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 #65764 – Physics – Mechanics | Relativity

Question:

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:

We need to find moments of inertia of the system with insect in center $I_i$ and on the outside edge of a disk $I_f$:

$$I_i = \frac{m_{\text{disc}} r^2}{2} = \frac{1}{2} \cdot 0.2 \cdot 0.04 = 0.004\, \mathrm{kg} \cdot \mathrm{m}^2;$$ $$I_f = \frac{m_{\text{disc}} r^2}{2} + m_{\text{insect}} r^2 = \frac{1}{2} \cdot 0.2 \cdot 0.04 + 0.02 \cdot 0.04 = 0.0048\, \mathrm{kg} \cdot \mathrm{m}^2;$$

Angular momentum remains, so, we can find final angular velocity:

$$\omega_i I_i = \omega_f I_f \Rightarrow \omega_f = \frac{\omega_i I_i}{I_f} = \frac{22 \cdot 0.004}{0.0048} = \frac{55}{3} \approx 18.3\,\mathrm{rad/s};$$

The change in kinetic energy of the system is:

$$\Delta W = \frac{I_f \omega_f^2}{2} - \frac{I_i \omega_i^2}{2} = -0.164\, \mathrm{J};$$

Answer:

$$\omega_f = 18.3\,\mathrm{rad/s}, \Delta W = -0.164\, \mathrm{J}.$$

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