# 10.4 Moment of inertia and rotational kinetic energy

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## Energy in a boomerang

A person hurls a boomerang into the air with a velocity of 30.0 m/s at an angle of $40.0\text{°}$ with respect to the horizontal ( [link] ). It has a mass of 1.0 kg and is rotating at 10.0 rev/s. The moment of inertia of the boomerang is given as $I=\frac{1}{12}m{L}^{2}$ where $L=0.7\phantom{\rule{0.2em}{0ex}}\text{m}$ . (a) What is the total energy of the boomerang when it leaves the hand? (b) How high does the boomerang go from the elevation of the hand, neglecting air resistance?

## Strategy

We use the definitions of rotational and linear kinetic energy to find the total energy of the system. The problem states to neglect air resistance, so we don’t have to worry about energy loss. In part (b), we use conservation of mechanical energy to find the maximum height of the boomerang.

## Solution

1. Moment of inertia: $I=\frac{1}{12}m{L}^{2}=\frac{1}{12}\left(1.0\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(0.7{\text{m}\right)}^{2}=0.041\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}$ .
Angular velocity: $\omega =\left(10.0\phantom{\rule{0.2em}{0ex}}\text{rev}\text{/}\text{s}\right)\left(2\pi \right)=62.83\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}$ .
The rotational kinetic energy is therefore
${K}_{\text{R}}=\frac{1}{2}\left(0.041\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}\right){\left(62.83\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}\right)}^{2}=80.93\phantom{\rule{0.2em}{0ex}}\text{J}.$

The translational kinetic energy is
${K}_{\text{T}}=\frac{1}{2}m{v}^{2}=\frac{1}{2}\left(1.0\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(30.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}{\right)}^{2}=450.0\phantom{\rule{0.2em}{0ex}}\text{J}.$

Thus, the total energy in the boomerang is
${K}_{\text{Total}}={K}_{\text{R}}+{K}_{\text{T}}=80.93+450.0=530.93\phantom{\rule{0.2em}{0ex}}\text{J}.$
2. We use conservation of mechanical energy. Since the boomerang is launched at an angle, we need to write the total energies of the system in terms of its linear kinetic energies using the velocity in the x - and y -directions. The total energy when the boomerang leaves the hand is
${E}_{\text{Before}}=\frac{1}{2}m{v}_{x}^{2}+\frac{1}{2}m{v}_{y}^{2}+\frac{1}{2}I{\omega }^{2}.$

The total energy at maximum height is
${E}_{\text{Final}}=\frac{1}{2}m{v}_{x}^{2}+\frac{1}{2}I{\omega }^{2}+mgh.$

By conservation of mechanical energy, ${E}_{\text{Before}}={E}_{\text{Final}}$ so we have, after canceling like terms,
$\frac{1}{2}m{v}_{y}^{2}=mgh.$

Since ${v}_{y}=30.0\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}\left(\text{sin}\phantom{\rule{0.2em}{0ex}}40\text{°}\right)=19.28\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}$ , we find
$h=\frac{{\left(19.28\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}\text{s}\right)}^{2}}{2\left(9.8\phantom{\rule{0.2em}{0ex}}\text{m}\text{/}{\text{s}}^{2}\right)}=18.97\phantom{\rule{0.2em}{0ex}}\text{m}.$

## Significance

In part (b), the solution demonstrates how energy conservation is an alternative method to solve a problem that normally would be solved using kinematics. In the absence of air resistance, the rotational kinetic energy was not a factor in the solution for the maximum height.

Check Your Understanding A nuclear submarine propeller has a moment of inertia of $800.0\phantom{\rule{0.2em}{0ex}}\text{kg}·{\text{m}}^{2}$ . If the submerged propeller has a rotation rate of 4.0 rev/s when the engine is cut, what is the rotation rate of the propeller after 5.0 s when water resistance has taken 50,000 J out of the system?

The initial rotational kinetic energy of the propeller is
${K}_{0}=\frac{1}{2}I{\omega }^{2}=\frac{1}{2}\left(800.0{\phantom{\rule{0.2em}{0ex}}\text{kg-m}}^{2}\right)\left(4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2\pi \phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}{\text{s}\right)}^{2}=2.53\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{J}$ .
At 5.0 s the new rotational kinetic energy of the propeller is
${K}_{\text{f}}=2.03\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{J}$ .
and the new angular velocity is
$\omega =\sqrt{\frac{2\left(2.03\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{J}\right)}{800.0{\phantom{\rule{0.2em}{0ex}}\text{kg-m}}^{2}}}=22.53\phantom{\rule{0.2em}{0ex}}\text{rad}\text{/}\text{s}$
which is 3.58 rev/s.

## Summary

• The rotational kinetic energy is the kinetic energy of rotation of a rotating rigid body or system of particles, and is given by $K=\frac{1}{2}I{\omega }^{2}$ , where I is the moment of inertia, or “rotational mass” of the rigid body or system of particles.
• The moment of inertia for a system of point particles rotating about a fixed axis is $I=\sum _{j}{m}_{j}{r}_{j}^{2}$ , where ${m}_{j}$ is the mass of the point particle and ${r}_{j}$ is the distance of the point particle to the rotation axis. Because of the ${r}^{2}$ term, the moment of inertia increases as the square of the distance to the fixed rotational axis. The moment of inertia is the rotational counterpart to the mass in linear motion.
• In systems that are both rotating and translating, conservation of mechanical energy can be used if there are no nonconservative forces at work. The total mechanical energy is then conserved and is the sum of the rotational and translational kinetic energies, and the gravitational potential energy.

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