# 0.5 8.6 - angular momentum and its conservation  (Page 4/6)

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Solution for (b)

Rotational kinetic energy is given by

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

The initial value is found by substituting known values into the equation and converting the angular velocity to rad/s:

$\begin{array}{lll}{\text{KE}}_{\text{rot}}& =& \left(0\text{.}5\right)\left(2\text{.}\text{34}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot {\text{m}}^{2}\right){\left(\left(0\text{.}\text{800}\phantom{\rule{0.25em}{0ex}}\text{rev/s}\right)\left(2\pi \phantom{\rule{0.25em}{0ex}}\text{rad/rev}\right)\right)}^{2}\\ & =& 29.6\phantom{\rule{0.25em}{0ex}}\text{J.}\end{array}$

The final rotational kinetic energy is

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

Substituting known values into this equation gives

$\begin{array}{lll}K{E}_{\text{rot}}\prime & =& \left(0\text{.}5\right)\left(0\text{.363 kg}\cdot {m}^{2}\right){\left[\left(5\text{.}\text{16 rev/s}\right)\left(2\pi rad/rev\right)\right]}^{2}\\ & =& \text{191 J.}\end{array}$

Discussion

In both parts, there is an impressive increase. First, the final angular velocity is large, although most world-class skaters can achieve spin rates about this great. Second, the final kinetic energy is much greater than the initial kinetic energy. The increase in rotational kinetic energy comes from work done by the skater in pulling in her arms. This work is internal work that depletes some of the skater’s food energy.

There are several other examples of objects that increase their rate of spin because something reduced their moment of inertia. Tornadoes are one example. Storm systems that create tornadoes are slowly rotating. When the radius of rotation narrows, even in a local region, angular velocity increases, sometimes to the furious level of a tornado. Earth is another example. Our planet was born from a huge cloud of gas and dust, the rotation of which came from turbulence in an even larger cloud. Gravitational forces caused the cloud to contract, and the rotation rate increased as a result. (See [link] .)

In case of human motion, one would not expect angular momentum to be conserved when a body interacts with the environment as its foot pushes off the ground. Astronauts floating in space aboard the International Space Station have no angular momentum relative to the inside of the ship if they are motionless. Their bodies will continue to have this zero value no matter how they twist about as long as they do not give themselves a push off the side of the vessel.

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

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.

## Section summary

• Every rotational phenomenon has a direct translational analog , likewise angular momentum $L$ can be defined as $L=\mathrm{I\omega }.$
• This equation is an analog to the definition of linear momentum as $p=\text{mv}$ . The relationship between torque and angular momentum is $\text{net}\phantom{\rule{0.25em}{0ex}}\tau =\frac{\text{Δ}L}{\text{Δ}t}.$
• 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.

## Conceptual questions

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.

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 [link] ).

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?

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.

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.

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.

## Problems&Exercises

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?

$\text{22}\text{.}\text{5 kg}\cdot {\text{m}}^{2}\text{/s}$

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.

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?

25.3 rpm

(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is $0\text{.}\text{400}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot {\text{m}}^{2}$ . (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?

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