Question
A solid sphere rolls down without slipping on a $30^o$ inclined plane. If $g = 10\,m/s^2$, the acceleration of the rolling sphere is
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Answer
$a=\frac{g \sin \theta}{\beta}=\frac{g \sin \theta}{1+\frac{I}{M R^{2}}}$
For a solid sphere $:$ $I=\frac{2}{5} M R^{2}$
$\therefore a=\frac{g \sin 30^{\circ}}{1=\frac{2}{5}}=\frac{10 \times \frac{1}{2}}{\frac{7}{5}}$
$=\frac{5}{7} \times 5=\frac{25}{7} \mathrm{m} \mathrm{s}^{-2}$
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