Question

An insect of mass 20 g crawls from the centre to the outside edge of a rotating disc of mass 200g and radius 20 cm. The disk was initially rotating at 22.0 rads-1. What will be its final angular velocity? What is the change in the kinetic energy of the system?

Answer 1

Answer on Question #65495-Mechanics - Relativity

An insect of mass $m = 20$ g crawls from the centre to the outside edge of a rotating disc of mass $M = 200$ g and radius $r = 20$ cm. The disk was initially rotating at $\omega_{1} = 22.0$ rads-1. What will be its final angular velocity? What is the change in the kinetic energy of the system?

Solution

Using a conservation law of the angular momentum, we obtain

$$J _ {1} \omega_ {1} = J _ {2} \omega_ {2}.$$

Here $J_{1} = \frac{Mr^{2}}{2} = \frac{0.2\times 0.2^{2}}{2} = 4\times 10^{-3}\mathrm{kg}\cdot \mathrm{m}^{2}$ is the initial moment of inertia, $J_{2} = \frac{Mr^{2}}{2} +mr^{2} = \frac{0.2\times 0.2^{2}}{2} +0.02\times 0.2^{2} = 4.8\times 10^{-3}\mathrm{kg}\cdot \mathrm{m}^{2}$ — the final moment of inertia.

So

$$\omega_ {2} = \frac {J _ {1}}{J _ {2}} \omega_ {1} = \frac {\frac {M r ^ {2}}{2}}{\frac {M r ^ {2}}{2} + m r ^ {2}} \omega_ {1} = \frac {M}{M + 2 m} \omega_ {1} = \frac {2 0 0}{2 0 0 + 2 \times 2 0} \times 2 2 = 1 8. 3 \frac {\mathrm {r a d}}{\mathrm {s}}.$$

The change in kinetic energy

$$\Delta W = \frac {J _ {2} \omega_ {2} ^ {2}}{2} - \frac {J _ {1} \omega_ {1} ^ {2}}{2} = \frac {4 . 8 \times 1 0 ^ {- 3} \times 1 8 . 3 ^ {2}}{2} - \frac {4 \times 1 0 ^ {- 3} \times 2 2 ^ {2}}{2} = - 1 6 4. 3 \mathrm {J}.$$

Answer $\omega_{2} = 18.3\frac{\mathrm{rad}}{\mathrm{s}},\Delta W = -164.3\mathrm{J}.$

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