Question

Consider a disc of mass 5 kg, radius 2 m, rotating with angular velocity of 10 rad/s about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is \\_\\_\\_\\_ J.

Answer 1

Step 1. Calculate Initial Moment of Inertia and Kinetic Energy:

For the first disc:

$I_1 = \frac{1}{2}MR^2 = \frac{1}{2} \times 5 \times (2)^2 = 10 \, \text{kg} \cdot \text{m}^2$

Initial angular velocity $\omega = 10 \, \text{rad/s}$. Initial kinetic energy:

$E_i = \frac{1}{2}I_1 \omega^2 = \frac{1}{2} \times 10 \times (10)^2 = 500 \, \text{J}$

Step 2. Final Moment of Inertia and Angular Velocity:

When the second disc is placed on top, the combined moment of inertia becomes:

$I_f = I_1 + I_2 = 10 + 10 = 20 \, \text{kg} \cdot \text{m}^2$

Using conservation of angular momentum:

$I_1 \omega = I_f \omega_f \Rightarrow 10 \times 10 = 20 \times \omega_f$

$\omega_f = 5 \, \text{rad/s}$

Step 3. Calculate Final Kinetic Energy:

$E_f = \frac{1}{2} I_f \omega_f^2 = \frac{1}{2} \times 20 \times (5)^2 = 250 \, \text{J}$

Step 4. Energy Dissipated:

Energy dissipated $\Delta E = E_i - E_f$:

$\Delta E = 500 - 250 = 250 \, \text{J}$