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The ratio of the accelerations for a solid sphere (mass 'm' and radius 'R') rolling down an incline of angle ' $\theta$ ' without slipping and slipping down the incline without rolling is:
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1908 Upvotes
Verified Answer
The correct answer is:
5: 7
For solid sphere rolling without slipping on inclined plane, acceleration
$$
a_{1}=\frac{g \sin \theta}{1+\frac{K^{2}}{R^{2}}}
$$
For solid sphere slipping on inclined plane without rolling, acceleration
$$
a_{2}=g \sin \theta
$$
Therefore required ratio $=\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}$
$$
=\frac{1}{1+\frac{K^{2}}{R^{2}}}=\frac{1}{1+\frac{2}{5}}=\frac{5}{7}
$$
$$
a_{1}=\frac{g \sin \theta}{1+\frac{K^{2}}{R^{2}}}
$$
For solid sphere slipping on inclined plane without rolling, acceleration
$$
a_{2}=g \sin \theta
$$
Therefore required ratio $=\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}$
$$
=\frac{1}{1+\frac{K^{2}}{R^{2}}}=\frac{1}{1+\frac{2}{5}}=\frac{5}{7}
$$
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