Question 652: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

Acceleration of the solid sphere slipping
down the incline without rolling is
$${a_{slipping}} = g\sin 	heta $$       ....(i)
Acceleration of the solid sphere rolling down the
incline without slipping is
$${a_{rolling}} = {{g\sin 	heta } \over {1 + {{{k^2}} \over {{R^2}}}}} = {{g\sin 	heta } \over {1 + {2 \over 5}}}$$
$$\left( \because{For\,\,solid\,\,sphere,{{{k^2}} \over {{R^2}}} = {2 \over 5}} ight)$$
= $${5 \over 7}g\sin 	heta $$   ... (ii)
Divide eqn. (ii) by eqn. (i), we get
$${{{a_{rolling}}} \over {{a_{slipping}}}} = {5 \over 7}$$

Answer

5 : 7

Answer

2 : 3

Answer

2 : 5

Answer

7 : 5

Centre of Mass and Rotational Motion

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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