Question:
A circular disc of mass M and radius R is rotating about its axis with angular speed \(\omega_1\). If another stationary disc having radius \(\frac{R}{2}\) and same mass M is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed \(\omega_2\). The energy lost in the process is p% of the initial energy. Value of p is __________.
A circular disc of mass M and radius R is rotating about its axis with angular speed \(\omega_1\). If another stationary disc having radius \(\frac{R}{2}\) and same mass M is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed \(\omega_2\). The energy lost in the process is p% of the initial energy. Value of p is __________.
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Angular momentum is conserved; compute I1 = MR^2/2 and I2 = M(R/2)^2/2, then the fractional energy loss is I2/(I1+I2).
Updated On: Nov 5, 2026
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Correct Answer: 20
Solution and Explanation
\(I_1 = \frac{1}{2}MR^2\), \(I_2 = \frac{1}{2}M\left(\frac{R}{2}\right)^2 = \frac{MR^2}{8}\).
Conservation of angular momentum: \(I_1\omega_1 = (I_1+I_2)\omega_2\) so \(\omega_2 = \frac{\frac{1}{2}}{\frac{5}{8}}\omega_1 = \frac{4}{5}\omega_1\).
Initial energy: \(E_i = \frac{1}{2}I_1\omega_1^2 = \frac{1}{4}MR^2\omega_1^2\).
Final energy: \(E_f = \frac{1}{2}(I_1+I_2)\omega_2^2 = \frac{1}{2}\cdot\frac{5MR^2}{8}\cdot\frac{16}{25}\omega_1^2 = \frac{1}{5}MR^2\omega_1^2\).
Loss fraction: \(\frac{E_i - E_f}{E_i} = 1 - \frac{1/5}{1/4} = 1 - \frac{4}{5} = \frac{1}{5}\), i.e. \(p = 20\).
Answer: 20
Conservation of angular momentum: \(I_1\omega_1 = (I_1+I_2)\omega_2\) so \(\omega_2 = \frac{\frac{1}{2}}{\frac{5}{8}}\omega_1 = \frac{4}{5}\omega_1\).
Initial energy: \(E_i = \frac{1}{2}I_1\omega_1^2 = \frac{1}{4}MR^2\omega_1^2\).
Final energy: \(E_f = \frac{1}{2}(I_1+I_2)\omega_2^2 = \frac{1}{2}\cdot\frac{5MR^2}{8}\cdot\frac{16}{25}\omega_1^2 = \frac{1}{5}MR^2\omega_1^2\).
Loss fraction: \(\frac{E_i - E_f}{E_i} = 1 - \frac{1/5}{1/4} = 1 - \frac{4}{5} = \frac{1}{5}\), i.e. \(p = 20\).
Answer: 20
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