Rolling motion

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Question

A circular object is gently released on a rough inclined plane as shown in figure and the object starts rolling down the plane without slipping with acceleration of its centre $\frac{10}{3} m / s^{2}$ . Then the object must be a

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Solution

$\begin{array}{l} \frac{g s i n θ}{1 + (\frac{I}{m r^{2}})} = a \Rightarrow \frac{10 \times s i n 30^{0}}{1 + \frac{I}{m r^{2}}} = \frac{10}{3} \\ \Rightarrow I = \frac{1}{2} m r^{2} \\ I t s h o u l d b e d i s c o r s o l i d c y l i n d e r \end{array}$

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