Question 358: Centre of Mass and Rotational Motion

Question

A mass $$m$$ hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass $$m$$ and radius $$R.$$ Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass $$m,$$ if the string does not slip on the pulley, is:

Accepted Answer

This is the free body diagram of pulley and mass
For translation motion of the block,
$$mg - T = ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 ight)$$
For rotational motion of the pulley,
$$T	imes R = I\alpha  = I{a \over R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 ight)$$
where $$\alpha  = $$ angular acceleration of disc = $$ {a \over R}$$
and $$I = {1 \over 2}m{R^2}$$ (For circular disc)
Solving $$(1)$$ & $$(2),$$
$$a = {{mg} \over {\left( {m + {I \over {{R^2}}}} ight)}} = {{mg} \over {m + {{m{R^2}} \over {2{R^2}}}}}$$
$$ = {{2mg} \over {3m}} = {{2g} \over 3}$$

Answer

$$g$$

Answer

$${2 \over 3}g$$

Answer

$${g \over 3}$$

Answer

$${3 \over 2}g$$

Centre of Mass and Rotational Motion

A mass $m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R. Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m,$ if the string does not slip on the pulley, is:

Choose the correct answer: