small mass Mis

attached to a string wound around the cylinder and dropped. The mass Mfalls

a distance hbefore striking the ground. Determine the final angular velocity of

the cylinder after the mass Mfalls. Assume that the string does not slip on the

cylinder.

Solution

Step 1: Determine the potential energy of the mass Minitially. Initially, the

mass Mis at height habove the ground. The potential energy of the mass M

initially is given by

P Einitial =Mgh

Step 2: Determine the kinetic energy of the mass Mjust before striking the

ground. Just before striking the ground, the mass Mhas a kinetic energy equal

to its initial potential energy, since all the potential energy has been converted

to kinetic energy. Therefore,

KEfinal =Mgh

Step 3: Use conservation of energy to find the final angular velocity of the

cylinder. The total initial energy is zero since the cylinder is not initially rotat-

ing. The total final energy is the sum of the final kinetic energy of the mass M

and the rotational kinetic energy of the cylinder. Thus,

KEfinal =1

2Iω2+M gh

where Iis the moment of inertia of the cylinder and ωis the final angular

velocity.

Step 4: Substitute the expressions for the kinetic energy of the mass Mand

the moment of inertia of the cylinder. The moment of inertia of a solid cylinder

about its axis of rotation is I=1

2mR2. Substituting into the equation from

Step 3,

Mgh =1

2(1

2mR2)ω2+Mgh

Step 5: Solve for the final angular velocity, ω. Solving for ω, we have

4mR2ω2=Mgh

ω2=4Mgh

mR2

ω= 2√Mgh

mR2

Therefore, the final angular velocity of the cylinder after the mass Mfalls

is ω= 2√Mgh

mR2.

Question 4

Question

A solid cylindrical gyroscope has a radius of 0.1 m and a mass of 2 kg. It is

spinning at a rate of 200 revolutions per minute. If the gyroscope is mounted

on a frictionless axle and the axle is tilted at an angle of 30 degrees from the

vertical, what is the gyroscope’s precessional angular velocity?

Solution

Step 1: Calculate the initial angular velocity of the gyroscope in rad/s Given

that the gyroscope is spinning at a rate of 200 revolutions per minute, we first

need to convert this to radians per second.

Angular velocity =200 rev/min ×2πrad/rev

60 s/min =400π

60 rad/s =20π

3rad/s

Step 2: Calculate the angular momentum of the gyroscope The angular

momentum of the gyroscope can be calculated using the formula:

L=Iω

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

inertia of a solid cylinder is given by I=1

2mR2. Substitute the given values:

I=1

2×2×(0.1)2= 0.01 kg m2

L= 0.01 ×20π

3=20π

300 kg m2/s

Step 3: Calculate the torque due to the weight of the gyroscope The weight

of the gyroscope exerts a torque which causes precession. The torque can be

calculated as:

τ=mgh

where mis the mass, gis acceleration due to gravity, and his the vertical

distance between the point of contact and the center of mass of the gyroscope.

Given that the axle is tilted at 30 degrees from the vertical, h=Rsin(30◦) =

0.1×sin(30◦). Substitute the values:

h= 0.1×1

2= 0.05 m

τ= 2 ×9.81 ×0.05 = 0.981 Nm

Step 4: Calculate the precessional angular velocity The precessional angular

velocity can be calculated using the formula:

ωp=τ

Substitute the values:

ωp=0.981

20π

300

=300 ×0.981

20π≈4.66 rad/s

Therefore, the gyroscope’s precessional angular velocity is approximately

4.66 rad/s.

Question 5

Question

A solid cylinder of mass mand radius ris rotating with an angular speed ω

about a horizontal axis passing through its center. The cylinder is mounted on

a vertical axle that is hinged at one end. Suddenly, a small mass ∆mis attached

to the edge of the cylinder at a distance dfrom its axis. Determine the angular

speed of the system after the mass is attached.

Solution

Step 1: Before the mass is attached, the angular momentum of the cylinder is

given by:

Linitial =Icylinderω

where Icylinder is the moment of inertia of the cylinder. Since the cylinder is a

solid cylinder rotating about its axis, the moment of inertia of the cylinder is

Icylinder =1

2mr2.

Step 2: After the mass is attached, the angular momentum of the system is

given by:

Lfinal =Icylinderω+ ∆m(r+d)vedge

where vedge is the speed of the edge of the cylinder after the mass is attached.

From conservation of angular momentum, Linitial =Lfinal.

Step 3: Solve for the angular speed after the mass is attached:

2mr2ω=1

2mr2ω+ ∆m(r+d)vedge

Step 4: Since the system is at the verge of toppling, the speed of the edge

of the cylinder vedge is 0.

2mr2ω=1

2mr2ω

Step 5: Therefore, the angular speed of the system after the mass is attached

remains the same, ω.

Question 6

Question

A thin circular hoop of radius Rand mass Mis rolling without sliding on a

horizontal surface with a velocity v. The hoop then encounters a step of height

h. Find the angular velocity of the hoop just after leaving the step.

Solution

Step 1: We start by considering the conservation of energy as the hoop moves

from its initial position just before the step to its position just after leaving

the step. At the initial position, the hoop has kinetic energy 1

2Mv2due to its

linear velocity v. At the final position just after leaving the step, the hoop has

rotational kinetic energy and potential energy due to its height habove the

initial position. The initial kinetic energy of the hoop is 1

2Mv2.

Step 2: The final kinetic energy of the hoop just after leaving the step

consists of both rotational kinetic energy and potential energy. The rotational

kinetic energy is 1

2Iω2, where Iis the moment of inertia of the hoop and ωis

the angular velocity. The potential energy at this position is Mgh.

Step 3: The moment of inertia of a hoop rolling without slipping is I=M R2.

Setting initial kinetic energy equal to final kinetic and potential energy gives:

2Mv2=1

2MR2ω2+Mgh

Step 4: Solving for the angular velocity ω, we get:

ω=√v2

R2+ 2gh

Therefore, the angular velocity of the hoop just after leaving the step is

√v2

R2+2gh .

Question 7

Question

A solid cylinder of mass mand radius ris rolling without slipping on a horizontal

surface with angular velocity ω. The cylinder’s moment of inertia about its

central axis is I=1

2mr2. At a certain instant, a small object of mass m/2is

glued to the surface of the cylinder so that it remains at a fixed position on

the surface after the collision. Determine the angular velocity of the modified

system after the collision.

Solution

Step 1: Calculate the initial angular momentum of the system before the col-

lision. The initial angular momentum Linitial of the system before the collision

can be calculated as:

Linitial =Iω

Linitial =(1

2mr2)ω

Step 2: Calculate the change in angular momentum due to the collision.

Since the small object is glued to the surface of the cylinder and moves with it,

the change in angular momentum due to the collision is equal to the angular

momentum of the small object about the central axis of the cylinder. The

angular momentum of the small object is given by:

Lobject = (m/2)r2ωobject

Step 3: Calculate the final angular momentum of the modified system after

the collision. The final angular momentum Lfinal of the modified system after

the collision is given by:

Lfinal =Linitial +Lobject

Step 4: Set the initial angular momentum equal to the final angular momen-

tum to solve for the angular velocity of the modified system.

2mr2ω=1

2mr2ωobject +1

2mr2ωobject

ω=ωobject

Therefore, the angular velocity of the modified system after the collision is

the same as the angular velocity of the small object, which is ωobject .

Question 8

Question

A gyroscope consists of a wheel of mass mand radius rthat is spinning at a

constant angular velocity ω. The gyroscope is mounted on a frictionless pivot

at one end of a horizontal rod of length L. Initially, the rod is vertical and the

angular momentum of the entire system is entirely due to the rotation of the

wheel. Calculate the angular velocity of precession Ωwhen the rod makes an

angle θwith the vertical.

Solution

Step 1: Identify the known quantities and relevant formulas.

Given: - Mass of the wheel: m- Radius of the wheel: r- Angular velocity of

the wheel: ω- Length of the rod: L- Angle made by the rod with the vertical:

Formula: - Angular momentum: L=Iω, where Iis the moment of inertia

of the wheel about its center.

Step 2: Find the moment of inertia of the wheel about its center.

The moment of inertia for a solid disk rotating about its center is I=1

2mr2.

Step 3: Calculate the angular momentum of the system when the rod makes

an angle θwith the vertical.

The angular momentum Lis given by L=Iω.

Substitute the moment of inertia I=1

2mr2and angular velocity ωinto the

formula.

L=1

2mr2·ω.

Step 4: Find the angular velocity of precession Ω.

The torque about the pivot due to the weight of the wheel and rod provides

the change in angular momentum that causes precession. The torque due to the

weight is τ=mgL sin θ. This torque causes the angular velocity to change at a

rate dL

dt =τ.

Since the total angular momentum Lis constant, we have dL

dt =Idω

dt .

Setting τ=Idω

dt and substituting τ=mgL sin θ, we get mgL sin θ=Idω

dt .

Substitute the moment of inertia I=1

2mr2and simplify the expression to

solve for dω

dt .

Step 5: Integrate the expression to find the angular velocity of precession Ω.

Integrate the expression dω

dt =2gsin θ

rwith respect to time.

∫dω =∫2gsin θ

rdt.

ω=2gsin θ

rt+C, where Cis the constant of integration.

Since the system starts from rest, ω= 0 at t= 0, so C= 0.

Therefore, the angular velocity of precession Ω = 2gL sin θ

Question 9

Question

A uniform cylindrical gyroscope has a radius Rand mass M. It is mounted on

a frictionless pivot at one end of a rod of length L. The other end of the rod is

attached to a hinge on a vertical wall. The gyroscope is set into rapid precession

about a vertical axis. Calculate the radial acceleration arof the center of mass

of the gyroscope during precession.

Solution

Step 1: The radial acceleration of the center of mass of the gyroscope dur-

ing precession can be calculated using the formula for acceleration in circular

motion: ar=Rω2, where ωis the angular velocity of precession.

Step 2: The angular velocity of precession is related to the time period of

precession Tby ω=2π

Step 3: The time period of precession Tcan be calculated using the gyro-

scope’s moment of inertia I, the applied torque τ, and the precessional angular

velocity Ω(not to be confused with the angular velocity ω).

Step 4: The moment of inertia Ifor a cylindrical gyroscope is I=1

2MR2.

Step 5: The torque τrequired to maintain the precession of a gyroscope is

τ=IΩ.

Step 6: Combining steps 3, 4, and 5, we have Ω = τ

Iand T=2π

Ω.

Step 7: Substituting I=1

2MR2into the expression for Ω, we get Ω = 2τ

MR2.

Step 8: Substituting Ω = 2τ

MR2into the expression for T, we have T=M R2

τ.

Step 9: Substituting ω=2π

Tinto the expression for ar=Rω2, we find

ar=4π2R

T2or ar=4π2Rτ 2

M2R4.

Therefore, the radial acceleration arof the center of mass of the gyroscope

during precession is 4π2Rτ 2

M2R4.

Question 10

Question

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

rotating about a vertical axis at a constant angular speed ω. The other end of

the rod carries a mass m. Determine the precession angular speed of the rod

about the vertical axis.

Solution

Let’s denote the precession angular speed of the rod as Ω.

Step 1: The angular momentum of the rod about the pivot is given by

L=Iω, where Iis the moment of inertia of the rod about the pivot. The

moment of inertia of the rod about the pivot can be approximated as I≈1

3ML2

for a thin uniform rod.

Step 2: The torque acting on the system about the pivot is due to the

force of gravity acting on the mass m. The torque can be expressed as τ=

r×F=rmg sin(θ), where ris the distance from the pivot to the mass m,gis

the acceleration due to gravity, and θis the angle between the position vector

and the force vector.

Step 3: Since the torque causes a change in angular momentum over time,

we can express the relation as dL

dt =τ. The rate of change of angular momentum

is given by dL

dt =Idω

dt .

Step 4: Equating the torque and the rate of change of angular momentum,

we have:

Idω

dt =rmg sin(θ)

Step 5: The rate of precession, Ω, is related to the rate of change of the

angle θwith respect to time, dθ

dt . We have the relation Ω = rdθ

dt .

Step 6: Since the rod is rotating steadily at an angular speed ω, the rate

at which θis changing is equal to ω. Therefore, dθ

dt =ω.

Step 7: Since rsin(θ)is equal to L, we can substitute these relations back

into the equation from Step 4 to get:

3ML2dω

dt =Lmg

Step 8: Solving for dω

dt , we find:

dω

dt =3mg

Step 9: Finally, we plug in the expression for dω

dt into the expression for Ω:

Ω = rdθ

dt =Ldθ

dt =Lω =3mgL

L= 3g

Step 10: Therefore, the precession angular speed of the rod about the

vertical axis is 3g.

Question 11

Question

A disc of radius Rand mass Mis spinning at an angular speed ωabout an axis

through its center. The disc is then placed inside a hoop of radius rwhich is

attached to a vertical pole. The disc starts precessing about the vertical axis

with an angular speed Ω. If the moment of inertia of the hoop is Ihoop =αMr2

and the moment of inertia of the disc about its center of mass is Idisc =βMR2,

find the ratio α

βin terms of ω,Ω,r, and R.

Solution

Step 1: The total angular momentum should be conserved. The initial angular

momentum of the disc is given by Linitial =Idiscωand the final angular mo-

mentum of the system is Lfinal =IdiscΩ + IhoopΩ. Setting Linitial =Lfinal, we

have:

Idiscω=IdiscΩ + IhoopΩ

Step 2: Substituting Idisc =βMR2and Ihoop =αM r2into the equation

gives:

βM R2ω=βMR2Ω + αMr2Ω

Step 3: Dividing both sides by MR2and simplifying, we get:

βω =βΩ + αr2

R2Ω

Step 4: Solving for α

β, we have:

β=βω −βΩ

Ωr2

Step 5: Multiplying the numerator and denominator by 1

βΩR2yields:

β=ω

Ω(R

r)2−1

Therefore, the ratio α

βin terms of ω,Ω,r, and Ris ω

Ω(R

r)2−1.

Question 12

Question

A uniform disk of radius Rand mass Mis mounted on a horizontal frictionless

axle. The disk is spinning counterclockwise with an angular velocity ω0. A

small mass mis attached at a distance rfrom the center of the disk along a

radial line. Determine the angular velocity of the disk and the mass of the disk

needed to keep the mass mstationary relative to the disk.

Solution

Step 1: The moment of inertia of the disk about its center is given by Idisk =

2MR2.

Step 2: The moment of inertia of the disk about the point where mass mis

attached is Idisk+m=1

2MR2+mr2.

Step 3: Conservation of angular momentum gives us: Idiskωf=Idisk+m(ωf+

ω), where ωfis the final angular velocity of the disk and mass m, and ωis the

angular velocity of mass mrelative to the disk.

Step 4: Substituting the expressions for Idisk and Idisk+m, we get 1

2MR2ωf=

2MR2+mr2)(ωf+ω).

Step 5: Solving for ωf, we find: ωf=mr2

MR2−mr2ω.

Step 6: For mass mto be stationary relative to the disk, ωmust be such

that ωf= 0.

Step 7: Therefore, setting ωfto zero in the equation derived in Step 5 gives

us: mr2

MR2−mr2ω= 0.

Step 8: Since ω= 0, we must have r= 0 in order to keep mass mstationary

relative to the disk.

Hence, the angular velocity of the disk is given by ωf=mr2

MR2−mr2ωand the

mass of the disk needed to keep mass mstationary relative to the disk is zero.

Question 13

Question

A uniform square plate of side aand mass mis rotating with an angular speed

ωabout an axis normal to the plate through a corner. Determine the gyroscopic

stability of this system.

Solution

Step 1: The angular momentum of the plate is given by L=Iω, where Iis the

moment of inertia of the plate. The moment of inertia of a square plate rotating

about an axis through its vertex is given by I=1

3ma2.

Step 2: The precession frequency, Ω, is given by Ω = Mg sin θ

L, where Mis

the mass of the system, gis the acceleration due to gravity, and θis the angle

between the angular velocity vector and the vertical axis.

Step 3: To determine the stability of the system, we look at the sign of dΩ

dθ .

If dΩ

dθ >0, the system is stable, and if dΩ

dθ <0, the system is unstable.

Step 4: Taking the derivative of Ωwith respect to θ, we get dΩ

dθ =Mg cos θ

Step 5: Substituting the expressions for Land Iinto dΩ

dθ , we have dΩ

dθ =

3 cos θ

sin θ.

Step 6: The sign of dΩ

dθ is positive when θ < π

2and negative when θ > π

Therefore, the system is stable for small angles and unstable for large angles.

Question 14

Question

A uniform disk of radius Rand mass Mis mounted on a frictionless horizontal

axle that is aligned along the z-axis. The disk is rotating at an angular speed

ωabout the z-axis. Calculate the magnitude of the angular momentum of the

disk about the axle.

Solution

Step 1: The angular momentum of an object about a given axis is given by the

formula: L=Iω, where Iis the moment of inertia about the axis and ωis the

angular speed.

Step 2: The moment of inertia for a uniform disk rotating about an axis that

is perpendicular to the disk and passes through its center is given by I=1

2MR2.

Step 3: Substitute the given moment of inertia into the formula for angular

momentum: L=1

2MR2ω.

Step 4: Finally, the magnitude of the angular momentum of the disk about

the axle is 1

2MR2ω.

Question 15

Question

A solid sphere of radius Rand mass Mis spinning about an axis through its

center with an angular speed ω. The sphere is then placed on a rough horizontal

surface. Determine the minimum coeﬀicient of static friction needed to prevent

slipping as the sphere rolls without slipping along the surface.

Solution

Step 1: Identify the forces acting on the sphere.

When the sphere is rolling without slipping, there are two forces acting on

it: the gravitational force, mg, and the normal force, N, exerted by the surface

upwards. Additionally, the static friction force, fs, acts in the direction opposite

to the motion that would result in slipping.

Step 2: Analyze the rotation of the sphere.

Since the sphere is both spinning about its axis and rolling without slipping,

its kinetic energy consists of both translational and rotational components. The

kinetic energy of the sphere is given by:

KE =1

2Mv2+1

2Iω2

where vis the speed of the sphere’s center of mass and I=2

5MR2is the moment

of inertia for a solid sphere.

Step 3: Determine the speed of the sphere’s center of mass.

Since the sphere is rolling without slipping, its linear speed is related to its

angular speed by v=Rω.

Step 4: Write the expression for the kinetic energy.

Substitute v=Rω and I=2

5MR2into the expression for kinetic energy to

get:

KE =1

2M(Rω)2+1

2(2

5MR2)ω2

Step 5: Find the minimum coeﬀicient of static friction.

To prevent slipping, the static friction force must provide enough torque to

prevent the sphere from slipping. The torque required to stop slipping without

acceleration is fsRand is equal to Iα, where αis the angular acceleration. We

can relate αto the translational acceleration ausing the relation a=Rα.

Step 6: Find the translational acceleration of the sphere.

The translational acceleration ais related to the net force acting on the

sphere by Fnet =Ma, where Fnet =N−mg.

Step 7: Set up the equations for rotational and translational acceleration.

Equating the torque required to stop slipping with fsRand the net force

with Ma, we get:

fsR=2

5MR2·a

fs=2

5Ma

Step 8: Express the acceleration in terms of ω.

Since the sphere is rolling without slipping, a=Rα =R(a

R)=α. From the

kinematics of rolling motion, we know that a=Rα =Rv

R=v. Thus, a=v.

Step 9: Substitute a=vinto the expression for fs.

Substitute a=vinto the expression fs=2

5Ma:

fs=2

5Mv

Step 10: Find the minimum coeﬀicient of static friction in terms of ω.

Finally, express vin terms of ωusing v=Rω, and substitute it into the

expression for fs:

fs=2

5MRω

Therefore, the minimum coeﬀicient of static friction needed to prevent slip-

ping as the sphere rolls without slipping along the surface is 2

5Rω.

Question 16

Question

A uniform rod of length Land mass Mis pivoted at one end to a fixed point.

A small object of mass mis attached to the free end. The system is set into

rotational motion about the pivot point with an angular velocity ω. Calculate

the minimum coeﬀicient of static friction required between the pivot point and

the rod to prevent the object from slipping.

Solution

Step 1: Identify the forces acting on the system. The forces acting on the

system are the gravitational force (mg) acting on the object at the end of the

rod and the normal force (N) acting at the pivot point in the vertical direction.

Additionally, there is static friction force (fs) acting at the pivot point in the

horizontal direction.

Step 2: Write the equation for rotational equilibrium. The condition for

rotational equilibrium is ∑τ= 0, where ∑τis the sum of the torques acting

on the system.

Step 3: Calculate the torque due to the gravitational force. The torque due

to the gravitational force about the pivot point is τgravity =mgL.

Step 4: Calculate the torque due to the static friction force. The torque due

to the static friction force about the pivot point is τfriction =fs·L.

Step 5: Set up the equation for rotational equilibrium. In rotational equi-

librium, the sum of the torques is zero:

τgravity +τfriction = 0

mgL +fs·L= 0

Step 6: Solve for the static friction force. Solving for the static friction force

gives:

fs=−mg

Step 7: Calculate the minimum coeﬀicient of static friction. The coeﬀicient

of static friction is given by:

µs=fs

Since fs=−mg and N=mg, we get:

µs=−mg

mg =−1

Therefore, the minimum coeﬀicient of static friction required between the

pivot point and the rod to prevent the object from slipping is 1. Note that this

negative sign indicates that the direction of the static friction force is opposite

to what was assumed.

Question 17

Question

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

rotational motion about a vertical axis at an angular speed ω. Determine the

minimum angular speed ωmin that the rod must be rotated to keep its other end

from falling under gravity due to its own weight. Take the acceleration due to

gravity as g.

Solution

Step 1: Consider the forces acting on the rod when it is rotated at an angular

speed ω: - Weight of the rod, W=Mg, acts downward at the center of mass.

- Centripetal force, Fcp =Mω2L, acts towards the pivot point to keep the rod

in circular motion.

Step 2: To find the minimum angular speed for the rod not to fall, the

centripetal force must be greater or equal to the weight of the rod, i.e.,

Fcp ≥W

Step 3: Substitute the expressions for Fcp and W:

Mω2L≥Mg

Step 4: Solve for the minimum angular speed ωmin:

ω2≥g

ωmin ≥√g

Step 5: Thus, the minimum angular speed that the rod must be rotated to

keep its other end from falling under gravity is ωmin =√g

Question 18

Question

A thin circular hoop of radius Rand mass Mis rolling without slipping on a

horizontal surface with a speed v. Calculate the rotational kinetic energy of the

hoop.

Solution

Step 1: The rotational kinetic energy of the hoop is given by the formula

KErot =1

2Iω2, where Iis the moment of inertia of the hoop and ωis the

angular velocity of the hoop.

Step 2: The moment of inertia of a hoop about its symmetry axis is I=

MR2.

Step 3: Since the hoop is rolling without slipping, the relationship between

the linear speed vand the angular speed ωis ω=v

Step 4: Substituting I=MR2and ω=v

Rinto the formula for rotational

kinetic energy, we have KErot =1

2MR2(v

R)2=1

2Mv2.

Step 5: Therefore, the rotational kinetic energy of the hoop is 1

2Mv2.

Question 19

Question

A solid cylinder of radius Rand mass Mis initially at rest on a horizontal

frictionless surface. A light string is wound around the cylinder and a block of

mass mis attached to the other end of the string. The block is released from

rest a height habove the floor. Assuming the cylinder rolls without slipping,

find the speed of the block just before it hits the floor. Take the acceleration

due to gravity as g.

Solution

Step 1: Find the acceleration of the block just before it hits the floor. Let Tbe

the tension in the string, αbe the angular acceleration of the cylinder, and a

be the acceleration of the block.

The forces acting on the block are its weight mg (downward) and the tension

Tin the string (upward). The net force on the block is the difference between

these two forces, so we have:

ma =T−mg

The forces acting on the cylinder are its weight Mg (downward) and the

tension Tin the string (upward). The net torque on the cylinder can be written

as:

T R =Iα

where I=1

2MR2is the moment of inertia of the cylinder about its center of

mass. Since the cylinder is rolling without slipping, the linear acceleration aof

the block is related to the angular acceleration αof the cylinder by a=Rα.

Thus, we have:

T R =1

2MR2·a

Substituting T=ma +mg into the torque equation gives:

(ma +mg)R=1

2MRa

Solving for agives:

a=2m+M

2m+ 2Mg

Step 2: Find the speed of the block just before it hits the floor. Using the

kinematic equation:

v2=u2+ 2as

where uis the initial velocity (zero in this case), ais the acceleration of the

block, and sis the distance the block falls (equal to h), we get:

v2= 0 + 2 (2m+M

2m+ 2Mg)h

v=√4(2m+M)gh

2m+ 2M

v=√4gh

2 + M

Therefore, the speed of the block just before it hits the floor is √4gh

2+ M

Question 20

Question

A solid sphere of mass mand radius Rrolls without slipping along a horizontal

surface with a speed v. The sphere then encounters a frictionless incline plane

that makes an angle θwith the horizontal. What will be the angular speed of

the sphere when it reaches the top of the incline?

Solution

Step 1: We start by determining the translational kinetic energy of the sphere

when it reaches the top of the incline. At that point, the sphere is rotating

about its center of mass while also moving up the incline. The total kinetic

energy is the sum of the rotational kinetic energy and the translational kinetic

energy:

KEtotal =KErotational +KEtranslational

Step 2: The rotational kinetic energy of the sphere can be calculated using

the equation:

KErotational =1

2Iω2

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

Step 3: For a solid sphere rolling without slipping, the moment of inertia

about its center of mass is I=2

5mR2.

Step 4: The translational kinetic energy of the sphere when it reaches the

top of the incline is:

KEtranslational =1

2mv2

Step 5: Since energy is conserved, the total energy at the bottom (purely

translational kinetic energy) is equal to the total energy at the top of the incline:

2mv2=1

2Iω2+1

2mv2

top

Step 6: We know that vtop =Rω. Substituting this into the energy equation

and using the expressions for Iand vtop, we get:

2mv2=1

5mR2ω2+1

2mR2ω2

Step 7: Simplify the equation to solve for ω:

2mv2=7

10mR2ω2

Step 8: Solve for ω:

ω=√2v2

7R2

Therefore, the angular speed of the sphere when it reaches the top of the

incline is √2v2

7R2.

Question 21

Question

A uniform square plate of side length aand mass mis pivoted about an axis

through its center and perpendicular to its plane. The plate is set spinning at

an angular velocity ω0about this axis. An insect of mass m′lands at one of

the corners of the plate when the plate is spinning. If the insect slides to the

opposite corner, what is its new kinetic energy? Assume the insect walks in a

straight line across the plate.

Solution

Step 1: When the insect lands at one corner, it will have the same angular

velocity ω0as the plate, but a different linear velocity vdepending on the

distance rof the insect from the center of the plate. The kinetic energy of the

insect when it lands is given by:

KE1=1

2m′v2

Step 2: The angular velocity ω0of the plate can be related to the linear

velocity vof the insect using the formula:

v=ω0r

Step 3: The distance rof the insect from the center of the plate is given by

r=a√2

2. Therefore, the linear velocity vof the insect when it lands is:

v=ω0·a√2

Step 4: Substituting the expression for vback into the kinetic energy formula

and simplifying, we find:

KE1=1

2m′(ω0·a√2

2)2

2m′·a2

2·ω2

Step 5: When the insect slides to the opposite corner, it will have the same

linear velocity vas before, but a different angular velocity ω1as it will be farther

from the axis of rotation. The kinetic energy of the insect after sliding to the

opposite corner is given by:

KE2=1

2m′v2

Step 6: The new distance r′of the insect from the center of the plate is

r′=a. Therefore, the linear velocity vof the insect after sliding to the opposite

corner is:

v=ω1·a

Step 7: Equating the kinetic energies before and after sliding, we have:

2m′·a2

2·ω2

0=1

2m′·a2·ω2

Step 8: Solving for ω1, we find:

ω1=1

√2ω0

Step 9: Substituting ω1back into the expression for kinetic energy after

sliding and simplifying, we find the new kinetic energy KE2to be:

KE2=1

2m′·a2·(1

√2ω0)2

4m′a2ω2

Therefore, the new kinetic energy of the insect after sliding to the opposite

corner is 1

4m′a2ω2

Question 22

Question

A solid cylinder of radius Rand mass Mrolls without slipping along a horizontal

surface. The cylinder is given an initial angular velocity ω0about its symmetry

axis. At a certain point, a small metal bead of mass mis dropped from a height

honto the surface inside the cylinder at a distance rfrom the axis. The bead

strikes the surface impulsively (i.e., without bouncing) and sticks. Calculate the

subsequent angular velocity of the cylinder.

Solution

Step 1: Calculate the initial angular momentum of the cylinder-bead system.

The initial angular momentum of the cylinder-bead system is given by:

Linitial =Icylinderω0+m×(R−r)×ω0

Where: - Icylinder =1

2MR2is the moment of inertia of the cylinder, - m×(R−r)

is the distance of the bead from the symmetry axis.

Step 2: Calculate the final angular momentum of the system. The final

angular momentum of the system is given by:

Lfinal = (Icylinder +m(R−r)2)ωfinal

Where ωfinal is the final angular velocity of the cylinder-bead system.

Step 3: Apply the principle of conservation of angular momentum. Accord-

ing to the principle of conservation of angular momentum, the initial angular

momentum is equal to the final angular momentum. Therefore,

Icylinderω0+m×(R−r)ω0= (Icylinder +m(R−r)2)ωfinal

Step 4: Solve for final angular velocity. Substitute the expressions for Icylinder

and simplify:

2MR2ω0+mRω0−mrω0=(1

2MR2+m(R2−2rR +r2))ωfinal

Therefore,

ωfinal =

2MR2ω0+mRω0−mrω0

2MR2+m(R2−2rR +r2)

Question 23

Question

A gyroscope consists of a uniform disk of mass Mand radius Rmounted on a

horizontal axle. The disk is set spinning at an angular speed ω. A small mass

mis fixed to the edge of the disk at a distance rfrom the center. Determine

the angular momentum of the gyroscope about the axle when the small mass m

is at the lowest point of its circular path.

Solution

Step 1: We can calculate the angular momentum of the gyroscope using the

formula L=Iω, where Iis the moment of inertia of the gyroscope about the

axle and ωis the angular speed.

Step 2: First, we calculate the moment of inertia of the gyroscope. The

moment of inertia of a disk rotating about an axis through its center is given

by Idisk =1

2MR2. The moment of inertia of the small mass mabout the axis

is Im=mr2.

Step 3: The total moment of inertia of the gyroscope about the axle is the

sum of the moments of inertia of the disk and the small mass:

Itotal =Idisk +Im=1

2MR2+mr2

Step 4: Now we substitute the total moment of inertia Itotal into the formula

for angular momentum:

L=Itotalω=(1

2MR2+mr2)ω

Step 5: The small mass mis at the lowest point of its circular path, so its

speed is maximum. The angular speed of the gyroscope is related to the linear

speed of the small mass vby ω=v

r. The linear speed vcan be determined

using conservation of energy: mgh =1

2mv2, where his the maximum vertical

distance from the lowest point.

Step 6: Substituting ω=v

rinto the angular momentum equation gives:

L=(1

2MR2+mr2)(v

r)=1

2MRv +mv2

Step 7: Solving for vusing the conservation of energy equation gives v=

√2gh. Substitute this back into the angular momentum equation to get the

final result for the angular momentum of the gyroscope:

L=1

2MR√2gh +m(2gh)

Question 24

Question

A disk of mass Mand radius Ris rotating about an axis perpendicular to the

disk and passing through its center with an angular velocity ω. The disk is

now set in gyroscopic motion by an externally applied couple of magnitude C.

Determine the rate of precession ωpand the angle through which the disk has

turned after a time tunder the action of the couple.

Given: I=1

2MR2,C=Iωp

Solution

Step 1: Let’s determine the rate of precession ωpusing the given formula for

the couple C.

C=Iωp

C=(1

2MR2)ωp

ωp=2Ct

MR2

Step 2: Now, let’s determine the angle θthrough which the disk has turned

after time t.

θ=ωpt

θ=(2Ct

MR2)t

θ=2Ct2

MR2

Therefore, the rate of precession ωpis 2Ct

MR2and the angle θthrough which

the disk has turned after time tis 2Ct2

MR2.

Question 25

Question

A uniform thin circular hoop of mass Mand radius Rrotates about a vertical

axis passing through its center with an angular speed ω. A light string is wound

around the hoop and a mass mis suspended from its free end. The mass is

released from rest at the same level as the center of the hoop. What will be the

speed of the mass when it has fallen through a height h? (Assume the string is

vertical and the hoop can rotate freely about its axis.)

Solution

Step 1: The total energy of the mass mat height his given by the sum of its

kinetic and potential energy:

E=K+U

Step 2: The kinetic energy Kof the mass mis given by 1

2mV 2, where Vis

the speed of the mass.

Step 3: The potential energy Uof the mass mat height his mgh, where g

is the acceleration due to gravity.

Step 4: At the initial position, all of the potential energy mgh is converted

to the kinetic energy 1

2mV 2at the final position.

Step 5: Setting the initial potential energy equal to the final kinetic energy,

we have:

mgh =1

2mV 2

Step 6: Solving for the speed V, we get:

V=√2gh

Step 7: Thus, the speed of the mass mwhen it has fallen through a height

his √2gh.

PHYS 201 - GENERAL PHYSICS I Gyroscopic Motion and Stability Question Bank Set 3

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