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Question

(A) For the solid sphere to climb the inclined surface to a height $h$,its total initial kinetic energy must be greater than or equal to the potential

Vedclass · Accepted Answer

$v \ge \sqrt {\frac{10}{7}gh}$

Vedclass · Answer

(A) For the solid sphere to climb the inclined surface to a height $h$,its total initial kinetic energy must be greater than or equal to the potential energy gained at height $h$.The total kinetic energy $(K.E.)_{total}$ is the sum of translational kinetic energy $(K.E.)_{T}$ and rotational kinetic energy $(K.E.)_{R}$:$(K.E.)_{total} = (K.E.)_{T} + (K.E.)_{R} \ge P.E.$$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \ge mgh$For a solid sphere,the moment of inertia $I = \frac{2}{5}mr^2$ and for pure rolling,$\omega = \frac{v}{r}$.Substituting these values:$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 \ge mgh$$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 \ge mgh$$\frac{7}{10}mv^2 \ge mgh$$v^2 \ge \frac{10gh}{7}$$v \ge \sqrt{\frac{10gh}{7}}$

$A$ solid sphere is rolling on a frictionless surface with a translational velocity $v \ m/s$,as shown in the figure. If it is to climb the inclined surface to a height $h$,then $v$ should be:

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