Question 702: Centre of Mass and Rotational Motion

Question

The ratio of the acceleration for a solid sphere (mass $$m$$ and radius $$R$$) rolling down an incline of angle '$$	heta$$' without slipping down the incline without rolling is.

Accepted Answer

For solid sphere rolling without slipping on inclined plane, acceleration is given as,
$$a_1=\frac{g \sin 	heta}{1+k^2 / R^2}$$
For solid sphere slipping on inclined plane without rolling, acceleration is as follows.
$$a_2=g \sin 	heta$$
Therefore, required ratio $$=a_1 / a_2$$
$$=\frac{1}{1+\frac{k^2}{R^2}}=\frac{1}{1+\frac{2}{5}}=\frac{5}{7}$$

Answer

5 : 7

Answer

2 : 3

Answer

2 : 5

Answer

7 : 5

Centre of Mass and Rotational Motion

The ratio of the acceleration for a solid sphere (mass $m and radius R) rolling down an incline of angle '\theta$' without slipping down the incline without rolling is.

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