A solid sphere at rest rolls down an inclined | vedclass

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(B) For a body rolling down an inclined plane without slipping,mechanical energy is conserved.Initial potential energy at the top is $PE = mgh$.Initia

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$\left(\frac{10gh}{7}\right)^{1/2}$

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(B) For a body rolling down an inclined plane without slipping,mechanical energy is conserved.Initial potential energy at the top is $PE = mgh$.Initial kinetic energy at the top is $KE = 0$ (since it starts from rest).At the bottom,the total kinetic energy is the sum of translational and rotational kinetic energy: $KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$.Since it rolls without slipping,$v = r\omega$,so $\omega = v/r$.Substituting these into the kinetic energy equation: $KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$.By conservation of energy,$mgh = \frac{7}{10}mv^2$.Solving for $v$,we get $v^2 = \frac{10gh}{7}$,which implies $v = \left(\frac{10gh}{7}\right)^{1/2}$.

$A$ solid sphere at rest rolls down an inclined plane of vertical height $h$ without sliding. Its speed on reaching the bottom of the plane is ($g=$ acceleration due to gravity).

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