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Physics of Hearing

- Understand the analogy between angular momentum and linear momentum.
- Observe the relationship between torque and angular momentum.
- Apply the law of conservation of angular momentum.

Angular momentum is completely analogous to linear momentum, first presented in Uniform Circular Motion and Gravitation. It has the same implications in terms of carrying rotation forward, and it is conserved when the net external torque is zero. Angular momentum, like linear momentum, is also a property of the atoms and subatomic particles.

$I=\frac{2{MR}^{2}}{5}\\$

so that$L=I\omega=\frac{2{MR}^{2}\omega}{5}\\$

.
Earth’s mass $\begin{array}{lll}L& =& 0.4{\left(5.979\times 10^{24}\text{ kg}\right)}{\left(6.376\times 10^{6}\text{ m}\right)}^{2}\left(\frac{1\text{ rev}}{\text{d}}\right)\\ & =& 9.72\times {10}^{37}\text{ kg}\cdot {\text{m}}^{2}\cdot \text{rev/d}\end{array}\\$

.
Substituting 2π rad for 1 rev and 8.64 × 10$\begin{array}{lll}L& =& {\left(9.72\times {10}^{37}\text{ kg}\cdot {\text{m}}^{2}\right)}\left(\frac{2\pi \text{ rad/rev}}{8.64\times {10}^{4}\text{ s/d}}\right)\left(1\text{ rev/d}\right)\\ & =& 7.07 \times {10}^{33}\text{ kg}\cdot {\text{ m}}^{2}\text{/s}\end{array}\\$

.
When you push a merry-go-round, spin a bike wheel, or open a door, you exert a torque. If the torque you exert is greater than opposing torques, then the rotation accelerates, and angular momentum increases. The greater the net torque, the more rapid the increase in *L*. The relationship between torque and angular momentum is

$\text{net }\tau =\frac{\Delta L}{\Delta t}\\$

.$\text{net }\tau =\frac{\Delta L}{\Delta t}\\$

is very fundamental and broadly applicable. It is, in fact, the rotational form of Newton’s second law.
Figure 2 shows a Lazy Susan food tray being rotated by a person in quest of sustenance. Suppose the person exerts a 2.50 N force perpendicular to the lazy Susan’s 0.260-m radius for 0.150 s. (a) What is the final angular momentum of the lazy Susan if it starts from rest, assuming friction is negligible? (b) What is the final angular velocity of the lazy Susan, given that its mass is 4.00 kg and assuming its moment of inertia is that of a disk?

**Strategy**

We can find the angular momentum by solving*L*, and using the given information to calculate the torque. The final angular momentum equals the change in angular momentum, because the lazy Susan starts from rest. That is, Δ*L *= *L*. To find the final velocity, we must calculate *ω* from the definition of *L* in *L *= *Iω*.

**Solution for (a)**

Solving *L* gives

*r*, we see that

**Solution for (b)**

The final angular velocity can be calculated from the definition of angular momentum,

*L *= *Iω*.
Solving for *ω* and substituting the formula for the moment of inertia of a disk into the resulting equation gives

**Discussion**

Note that the imparted angular momentum does not depend on any property of the object but only on torque and time. The final angular velocity is equivalent to one revolution in 8.71 s (determination of the time period is left as an exercise for the reader), which is about right for a lazy Susan.

$\text{net }\tau =\frac{\Delta L}{\Delta t}\\$

for $\Delta L\\$

for Δ$\text{net}\tau =\frac{\Delta L}{\Delta t}\\$

for Δ$\Delta L=\left(\text{net}\tau\right){\Delta t}\\$

Because the force is perpendicular to $\text{net }\tau ={rF}\\$

, so that$\begin{array}{lll}L& =& {rF}\Delta t=\left(0.260 \text{ m}\right)\left(2.50 \text{ N}\right)\left(0.150 \text{s}\right)\\ & =& 9.75\times {10}^{-2}\text{ kg}\cdot {\text{m}}^{2}/\text{s}\end{array}\\$

.
$\omega =\frac{L}{I}=\frac{L}{\frac{1}{2}{{MR}}^{2}}\\$

.
And substituting known values into the preceding equation yields$\omega =\frac{9.75\times {10}^{-2}\text{ kg}\cdot{\text{ m}}^{2}\text{/s}}{\left(0.500\right)\left(4.00\text{ kg}\right)\left(0.260\text{ m}\right)}=0.721\text{ rad/s}\\$

.
The person whose leg is shown in Figure 3 kicks his leg by exerting a 2000-N force with his upper leg muscle. The effective perpendicular lever arm is 2.20 cm. Given the moment of inertia of the lower leg is 1.25 kg⋅m^{2}, (a) find the angular acceleration of the leg. (b) Neglecting the gravitational force, what is the rotational kinetic energy of the leg after it has rotated through 57.3º (1.00 rad)?

**Strategy**

The angular acceleration can be found using the rotational analog to Newton’s second law, or *I* is given and the torque can be found easily from the given force and perpendicular lever arm. Once the angular acceleration *α* is known, the final angular velocity and rotational kinetic energy can be calculated.

**Solution to (a)**

From the rotational analog to Newton’s second law, the angular acceleration *α* is

*α* gives

**Solution to (b)**

The final angular velocity can be calculated from the kinematic expression

*ω*^{2} just found and the given value for the moment of inertia. The kinetic energy is then

**Discussion**

These values are reasonable for a person kicking his leg starting from the position shown. The weight of the leg can be neglected in part (a) because it exerts no torque when the center of gravity of the lower leg is directly beneath the pivot in the knee. In part (b), the force exerted by the upper leg is so large that its torque is much greater than that created by the weight of the lower leg as it rotates. The rotational kinetic energy given to the lower leg is enough that it could give a ball a significant velocity by transferring some of this energy in a kick.

$\alpha =\text{net }\tau/I\\$

. The moment of inertia $\alpha =\frac{\text{net}\tau }{I}\\$

.
Because the force and the perpendicular lever arm are given and the leg is vertical so that its weight does not create a torque, the net torque is thus$\begin{array}{lll}\text{net}\tau & =& {r}_{\perp }F\\ & =& \left(0\text{.}\text{0220 m}\right)\left(\text{2000}\text{N}\right)\\ & =& 44.0\text{ N}\cdot \text{m}\end{array}\\$

.
Substituting this value for the torque and the given value for the moment of inertia into the expression for $\alpha =\frac{{44.0}\text{ N}\cdot\text{ m}}{{1.25}\text{ kg}\cdot\text{ m}^{2}}=35.2{\text{ rad/s}}^{2}\\$

.
${\omega}^{2}={\omega_{0}}^{2}+2\alpha\theta\\$

or${\omega}^{2}=2\alpha\theta\\$

because the initial angular velocity is zero. The kinetic energy of rotation is${\text{KE}}_{\text{rot}}=\frac{1}{2}{{I\omega}}^{2}\\$

so it is most convenient to use the value of $\begin{array}{lll}{\text{KE}}_{\text{rot}}& =& 0.5\left(1\text{.25}\text{ kg}\cdot {\text{m}}^{2}\right)\left(\text{70.}4{\text{ rad}}^{2}/{\text{s}}^{2}\right)\\ & =& \text{44}\text{.}0\text{ J}\end{array}\\$

.
Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero.

$\Delta L=\left(\text{net }\tau\right)Delta t\\$

. This equation means that, to change angular momentum, a torque must act over some period of time. Because Earth has a large angular momentum, a large torque acting over a long time is needed to change its rate of spin. So what external torques are there? Tidal friction exerts torque that is slowing Earth’s rotation, but tens of millions of years must pass before the change is very significant. Recent research indicates the length of the day was 18 h some 900 million years ago. Only the tides exert significant retarding torques on Earth, and so it will continue to spin, although ever more slowly, for many billions of years.What we have here is, in fact, another conservation law. If the net torque is

$\text{net }\tau =\frac{\Delta L}{\Delta t}\\$

for the situation in which the net torque is zero. In that case,net *τ *= 0

$\frac{\Delta L}{\Delta t}=0\\$

.
If the change in angular momentum Δ*L* = constant (net *τ* = 0)

*L* = *L*′ (net *τ* = 0).

*L *= *L*′.

*Iω *= *I*′*ω*′,

${\text{KE}}_{\text{rot}}=\frac{1}{2}{{I\omega }}^{2}\\$

.
$\begin{array}{lll}\omega′ & =& \frac{I}{I′}\omega =\left(\frac{\text{2.34 kg}\cdot {m}^{2}}{0\text{.363 kg}\cdot {m}^{2}}\right)\left(\text{0.800 rev/s}\right)\\ & =& \text{}\text{5.16 rev/s}\end{array}\\$

.
${\text{KE}}_{\text{rot}}=\frac{1}{2}{{I\omega }}^{2}\\$

.
The initial value is found by substituting known values into the equation and converting the angular velocity to rad/s:$\begin{array}{lll}{\text{KE}}_{\text{rot}}& =& \left(0.5\right)\left(2.34\text{ kg}\cdot{\text{m}}^{2}\right){\left(\left(0.800\text{rev/s}\right)\left(2\pi \text{ rad/rev}\right)\right)}^{2}\\ & =& 29.6\text{ J}\end{array}\\$

.
The final rotational kinetic energy is${\text{KE}}_{\text{rot}}′ =\frac{1}{2}I′ {\omega′}^{2}\\$

.
Substituting known values into this equation gives$\begin{array}{lll}{KE_{\text{rot}}}′ & =& \left(0\text{.}5\right)\left(0\text{.363 kg}\cdot {m}^{2}\right){\left[\left(5\text{.}\text{16 rev/s}\right)\left(2\pi \text{ rad/rev}\right)\right]}^{2}\\ & =&\text{191 J.}\end{array}\\$

Is angular momentum completely analogous to linear momentum? What, if any, are their differences?

**Solution**

Yes, angular and linear momentums are completely analogous. While they are exact analogs they have different units and are not directly inter-convertible like forms of energy are.

- Every rotational phenomenon has a direct translational analog , likewise angular momentum
*L*can be defined as*L = Iω*. - This equation is an analog to the definition of linear momentum as
*p = mv*. The relationship between torque and angular momentum is$\text{net }\tau =\frac{\Delta L}{\Delta t}\\$. - Angular momentum, like energy and linear momentum, is conserved. This universally applicable law is another sign of underlying unity in physical laws. Angular momentum is conserved when net external torque is zero, just as linear momentum is conserved when the net external force is zero.

1. When you start the engine of your car with the transmission in neutral, you notice that the car rocks in the opposite sense of the engine’s rotation. Explain in terms of conservation of angular momentum. Is the angular momentum of the car conserved for long (for more than a few seconds)?

2. Suppose a child walks from the outer edge of a rotating merry-go round to the inside. Does the angular velocity of the merry-go-round increase, decrease, or remain the same? Explain your answer.

3. Suppose a child gets off a rotating merry-go-round. Does the angular velocity of the merry-go-round increase, decrease, or remain the same if: (a) He jumps off radially? (b) He jumps backward to land motionless? (c) He jumps straight up and hangs onto an overhead tree branch? (d) He jumps off forward, tangential to the edge? Explain your answers. (Refer to Figure 6).

4. Helicopters have a small propeller on their tail to keep them from rotating in the opposite direction of their main lifting blades. Explain in terms of Newton’s third law why the helicopter body rotates in the opposite direction to the blades.

5. Whenever a helicopter has two sets of lifting blades, they rotate in opposite directions (and there will be no tail propeller). Explain why it is best to have the blades rotate in opposite directions.

6. Describe how work is done by a skater pulling in her arms during a spin. In particular, identify the force she exerts on each arm to pull it in and the distance each moves, noting that a component of the force is in the direction moved. Why is angular momentum not increased by this action?

7. When there is a global heating trend on Earth, the atmosphere expands and the length of the day increases very slightly. Explain why the length of a day increases.

8. Nearly all conventional piston engines have flywheels on them to smooth out engine vibrations caused by the thrust of individual piston firings. Why does the flywheel have this effect?

9. Jet turbines spin rapidly. They are designed to fly apart if something makes them seize suddenly, rather than transfer angular momentum to the plane’s wing, possibly tearing it off. Explain how flying apart conserves angular momentum without transferring it to the wing.

10. An astronaut tightens a bolt on a satellite in orbit. He rotates in a direction opposite to that of the bolt, and the satellite rotates in the same direction as the bolt. Explain why. If a handhold is available on the satellite, can this counter-rotation be prevented? Explain your answer.

11. Competitive divers pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down. Explain the effect of both actions on their angular velocities. Also explain the effect on their angular momenta.

12. Draw a free body diagram to show how a diver gains angular momentum when leaving the diving board.

13. In terms of angular momentum, what is the advantage of giving a football or a rifle bullet a spin when throwing or releasing it?

(b) Compare this angular momentum with the angular momentum of Earth on its axis.

2. (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.

(c) Discuss whether the values found in parts (a) and (b) seem consistent with the fact that tidal effects with Earth have caused the Moon to rotate with one side always facing Earth.

3. Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s. What angular momentum is given to the engine if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time?

4. A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge? The child is initially at rest.

5. 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?

6. (a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.400 kg. (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 15.0 s?

7.

- angular momentum:
- the product of moment of inertia and angular velocity

- law of conservation of angular momentum:
- angular momentum is conserved, i.e., the initial angular momentum is equal to the final angular momentum when no external torque is applied to the system

The angular momentum of the Earth in its orbit around the Sun 3.77 × 10

3. 22.5 kg ⋅ m

5. 25.3 rpm