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A round uniform body of radius $\mathrm{R}$, mass $\mathrm{M}$ and moment of inertia '$\mathrm{I}$', rolls down (without slipping) an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is
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Verified Answer
The correct answer is:
$\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{M R}^2}}$
$\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{M R}^2}}$
$\mathrm{M g \sin \theta-f=M a}$
$\begin{aligned}
& \mathrm{f R=I \frac{a}{R}}\\
& \Rightarrow a=\frac{g \sin \theta}{\left(1+\frac{\mathrm{I}}{\mathrm{M}\mathrm{R^2}}\right)}
\end{aligned}$
$\begin{aligned}
& \mathrm{f R=I \frac{a}{R}}\\
& \Rightarrow a=\frac{g \sin \theta}{\left(1+\frac{\mathrm{I}}{\mathrm{M}\mathrm{R^2}}\right)}
\end{aligned}$
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