A circular disc of mass M and radius R is rotating about its axis… | JEE Main 2020 (4 Sep, Shift 1)

Q55JEE Main 2020 · 4 SEP · SHIFT 1Medium

Angular Momentum & its Conservation

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 __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 20

$I_1 = \frac{1}{2}MR^2$ , $I_2 = \frac{1}{2}M\left(\frac{R}{2}\right)^2 = \frac{1}{8}MR^2$ .
With $L$ conserved, $\frac{K_f}{K_i} = \frac{I_1}{I_1+I_2}$ , so fraction lost $= \frac{I_2}{I_1+I_2}$ .
$\frac{I_2}{I_1+I_2} = \frac{1/8}{5/8} = \frac{1}{5} = 20\%$
Answer: $p = 20$

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