A circular disc of mass M and radius R is rotating about its axis with angular speed $ω_{1}$ . 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 $ω_{2}$ . The energy lost in the process is $ρ %$ of the initial energy. Value of $ρ$ is _________

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answer is 20.

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Detailed Solution

$K = \frac{1}{2} {Iω}^{2} = \frac{1}{2} \frac{{(Iω)}^{2}}{I} = \frac{L^{2}}{2 I}$

$I_{i} = I_{1} = \frac{{MR}^{2}}{2}$ ; $I_{f} = I_{1} + I_{2} = \frac{{MR}^{2}}{2} + \frac{{MR}^{2}}{2 \times 4} = \frac{5 {MR}^{2}}{8}$

$\Rightarrow % change in energy = \frac{k_{f} - k_{i}}{k_{i}} \times 100$

$= \frac{\frac{L^{2}}{2 (\frac{{MR}^{2}}{2})} - \frac{L^{2}}{2 (\frac{5 {MR}^{2}}{8})}}{\frac{L^{2}}{2 (\frac{{MR}^{2}}{2})}} \times 100 = 20 %$

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