Content check — University Physics 1, Chapter 11: Angular Momentum

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 11: Angular Momentum

54 problems · 10 figures

Check Your Understanding 11.1 §11.1 source ↗

A hollow cylinder is on an incline at an angle of

60 ° .

The coefficient of static friction on the surface is

μ_{s} = 0.6

. (a) Does the cylinder roll without slipping? (b) Will a solid cylinder roll without slipping?

Answer

μ_{s} \geq \frac{tan θ}{1 + (m r^{2} / I_{CM})}

; inserting the angle and noting that for a hollow cylinder

I_{CM} = m r^{2},

we have

μ_{s} \geq \frac{tan 60 °}{1 + (m r^{2} / m r^{2})} =

\frac{1}{2} tan 60 ° =

0.87;

we are given a value of 0.6 for the coefficient of static friction, which is less than 0.87, so the condition isn’t satisfied and the hollow cylinder will slip; b. The solid cylinder obeys the condition

μ_{s} \geq \frac{1}{3} tan θ =

\frac{1}{3} tan 60 ° =

0.58 .

The value of 0.6 for

μ_{S}

satisfies this condition, so the solid cylinder will not slip.

Check Your Understanding 11.2 §11.2 source ↗

A proton spiraling around a magnetic field executes circular motion in the plane of the paper, as shown below. The circular path has a radius of 0.4 m and the proton has velocity $4.0 \times 10^{6} m / s$ . What is the angular momentum of the proton about the origin?

A proton moves in a counterclockwise circle. The circle has radius r. The proton is shown when it is to the right of the center of the circle, and its velocity is v sub perpendicular in the upward, positive y, direction.

Answer

From the figure, we see that the cross product of the radius vector with the momentum vector gives a vector directed out of the page. Inserting the radius and momentum into the expression for the angular momentum, we have

\vec{l} =

\vec{r} \times \vec{p} =

(0.4 m \hat{i}) \times (1.67 \times 10^{−27} kg (4.0 \times 10^{6} m / s) \hat{j}) =

2.7 \times 10^{−21} kg \cdot m^{2} / s \hat{k}

Check Your Understanding 11.3 §11.2 source ↗

Which has greater angular momentum: a solid sphere of mass m rotating at a constant angular frequency

ω_{0}

about the z-axis, or a solid cylinder of same mass and rotation rate about the z-axis?

Answer

I_{sphere} =

\frac{2}{5} m r^{2},

I_{cylinder} =

\frac{1}{2} m r^{2}

; Taking the ratio of the angular momenta, we have:

\frac{L_{cylinder}}{L_{sphere}} =

\frac{I_{cylinder} ω_{0}}{I_{sphere} ω_{0}} =

\frac{\frac{1}{2} m r^{2}}{\frac{2}{5} m r^{2}} =

\frac{5}{4}

. Thus, the cylinder has

25 %

more angular momentum. This is because the cylinder has more mass distributed farther from the axis of rotation.

Check Your Understanding 11.4 §11.3 source ↗

A merry-go-round at a playground is rotating at 4.0 rev/min. Three children jump on and increase the moment of inertia of the merry-go-round/children rotating system by

25 %

. What is the new rotation rate?

Answer

Using conservation of angular momentum, we have

I (4.0 rev/min) =

1.25 I ω_{f},

ω_{f} =

\frac{1.0}{1.25} (4.0 rev / min) =

3.2 rev / min

Check Your Understanding 11.5 §11.4 source ↗

A top has a precession frequency of 5.0 rad/s on Earth. What is its precession frequency on the Moon?

Answer

The Moon’s gravity is 1/6 that of Earth’s. By examining Equation 11.12, we see that the top’s precession frequency is linearly proportional to the acceleration of gravity. All other quantities, mass, moment of inertia, and spin rate are the same on the Moon. Thus, the precession frequency on the Moon is

ω_{P} (Moon) =

\frac{1}{6} ω_{P} (Earth) =

\frac{1}{6} (5.0 rad / s) =

0.83 rad / s .

11-1 §11.1 source ↗

Can a round object released from rest at the top of a frictionless incline undergo rolling motion?

Answer

No, the static friction force is zero.

11-3 §11.1 source ↗

A wheel is released from the top on an incline. Is the wheel most likely to slip if the incline is steep or gently sloped?

Answer

The wheel is more likely to slip on a steep incline since the coefficient of static friction must increase with the angle to keep rolling motion without slipping.

11-5 §11.1 source ↗

A hollow sphere and a hollow cylinder of the same radius and mass roll up an incline without slipping and have the same initial center of mass velocity. Which object reaches a greater height before stopping?

Answer

The cylinder reaches a greater height. By Equation 11.4, its acceleration in the direction down the incline would be less.

11-7 §11.2 source ↗

For a particle traveling in a straight line, are there any points about which the angular momentum is zero? Assume the line intersects the origin.

Answer

All points on the straight line will give zero angular momentum, because a vector crossed into a parallel vector is zero.

11-9 §11.2 source ↗

If a particle is moving with respect to a chosen origin it has linear momentum. What conditions must exist for this particle’s angular momentum to be zero about the chosen origin?

Answer

The particle must be moving on a straight line that passes through the chosen origin.

11-11 §11.3 source ↗

What is the purpose of the small propeller at the back of a helicopter that rotates in the plane perpendicular to the large propeller?

Answer

Without the small propeller, the body of the helicopter would rotate in the opposite sense to the large propeller in order to conserve angular momentum. The small propeller exerts a thrust at a distance R from the center of mass of the aircraft to prevent this from happening.

11-13 §11.3 source ↗

As the rope of a tethered ball winds around a pole, what happens to the angular velocity of the ball?

Answer

The angular velocity increases because the moment of inertia is decreasing.

11-15 §11.3 source ↗

Explain why stars spin faster when they collapse.

Answer

More mass is concentrated near the rotational axis, which decreases the moment of inertia causing the star to increase its angular velocity.

11-17 §11.4 source ↗

Gyroscopes used in guidance systems to indicate directions in space must have an angular momentum that does not change in direction. When placed in the vehicle, they are put in a compartment that is separated from the main fuselage, such that changes in the orientation of the fuselage does not affect the orientation of the gyroscope. If the space vehicle is subjected to large forces and accelerations how can the direction of the gyroscopes angular momentum be constant at all times?

Answer

A torque is needed in the direction perpendicular to the angular momentum vector in order to change its direction. These forces on the space vehicle are external to the container in which the gyroscope is mounted and do not impart torques to the gyroscope’s rotating disk.

11-19 §11.1 source ↗

What is the angular velocity of a 75.0-cm-diameter tire on an automobile traveling at 90.0 km/h?

Answer

v_{CM} =

R ω \Rightarrow ω =

66.7 rad/s

11-21 §11.1 source ↗

If the boy on the bicycle in the preceding problem accelerates from rest to a speed of 10.0 m/s in 10.0 s, what is the angular acceleration of the tires?

Answer

α = 3.3 rad / s^{2}

11-23 §11.1 source ↗

A marble rolls down an incline at

30 °

from rest. (a) What is its acceleration? (b) How far does it go in 3.0 s?

Answer

I_{CM} =

\frac{2}{5} m r^{2},

a_{CM} =

3.5 m / s^{2};

x =

15.75 m

11-25 §11.1 source ↗

A rigid body with a cylindrical cross-section is released from the top of a

30 °

incline. It rolls 10.0 m to the bottom in 2.60 s. Find the moment of inertia of the body in terms of its mass m and radius r.

Answer

positive is down the incline plane;

a_{CM} =

\frac{m g sin θ}{m + (I_{CM} / r^{2})} \Rightarrow I_{CM} =

r^{2} [\frac{m g sin 30}{a_{CM}} - m]

x -

x_{0} =

v_{0} t -

\frac{1}{2} a_{CM} t^{2} \Rightarrow a_{CM} =

2.96 {m/s}^{2},

I_{CM} = 0.66 m r^{2}

11-27 §11.1 source ↗

A solid cylinder of radius 10.0 cm rolls down an incline with slipping. The angle of the incline is

30 ° .

The coefficient of kinetic friction on the surface is 0.400. What is the angular acceleration of the solid cylinder? What is the linear acceleration?

Answer

α =

67.9 rad / s^{2}

{(a_{CM})}_{x} =

1.5 m / s^{2}

11-29 §11.1 source ↗

A 40.0-kg solid cylinder is rolling across a horizontal surface at a speed of 6.0 m/s. How much work is required to stop it?

Answer

W = −1080.0 J

11-31 §11.1 source ↗

A solid cylinder rolls up an incline at an angle of

20 ° .

If it starts at the bottom with a speed of 10 m/s, how far up the incline does it travel?

Answer

Mechanical energy at the bottom equals mechanical energy at the top;

\frac{1}{2} m v_{0}^{2} +

\frac{1}{2} (\frac{1}{2} m r^{2}) {(\frac{v_{0}}{r})}^{2} =

m g h \Rightarrow h =

\frac{1}{g} (\frac{1}{2} + \frac{1}{4}) v_{0}^{2}

h = 7.7 m,

so the distance up the incline is

22.5 m

11-33 §11.1 source ↗

A hollow cylinder that is rolling without slipping is given an initial velocity and rolls up an incline to a vertical height of 1.0 m. If a hollow sphere of the same mass and radius is given the same initial velocity, how high vertically does it roll up the incline?

Answer

Use energy conservation

\frac{1}{2} m v_{0}^{2} +

\frac{1}{2} I_{Cyl} ω_{0}^{2} =

m g h_{Cyl}

\frac{1}{2} m v_{0}^{2} +

\frac{1}{2} I_{Sph} ω_{0}^{2} =

m g h_{Sph}

.
Subtracting the two equations, eliminating the initial translational energy, we have

h_{Cyl} = \frac{v_{0}^{2}}{g}

h_{Sph} = \frac{5 v_{0}^{2}}{6 g}

,
The ratio of the height reached by the sphere to the height reached by the cylinder is

\frac{h_{Sph}}{h_{Cyl}} =

\frac{5}{6} .

So the sphere reaches a lower height of

\frac{5}{6} (1.0 m) =

0.83 m.

11-35 §11.2 source ↗

A bird flies overhead from where you stand at an altitude of 300.0 m and at a speed horizontal to the ground of 20.0 m/s. The bird has a mass of 2.0 kg. The radius vector to the bird makes an angle

θ

with respect to the ground. The radius vector to the bird and its momentum vector lie in the xy-plane. What is the bird’s angular momentum about the point where you are standing?

Answer

The magnitude of the cross product of the radius to the bird and its momentum vector yields

r p sin θ

, which gives

r sin θ

as the altitude of the bird h. The direction of the angular momentum is perpendicular to the radius and momentum vectors, which we choose arbitrarily as

\hat{k}

, which is in the plane of the ground:

\vec{L} =

\vec{r} \times \vec{p} =

h m v \hat{k} =

(300.0 m) (2.0 kg) (20.0 m / s) \hat{k} =

12,000.0 kg \cdot m^{2} / s \hat{k}

11-37 §11.2 source ↗

A particle of mass 5.0 kg has position vector

\vec{r} =

(2.0 \hat{i} - 3.0 \hat{j}) m

at a particular instant of time when its velocity is

\vec{v} =

(3.0 \hat{i}) m / s

with respect to the origin. (a) What is the angular momentum of the particle? (b) If a force

\vec{F} = 5.0 \hat{j} N

acts on the particle at this instant, what is the torque about the origin?

Answer

\vec{l} =

45.0 kg \cdot m^{2} / s \hat{k}

;
b.

\vec{τ} =

10.0 N \cdot m \hat{k}

11-39 §11.2 source ↗

Suppose the particles in the preceding problem have masses

m_{1} =

0.10 kg, m_{2} =

0.20 kg, m_{3} =

0.30 kg,

m_{4} = 0.40 kg

. The velocities of the particles are

v_{1} = 2.0 \hat{i} m / s

v_{2} =

(3.0 \hat{i} - 3.0 \hat{j}) m / s

v_{3} =

−1.5 \hat{j} m / s

v_{4} =

−4.0 \hat{i} m / s

. (a) Calculate the angular momentum of each particle about the origin. (b) What is the total angular momentum of the four-particle system about the origin?

Answer

{\vec{l}}_{1} =

−0.4 kg \cdot m^{2} / s \hat{k}

{\vec{l}}_{2} = {\vec{l}}_{4} = 0

{\vec{l}}_{3} =

1.35 kg \cdot m^{2} / s \hat{k}

; b.

\vec{L} =

0.95 kg \cdot m^{2} / s \hat{k}

11-41 §11.2 source ↗

An airplane of mass

4.0 \times 10^{4} kg

flies horizontally at an altitude of 10 km with a constant speed of 250 m/s relative to Earth. (a) What is the magnitude of the airplane’s angular momentum relative to a ground observer directly below the plane? (b) Does the angular momentum change as the airplane flies along a constant altitude?

Answer

L =

1.0 \times 10^{11} kg \cdot m^{2} / s

; b. No, the angular momentum stays the same since the cross-product involves only the perpendicular distance from the plane to the ground no matter where it is along its path.

11-43 §11.2 source ↗

A particle of mass m is dropped at the point $(− d, 0)$ and falls vertically in Earth’s gravitational field $− g \hat{j} .$ (a) What is the expression for the angular momentum of the particle around the z-axis, which points directly out of the page as shown below? (b) Calculate the torque on the particle around the z-axis. (c) Is the torque equal to the time rate of change of the angular momentum?

An x y coordinate system is shown, with positive x to the right and positive y up. A particle is shown on the x axis, to the left of the y axis, at location minus d comma zero. A force minus m g j hat acts downward on the particle.

Answer

\vec{v} =

− g t \hat{j},

{\vec{r}}_{⊥} =

− d \hat{i},

\vec{l} =

m d g t \hat{k}

;
b.

\vec{F} =

− m g \hat{j},

\sum \vec{τ} =

d m g \hat{k}

; c. yes

11-45 §11.2 source ↗

A boulder of mass 20 kg and radius 20 cm rolls down a hill 15 m high from rest. What is its angular momentum when it is half way down the hill? (b) At the bottom?

Answer

m g h =

\frac{1}{2} m {(r ω)}^{2} +

\frac{1}{2} \frac{2}{5} m r^{2} ω^{2}

;

ω = 51.2 rad / s

;

L =

16.4 kg \cdot m^{2} / s

;
b.

ω = 72.5 rad / s

;

L =

23.2 kg \cdot m^{2} / s

11-47 §11.2 source ↗

A propeller consists of two blades each 3.0 m in length and mass 120 kg each. The propeller can be approximated by a single rod rotating about its center of mass. The propeller starts from rest and rotates up to 1200 rpm in 30 seconds at a constant rate. (a) What is the angular momentum of the propeller at

t =

10 s; t =

20 s?

(b) What is the torque on the propeller?

Answer

I =

720.0 kg \cdot m^{2}

;

α =

4.20 rad / s^{2}

;

ω (10 s) =

42.0 rad / s

;

L =

3.02 \times 10^{4} kg \cdot m^{2} / s

;

ω (20 s) =

84.0 rad / s

;
b.

τ =

3.03 \times 10^{3} N \cdot m

11-49 §11.2 source ↗

The blades of a wind turbine are 30 m in length and rotate at a maximum rotation rate of 20 rev/min. (a) If the blades are 6000 kg each and the rotor assembly has three blades, calculate the angular momentum of the turbine at this rotation rate. (b) What is the torque require to rotate the blades up to the maximum rotation rate in 5 minutes?

Answer

L =

1.131 \times 10^{7} kg \cdot m^{2} / s

;
b.

τ =

3.77 \times 10^{4} N \cdot m

11-51 §11.2 source ↗

A mountain biker takes a jump in a race and goes airborne. The mountain bike is travelling at 10.0 m/s before it goes airborne. If the mass of the front wheel on the bike is 750 g and has radius 35 cm, what is the angular momentum of the spinning wheel in the air the moment the bike leaves the ground?

Answer

ω =

28.6 rad / s \Rightarrow L =

2.6 kg \cdot m^{2} / s

11-53 §11.3 source ↗

The Sun’s mass is

2.0 \times 10^{30} kg,

its radius is

7.0 × 10^{5} km,

and it has a rotational period of approximately 28 days. If the Sun should collapse into a white dwarf of radius

3.5 × 10^{3} km,

what would its period be if no mass were ejected and a sphere of uniform density can model the Sun both before and after?

Answer

L_{f} =

\frac{2}{5} M_{S} {(3.5 × 10^{3} km)}^{2} \frac{2 π}{T_{f}}

\begin{matrix} {(7.0 × 10^{5} km)}^{2} \frac{2 π}{28 days} = {(3.5 × 10^{3} km)}^{2} \frac{2 π}{T_{f}} \Rightarrow T_{f} \\ = 28 days \frac{{(3.5 × 10^{3} km)}^{2}}{{(7.0 × 10^{5} km)}^{2}} = 7.0 × 10^{- 4} day = 60.5 s \end{matrix}

11-55 §11.3 source ↗

A diver off the high board imparts an initial rotation with their body fully extended before going into a tuck and executing three back somersaults before hitting the water. If their moment of inertia before the tuck is

16.9 kg \cdot m^{2}

and after the tuck during the somersaults is

4.2 kg \cdot m^{2}

, what rotation rate must the diver impart to their body directly off the board and before the tuck if they take 1.4 s to execute the somersaults before hitting the water?

Answer

f_{f} =

2.1 rev / s \Rightarrow f_{0} =

0.5 rev / s

11-57 §11.3 source ↗

A Molniya orbit is a highly eccentric orbit of a communication satellite so as to provide continuous communications coverage for Scandinavian countries and adjacent Russia. The orbit is positioned so that these countries have the satellite in view for extended periods in time (see below). If a satellite in such an orbit has an apogee at 40,000.0 km as measured from the center of Earth and a velocity of 1.68 km/s, what would be its velocity at perigee measured at 200.0 km altitude?

An highly eccentric elliptic orbit around the Earth is shown. The Earth is at one focal point of the ellipse. 11 points corresponding to time in hours are marked on the orbit. Time 0 is at the perigee (the point on the orbit that is closest to the earth, and point 6 is at the apogee, the point on the orbit farthest from the earth. The spacing of the points 0 through 6 along the orbit decreases with time, and the spacing from 6 to 11 and back to 0 increases.

Answer

r_{P} m v_{P} =

r_{A} m v_{A} \Rightarrow v_{P} =

10.2 km / s

11-59 §11.3 source ↗

A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk

(M = 0.10 kg, R = 0.10 m)

rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 10.0 rad/s. The bug crawls to the center of the disk. (a) What is the new angular velocity of the disk? (b) What is the change in the kinetic energy of the system? (c) If the bug crawls back to the outer edge of the disk, what is the angular velocity of the disk then? (d) What is the new kinetic energy of the system? (e) What is the cause of the increase and decrease of kinetic energy?

Answer

I_{disk} =

5.0 \times 10^{−4} kg \cdot m^{2}

I_{bug} =

2.0 × 10^{−4} kg \cdot m^{2}

\begin{matrix} (I_{disk} + I_{bug}) ω_{1} = I_{disk} ω_{2}, \\ ω_{2} = 14.0 rad / s \end{matrix}

Δ K = 0.014 J

;
c.

ω_{3} =

10.0 rad / s

back to the original value;
d.

\frac{1}{2} (I_{disk} + I_{bug}) ω_{3}^{2} =

0.035 J

back to the original value;
e. work of the bug crawling on the disk

11-61 §11.3 source ↗

A merry-go-round has a radius of 2.0 m and a moment of inertia

300 kg \cdot m^{2} .

A boy of mass 50 kg runs tangent to the rim at a speed of 4.0 m/s and jumps on. If the merry-go-round is initially at rest, what is the angular velocity after the boy jumps on?

Answer

L_{i} =

400.0 kg \cdot m^{2} / s

L_{f} =

500.0 kg \cdot m^{2} ω

ω = 0.80 rad / s

11-63 §11.3 source ↗

Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

Answer

I_{0} =

340.48 kg \cdot m^{2}

I_{f} =

268.8 kg \cdot m^{2}

ω_{f} = 25.33 rpm

11-65 §11.3 source ↗

Twin skaters approach one another as shown below and lock hands. (a) Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius. (b) Compare the initial kinetic energy and final kinetic energy.

Figure a, on the left, is a drawing of two ice skaters viewed from above moving with speed v toward each other along parallel lines. The upper one is skating to the right and the lower one to the left, and they are separated so that their hands will meet as they cross. Figure b, on the right, shows the skaters holding hands and moving together in a circle with angular velocity omega. Their motion is clockwise as viewed from above.

Answer

L =

280 kg \cdot m^{2} / s

I_{f} =

89.6 kg \cdot m^{2}

ω_{f} =

3.125 rad / s

; b.

K_{i} = 437.5 J

K_{f} = 437.5 J

11-67 §11.3 source ↗

In 2015, in Warsaw, Poland, Olivia Oliver of Nova Scotia broke the world record for being the fastest spinner on ice skates. She achieved a record 342 rev/min, beating the existing Guinness World Record by 34 rotations. If an ice skater extends her arms at that rotation rate, what would be her new rotation rate? Assume she can be approximated by a 45-kg rod that is 1.7 m tall with a radius of 15 cm in the record spin. With her arms stretched take the approximation of a rod of length 130 cm with

10 %

of her body mass aligned perpendicular to the spin axis. Neglect frictional forces.

Answer

Moment of inertia in the record spin:

I_{0} = 0.5 kg \cdot m^{2}

I_{f} = 1.1 kg \cdot m^{2}

ω_{f} =

\frac{I_{0}}{I_{f}} ω_{0} \Rightarrow f_{f} =

155.5 rev / min

11-69 §11.3 source ↗

A gymnast does cartwheels along the floor and then launches herself into the air and executes several flips in a tuck while she is airborne. If her moment of inertia when executing the cartwheels is

13.5 kg \cdot m^{2}

and her spin rate is 0.5 rev/s, how many revolutions does she do in the air if her moment of inertia in the tuck is

3.4 kg \cdot m^{2}

and she has 2.0 s to do the flips in the air?

Answer

Her spin rate in the air is:

f_{f} = 2.0 rev / s

;
She can do four flips in the air.

11-71 §11.3 source ↗

A ride at a carnival has four spokes to which pods are attached that can hold two people. The spokes are each 15 m long and are attached to a central axis. Each spoke has mass 200.0 kg, and the pods each have mass 100.0 kg. If the ride spins at 0.2 rev/s with each pod containing two 50.0-kg children, what is the new spin rate if all the children jump off the ride?

Answer

Moment of inertia with all children aboard:

I_{0} =

2.4 × 10^{5} kg \cdot m^{2}

;

I_{f} =

1.5 × 10^{5} kg \cdot m^{2}

;

f_{f} = 0.3 rev / s

11-73 §11.3 source ↗

A space station consists of a giant rotating hollow cylinder of mass

10^{6} kg

including people on the station and a radius of 100.00 m. It is rotating in space at 3.30 rev/min in order to produce artificial gravity. If 100 people of an average mass of 65.00 kg spacewalk to an awaiting spaceship, what is the new rotation rate when all the people are off the station?

Answer

I_{0} =

1.00 × 10^{10} kg \cdot m^{2}

I_{f} =

9.94 × 10^{9} kg \cdot m^{2}

f_{f} =

3.32 rev / min

11-75 §11.4 source ↗

A gyroscope has a 0.5-kg disk that spins at 40 rev/s. The center of mass of the disk is 15 cm from a pivot with a radius of the disk of 10 cm. What is the precession angular velocity?

Answer

I =

2.5 \times 10^{−3} kg \cdot m^{2}

ω_{P} =

1.17 rad / s

11-77 §11.4 source ↗

The axis of Earth makes a $23.5 °$ angle with a direction perpendicular to the plane of Earth’s orbit. As shown below, this axis precesses, making one complete rotation in 25,780 y.

(a) Calculate the change in angular momentum in half this time.

(b) What is the average torque producing this change in angular momentum?

(c) If this torque were created by a pair of forces acting at the most effective point on the equator, what would the magnitude of each force be?

In the figure, the Earth’s image is shown. The plane of the Earth’s orbit is shown as a horizontal line at the equator. The Earth’s north south axis is inclined at an angle of 23.5 degrees from the vertical. There are two vectors, L and L prime, inclined at an angle of twenty three point five degree to the vertical, starting from the center of the Earth. Vector L goes through the Earth’s north pole. At the heads of the two vectors there is a circle, directed in counter clockwise direction as viewed from above. An angular momentum vector, Delta L, directed toward left, along its diameter, is shown.

Answer

L_{Earth} =

7.06 × 10^{33} kg \cdot m^{2} / s

Δ L =

5.63 × 10^{33} kg \cdot m^{2} / s

;
b.

τ =

1.4 × 10^{22} N \cdot m

;
c. The two forces at the equator would have the same magnitude but different directions, one in the north direction and the other in the south direction on the opposite side of Earth. The angle between the forces and the lever arms to the center of Earth is

90 °

, so a given torque would have magnitude

τ =

F R_{E} sin 90 ° =

F R_{E}

. Both would provide a torque in the same direction:

τ =

2 F R_{E} \Rightarrow F =

1.1 × 10^{15} N

11-79 §11.AP source ↗

Repeat the preceding problem replacing the marble with a hollow sphere. Explain the new results.

Answer

a_{CM} = - \frac{3}{10} g

v^{2} =

v_{0}^{2} +

2 a_{CM} x \Rightarrow v^{2} =

(7.0 m / {s)}^{2} -

2 (\frac{3}{10} g) x,

v^{2} =

0 \Rightarrow x =

8.34 m;

t = \frac{v - v_{0}}{a_{CM}},

v =

v_{0} +

a_{CM} t \Rightarrow t =

2.38 s

;
The hollow sphere has a larger moment of inertia, and therefore is harder to bring to a rest than the marble, or solid sphere. The distance travelled is larger and the time elapsed is longer.

11-81 §11.AP source ↗

Repeat the preceding problem for a hollow sphere of the same radius and mass and initial speed. Explain the differences in the results.

Answer

W = −500.0 J

;
b.

K +

U_{grav} =

constant

500 J +

0 =

0 +

(6.0 kg) (9.8 m / s^{2}) h

h =

8.5 m, d =

17.0 m

;
The moment of inertia is less for the hollow sphere, therefore less work is required to stop it. Likewise it rolls up the incline a shorter distance than the hoop.

11-83 §11.AP source ↗

A 4.0-kg particle moves in a circle of radius 2.0 m. The angular momentum of the particle varies in time according to

l = 5.0 t^{2} .

(a) What is the torque on the particle about the center of the circle at

t = 3.4 s

? (b) What is the angular velocity of the particle at

t = 3.4 s

Answer

τ = 34.0 N \cdot m

;
b.

l =

m r^{2} ω \Rightarrow ω =

3.6 rad / s

11-85 §11.AP source ↗

(a) What is the angular momentum of the Moon in its orbit around Earth? (b) How does this angular momentum compare with the angular momentum of the Moon on its axis? Remember that the Moon keeps one side toward Earth at all times.

Answer

d_{M} =

3.85 × 10^{8} m

average distance to the Moon; orbital period

27.32 d =

2.36 × 10^{6} s

; speed of the Moon

\frac{2 π 3.85 × 10^{8} m}{2.36 × 10^{6} s} =

1.0 × 10^{3} m / s

; mass of the Moon

7.35 × 10^{22} kg

L =

2.90 × 10^{34} kg m^{2} / s

;
b. radius of the Moon

1.74 × 10^{6} m

; the orbital period is the same as (a):

ω =

2.66 × 10^{−6} rad / s

L =

2.37 × 10^{29} kg \cdot m^{2} / s

;
The orbital angular momentum is

1.22 \times 10^{5}

times larger than the rotational angular momentum for the Moon.

11-87 §11.AP source ↗

A potter’s disk spins from rest up to 10 rev/s in 15 s. The disk has a mass 3.0 kg and radius 30.0 cm. What is the angular momentum of the disk at

t = 5 s, t = 1 0 s

Answer

I =

0.135 kg \cdot m^{2}

α =

4.19 rad / s^{2},

ω = ω_{0} + α t

ω (5 s) =

21.0 rad / s,

L =

2.84 kg \cdot m^{2} / s

ω (10 s) =

41.9 rad / s,

L =

5.66 kg \cdot m / s^{2}

11-89 §11.AP source ↗

A solid cylinder of mass 2.0 kg and radius 20 cm is rotating counterclockwise around a vertical axis through its center at 600 rev/min. A second solid cylinder of the same mass and radius is rotating clockwise around the same vertical axis at 900 rev/min. If the cylinders couple so that they rotate about the same vertical axis, what is the angular velocity of the combination?

Answer

In the conservation of angular momentum equation, the rotation rate appears on both sides so we keep the (rev/min) notation as the angular velocity can be multiplied by a constant to get (rev/min):

\begin{matrix} L_{i} & = & −0.04 kg \cdot m^{2} (300.0 rev / min), \\ L_{f} & = & 0.08 kg \cdot m^{2} f_{f} \Rightarrow f_{f} = −150.0 rev / min clockwise \end{matrix}

11-91 §11.AP source ↗

Eight children, each of mass 40 kg, climb on a small merry-go-round. They position themselves evenly on the outer edge and join hands. The merry-go-round has a radius of 4.0 m and a moment of inertia

1000.0 kg \cdot m^{2}

. After the merry-go-round is given an angular velocity of 6.0 rev/min, the children walk inward and stop when they are 0.75 m from the axis of rotation. What is the new angular velocity of the merry-go-round? Assume there is negligible frictional torque on the structure.

Answer

I_{0} ω_{0} = I_{f} ω_{f}

I_{0} =

6120.0 kg \cdot m^{2}

I_{f} =

1180.0 kg \cdot m^{2}

ω_{f} =

31.1 rev / min

11-93 §11.AP source ↗

A satellite in the shape of a sphere of mass 20,000 kg and radius 5.0 m is spinning about an axis through its center of mass. It has a rotation rate of 8.0 rev/s. Two antennas deploy in the plane of rotation extending from the center of mass of the satellite. Each antenna can be approximated as a rod has mass 200.0 kg and length 7.0 m. What is the new rotation rate of the satellite?

Answer

L_{i} =

1.00 × 10^{7} kg \cdot m^{2} / s

I_{f} =

2.025 × 10^{5} kg \cdot m^{2}

ω_{f} =

7.86 rev / s

11-95 §11.CP source ↗

The truck shown below is initially at rest with solid cylindrical roll of paper sitting on its bed. If the truck moves forward with a uniform acceleration a, what distance s does it move before the paper rolls off its back end? (Hint: If the roll accelerates forward with $a^{'}$ , then is accelerates backward relative to the truck with an acceleration $a - a^{'}$ . Also, $R α = a - a^{'}$ .)

A drawing of a flatbed truck on a horizontal road. The truck is accelerating forward with acceleration a. The bed of the truck has a cylinder on it, a distance d from the back end of the bed.

Answer

Assume the roll accelerates forward with respect to the ground with an acceleration $a^{'}$ . Then it accelerates backwards relative to the truck with an acceleration $(a - a^{'})$ .

The forces on a cylinder on a horizontal surface are shown. The cylinder has radius R and moment of inertia one half m R squared and is centered on an x y coordinate system that has positive x to the right and positive y up. Force m g acts on the center of the cylinder and points down. Force N points up and acts at the contact point where the cylinder touches the surface. Force f sub s points to the right and acts at the contact point where the cylinder touches the surface.

Also, $R α =$ $a -$ $a^{'} I =$ $\frac{1}{2} m R^{2} \sum F_{x} = f_{s} = m a^{'}$ ,
$\sum τ =$ $f_{s} R =$ $I α =$ $I \frac{a - a^{'}}{R} f_{s} =$ $\frac{I}{R^{2}} (a - a^{'}) =$ $\frac{1}{2} m (a - a^{'})$ ,
Solving for $a^{'}$ : $f_{s} =$ $\frac{1}{2} m (a - a^{'})$ ; $a^{'} = \frac{a}{3}$ ,
$x -$ $x_{0} =$ $v_{0} t +$ $\frac{1}{2} a t^{2};$ $d =$ $\frac{1}{3} a t^{2};$ $t =$ $\sqrt{\frac{3 d}{a}};$
therefore, $s = 1.5 d$

11-97 §11.CP source ↗

A small ball of mass 0.50 kg is attached by a massless string to a vertical rod that is spinning as shown below. When the rod has an angular velocity of 6.0 rad/s, the string makes an angle of $30 °$ with respect to the vertical. (a) If the angular velocity is increased to 10.0 rad/s, what is the new angle of the string? (b) Calculate the initial and final angular momenta of the ball. (c) Can the rod spin fast enough so that the ball is horizontal?

A vertical rod is spinning on its axis. A string is attached to the top of the rod at one end and a ball at the other end. The string hangs down at an angle from the rod.

Answer

a. The tension in the string provides the centripetal force such that

T sin θ = m r_{⊥} ω^{2}

. The component of the tension that is vertical opposes the gravitational force such that

T cos θ = m g

. This gives

T = 5.7 N

. We solve for

r_{⊥} = 0.16 m

. This gives the length of the string as

r = 0.32 m

.
At

ω = 10.0 rad / s

, there is a new angle, tension, and perpendicular radius to the rod. Dividing the two equations involving the tension to eliminate it, we have

\frac{sin θ}{cos θ} =

\frac{(0.32 m sin θ) ω^{2}}{g} \Rightarrow \frac{1}{cos θ} =

\frac{0.32 m ω^{2}}{g}

;

cos θ =

0.31 \Rightarrow θ =

72.2 °

; b.

l_{initial} =

0.08 kg \cdot m^{2} / s

l_{final} =

0.46 kg \cdot m^{2} / s

; c. No, the cosine of the angle is inversely proportional to the square of the angular velocity, therefore in order for

θ \to 90 °, ω \to \infty

. The rod would have to spin infinitely fast.

Loading chapter 12…

Problems and answers are adapted from University Physics Volume 1 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-1/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.