A solid ball of mass m and radius r rolls wit | vedclass

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Question

(C) The total mechanical energy is conserved as the ball rolls without slipping.Initial energy at height $h = 8R$ is purely potential: $E_i = mgh = mg

Vedclass · Accepted Answer

$\sqrt{10gR}$

Vedclass · Answer

(C) The total mechanical energy is conserved as the ball rolls without slipping.Initial energy at height $h = 8R$ is purely potential: $E_i = mgh = mg(8R) = 8mgR$.At point $P$,the height of the ball is $R$ from the ground. The total energy at $P$ is the sum of potential energy and kinetic energy (translational + rotational).$E_P = mgR + \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a solid ball,the moment of inertia $I = \frac{2}{5}mr^2$. Since it rolls without slipping,$\omega = v/r$.Substituting these,$E_P = mgR + \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgR + \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgR + \frac{7}{10}mv^2$.Equating $E_i = E_P$: $8mgR = mgR + \frac{7}{10}mv^2$.$7mgR = \frac{7}{10}mv^2$.$v^2 = 10gR$.$v = \sqrt{10gR}$.

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