JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2020
- Hard
- Numerical
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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Answer & Solution
Correct Answer
(C) \(20\)
Step-by-step Solution
Detailed explanation
Let moment of inertia of bigger disc is \(I =\frac{ MR ^{2}}{2}\)
\(\Rightarrow\) \(MOI\) of small disc \(I_{2}=\frac{M\left(\frac{R}{2}\right)^{2}}{2}=\frac{I}{4}\)
by angular momentum conservation
\(I \omega_{1}+\frac{ I }{4}( D )= I \omega_{2}+\frac{ I }{4} \omega_{2} \Rightarrow \omega_{2}=\frac{4 \omega_{1}}{5}\)
initial kinetic energy \(K _{1}=\frac{1}{2} I \omega_{1}^{2}\)
final kinetic energy \(K _{2}\)
\(=\frac{1}{2}\left( I +\frac{ I }{4}\right)\left(\frac{4 \omega_{1}}{5}\right)^{2}=\frac{1}{2} I \omega_{1}^{2}\left(\frac{4}{5}\right)\)
\(P \%=\frac{ K _{1}- K _{2}}{ K _{1}} \times 100 \%=\frac{1-4 / 5}{1} \times 100=20 \%\)
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