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 #66017-Physics-Mechanics-Relativity

An insect of mass $20\,\mathrm{g}$ crawls from the center to the outside edge of a rotating disc of mass $200\,\mathrm{g}$ and radius $20\,\mathrm{cm}$. 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?

Solution

From the conservation of momentum:

$$I_1 \omega_1 = I_2 \omega_2$$ $$\omega_2 = \frac{I_1 \omega_1}{I_2} = \frac{\frac{1}{2} M r^2}{\frac{1}{2} M r^2 + m r^2} \omega_1 = \frac{M}{M + 2m} = \frac{200}{200 + 2(20)} 22 = 18.3\,\frac{\mathrm{rad}}{\mathrm{s}}.$$

The change in kinetic energy:

$$\Delta K = \frac{I_2 \omega_2^2}{2} - \frac{I_1 \omega_1^2}{2} = \frac{1}{2} \left( \left[ 0.02(0.2)^2 + \frac{0.2}{2}(0.2)^2 \right] (18.3)^2 - \frac{0.2}{2}(0.2)^2 (22)^2 \right) = -164\,J.$$

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