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(C) The total kinetic energy $K$ of a rolling ball is the sum of its translational and rotational kinetic energies:$K = \frac{1}{2}mv^2 + \frac{1}{2}I

Vedclass · Accepted Answer

$4$

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(C) The total kinetic energy $K$ of a rolling ball is the sum of its translational and rotational kinetic energies:$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a solid sphere (ball),the moment of inertia $I = \frac{2}{5}mr^2$. Since $v = r\omega$,we have $\omega = \frac{v}{r}$.$K = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$By the law of conservation of energy,the potential energy at the top equals the kinetic energy at the bottom:$mgh = \frac{7}{10}mv^2$Given $h = l \sin 	heta$,where $l = 2 \ m$ and $	heta = 35^{\circ}$:$v = \sqrt{\frac{10}{7}gh} = \sqrt{\frac{10}{7} \cdot g \cdot l \sin 35^{\circ}}$Using $g = 9.8 \ m/s^2$:$v = \sqrt{\frac{10}{7} \cdot 9.8 \cdot 2 \cdot 0.57} = \sqrt{14 \cdot 2 \cdot 0.57} = \sqrt{15.96} \approx 4 \ m/s$.

$A$ ball of radius $11 \ cm$ and mass $8 \ kg$ rolls from rest down a ramp of length $2 \ m$. The ramp is inclined at $35^{\circ}$ to the horizontal. When the ball reaches the bottom,its velocity is .......... $m/s$. $(\sin 35^{\circ} = 0.57)$

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