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Question

(B) When a solid sphere rolls down an inclined plane without slipping,the mechanical energy is conserved. The potential energy at the top is converted

Vedclass · Accepted Answer

$v_1 = \sqrt{\frac{5}{7}} v_2$

Vedclass · Answer

(B) When a solid sphere rolls down an inclined plane without slipping,the mechanical energy is conserved. The potential energy at the top is converted into both translational and rotational kinetic energy at the bottom: $mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}I\omega^2$. For a solid sphere,$I = \frac{2}{5}mr^2$ and $\omega = \frac{v_1}{r}$. Substituting these,$mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v_1^2}{r^2}) = \frac{1}{2}mv_1^2 + \frac{1}{5}mv_1^2 = \frac{7}{10}mv_1^2$. Thus,$v_1 = \sqrt{\frac{10}{7}gh}$. When the sphere slides down without friction,there is no rotation,so all potential energy converts to translational kinetic energy: $mgh = \frac{1}{2}mv_2^2$. Thus,$v_2 = \sqrt{2gh}$. Comparing the two,$\frac{v_1}{v_2} = \frac{\sqrt{10/7}}{\sqrt{2}} = \sqrt{\frac{5}{7}}$. Therefore,$v_1 = \sqrt{\frac{5}{7}} v_2$.

$A$ solid sphere rolls down an inclined plane and its velocity at the bottom is $v_1$. Then the same sphere slides down the plane (without friction) and let its velocity at the bottom be $v_2$. Which of the following relations is correct?

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