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 #65590-Physics-Mechanics-Relativity

An insect of mass 20 g crawls from the center 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?

Solution

From the conservation of momentum:

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

The change in kinetic energy:

$$\Delta K = \frac {I _ {2} \omega_ {2} ^ {2}}{2} - \frac {I _ {1} \omega_ {1} ^ {2}}{2} = \frac {1}{2} \left(\left[ 0. 0 2 (0. 2) ^ {2} + \frac {0 . 2}{2} (0. 2) ^ {2} \right] (1 8. 3) ^ {2} - \frac {0 . 2}{2} (0. 2) ^ {2} (2 2) ^ {2}\right) = - 1 6 4 J.$$

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