A metal sphere of radius r and specific heat | vedclass

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(A) The rotational kinetic energy of the sphere is $K = \frac{1}{2} I \omega^2$.For a solid sphere,the moment of inertia is $I = \frac{2}{5} m r^2$.Th

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$\frac{2\pi^2 f^2 r^2}{5S}$

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(A) The rotational kinetic energy of the sphere is $K = \frac{1}{2} I \omega^2$.For a solid sphere,the moment of inertia is $I = \frac{2}{5} m r^2$.The angular velocity is $\omega = 2\pi f$.Thus,$K = \frac{1}{2} \left( \frac{2}{5} m r^2 ight) (2\pi f)^2 = \frac{1}{5} m r^2 (4\pi^2 f^2) = \frac{4}{5} m r^2 \pi^2 f^2$.Given that $50\%$ of this energy is converted into heat,the heat energy $H$ is $H = 0.5 	imes K = \frac{1}{2} 	imes \frac{4}{5} m r^2 \pi^2 f^2 = \frac{2}{5} m r^2 \pi^2 f^2$.This heat energy causes a temperature rise $\Delta T$,where $H = m S \Delta T$.Equating the two expressions: $m S \Delta T = \frac{2}{5} m r^2 \pi^2 f^2$.Solving for $\Delta T$: $\Delta T = \frac{2\pi^2 f^2 r^2}{5S}$.

$A$ metal sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre at a speed of $f$ rotations per second. It is suddenly stopped,and $50\%$ of its rotational kinetic energy is used in increasing its temperature. The rise in temperature of the sphere is

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