JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 27)
_30th_January_Morning_Shift_en_27_1.png)
Consider a Disc of mass $$5 \mathrm{~kg}$$, radius $$2 \mathrm{~m}$$, rotating with angular velocity of $$10 \mathrm{~rad} / \mathrm{s}$$ about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is ________ J.
Answer
250
Explanation
$$\begin{aligned}
& \vec{L}_i=I \omega_i=\frac{M R^2}{2} \cdot \omega=100 \mathrm{~kgm}^2 / \mathrm{s} \\
& E_i=\frac{1}{2} \cdot \frac{M R^2}{2} \cdot \omega^2=500 \mathrm{~J} \\
& \vec{L}_i=\vec{L}_f \Rightarrow 100=2 I \omega_f \\
& \omega_f=5 \mathrm{~rad} / \mathrm{sec} \\
& E_f=2 \times \frac{1}{2} \cdot \frac{5(2)^2}{2} \cdot(5)^2=250 \mathrm{~J} \\
& \Delta E=250 \mathrm{~J}
\end{aligned}$$
Comments (0)
