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A solid sphere is rolling down a frictionless surface with translational velocity $\mathrm{V}$. It climbs the inclined plane from ‘A’ to ‘ $\mathrm{B}^{\prime}$ and then moves away from ‘ $\mathrm{B}^{\prime}$ on the smooth horizontal surface. The value of $V$ should beA. $\sqrt{g h}$B. $\geq\left[\frac{10 g h}{7}\right]^{1 / 2}$C. $\sqrt{2 g h}$D. $10 g h$
A solid sphere is rolling down a frictionless surface with translational velocity $\mathrm{V}$. It climbs the inclined plane from ‘A’ to ‘ $\mathrm{B}^{\prime}$ and then moves away from ‘ $\mathrm{B}^{\prime}$ on the smooth horizontal surface. The value of $V$ should be
A. $\sqrt{g h}$
B. $\geq\left[\frac{10 g h}{7}\right]^{1 / 2}$
C. $\sqrt{2 g h}$
D. $10 g h$
Physics
A solid sphere is rolling down a frictionless surface with translational velocity $\mathrm{V}$. It climbs the inclined plane from ‘A’ to ‘ $\mathrm{B}^{\prime}$ and then moves away from ‘ $\mathrm{B}^{\prime}$ on the smooth horizontal surface. The value of $V$ should be
A. $\sqrt{g h}$
B. $\geq\left[\frac{10 g h}{7}\right]^{1 / 2}$
C. $\sqrt{2 g h}$
D. $10 g h$

Correct answer is B.

Applying law of conservation of energy for rotating body,
$\begin{array}{l}
\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=\mathrm{mgh} \\
\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \frac{2}{5} \mathrm{mr}^{2} \times \frac{\mathrm{v}^{2}}{\mathrm{r}^{2}}=\mathrm{mgh} \\
\frac{\mathrm{v}^{2}}{2}+\frac{2 \mathrm{v}^{2}}{10}=\mathrm{gh} \\
\frac{5 \mathrm{v}^{2}+2 \mathrm{v}^{2}}{10}=\mathrm{gh} \Rightarrow \mathrm{v}^{2}=\frac{10}{7} \mathrm{gh} \\
\mathrm{v} \geq \sqrt{\frac{10}{7} \mathrm{gh}}
\end{array}$

i

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