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A circular disc of mass M and radius R is rotating about its axis with angular speed \omega _{1}. if the another stationary disc having radius \frac{R}{2} and same mass M is dropped co-axially on the rotating disc. Gradually both the discs attain constant angular speed \omega _{2}. The energy lost in the process is p% of the initial energy. Value of p is_________
Option: 1 10
Option: 2 20
Option: 3 30
Option: 4 40

Answers (1)


Let the moment of inertia of bigger disc is I=\frac{M R^{2}}{2}

\Rightarrow \text{ MOI of small disc} =I_{2}=\frac{M\left(\frac{R}{2}\right)^{2}}{2}=\frac{I}{4}
by angular momentum conservation
\mathrm{I} \omega_{1}+\frac{\mathrm{I}}{4}(0)=\mathrm{I} \omega_{2}+\frac{\mathrm{I}}{4} \omega_{2} \Rightarrow \omega_{2}=\frac{4 \omega_{1}}{5}
initial kinetic energy =\mathrm{K}_{1}=\frac{1}{2} \mathrm{I} \omega_{1}^{2}
final kinetic energy 

\mathrm{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) \\

 \mathrm{P} \%=\frac{\mathrm{K}_{1}-\mathrm{K}_{2}}{\mathrm{~K}_{1}} \times 100 \%=\frac{1-4 / 5}{1} \times 100=20 \%

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