Question 462: Centre of Mass and Rotational Motion

Question

A solid disc of radius 'a' and mass 'm' rolls down without slipping on an inclined plane making an angle $$	heta$$ with the horizontal. The acceleration of the disc will be $${2 \over b}$$g sin$$	heta$$ where b is ____________. (Round off to the Nearest Integer) (g = acceleration due to gravity, $$	heta$$ = angle as shown in figure)

Accepted Answer

We know that, on an inclined plane
Acceleration, $$a = {{g\sin 	heta } \over {1 + {I \over {m{R^2}}}}}$$
$$ \Rightarrow a = {{g\sin 	heta } \over {1 + {1 \over 2}}}$$ [$$\because$$ For disc, $$I = {{m{R^2}} \over 2}$$]
$$ \Rightarrow a = {2 \over 3}g\sin 	heta $$ ....... (i)
As per question, acceleration of the disc will be $${2 \over b}g\sin 	heta $$.
Comparing it with Eq. (i), we get
$$b = 3$$

Centre of Mass and Rotational Motion

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