Question

An insect of mass 20g crawls from the centre to the outside edge of a rotating disc of mass 200g and radius 20cm. The disc was initially rotating at 22.0 rad/s. What is the change in the kinetic energy of the system?

Answer 1

Answer Question #66347, Physics / Mechanics | Relativity

An insect of mass 20g crawls from the centre to the outside edge of a rotating disc of mass 200g and radius 20cm. The disc was initially rotating at 22.0 rad/s. What is the change in the kinetic energy of the system?

Solution:

Using Law of conservation of angular momentum

$$J_1 \omega_1 = J_2 \omega_2$$

We find is $J_1$

$$J_1 = \frac{M r^2}{2}$$ $$J_1 = 0.2 \times (0.2)^2 / 2 = 0.2 \times 0.2 \times 0.2 / 2 = 4 \times 10^{-3} \text{ kgm}^2$$

We find is $J_2$

$$J_2 = \frac{M r^2}{2} + m r^2$$ $$J_2 = 4 \times 10^{-3} + 0.02 \times (0.2)^2 = 4.8 \times 10^{-3} \text{ kgm}^2$$

Now, we find is $\omega_2$

$$\omega_2 = \left(J_1 / J_2\right) \omega_1$$ $$\omega_2 = \left(4 \times 10^{-3} / 4.8 \times 10^{-3}\right) \times 22 = 18.3 \text{ rad/s}$$

Change in K.E

$$\Delta K = \frac{J_2 \omega_2^2}{2} - \frac{J_1 \omega_1^2}{2}$$ $$\Delta K = \left[\left(4.8 \times 10^{-3} \times 18.3 \times 18.3\right) / 2\right] - \left[\left(4 \times 10^{-3} \times 22 \times 22\right) / 2\right] = -164.3 \text{ J}$$

Answer: -164.3 J

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