Question
One solid sphere and disc of same radius are falling along an inclined plane without slipping. One reaches earlier than the other due to
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Answer
(a)
$(i)$ For solid sphere, the moment of inertia about the diameter is $I _{ s }=\frac{2}{5} MR ^2$
Now $I = MK ^2$ for any body, where $K$ is radius of gyration of that body.
so $MK ^2=\frac{2}{5} MR ^2 \Rightarrow K = R \sqrt{2 / 5}$
$(ii)$ The moment of inertia of disc about an axis passing through its centre and perpendicular to plane is
$I _{ d }=\frac{ MR ^2}{2}= MK ^2 \Rightarrow K = R \sqrt{1 / 2}$
Now acceleration of any body which is rolling on an inclined plane is
$a=\frac{g \sin \theta}{1+K^2 / R^2}$
For same R, the acceleration of the body depends only on radius of gyration $K$, [see eq$(iii)$] so solid sphere will reach earlier to bottom of an inclined plane than disc.
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