Rotational Motion

JEE Main level, numerical answer, medium

253 of 280

Question 253 of 280

A uniform disc of mass 0.5kg0.5kg and radius rr is projected with velocity 18ms−118m{s}^{-1} at t=0st=0s on a rough horizontal surface. It starts off with a purely sliding motion at t=0st=0s. After 2s2s it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after 2s2s will be ______ JJ.
(given, coefficient of friction is 0.30.3 and g=10ms−2g=10m{s}^{-2}).

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Answer: 54

Acceleration of the disc, a=−μkg=−3ms−2a=-{\mu }_{k}g=-3m{s}^{-2}.
Now, velocity of the centre of mass
v=u+at=18−3×2v=u+at =18-3\times 2
⇒v=12ms−1\Rightarrow v=12m{s}^{-1}
Now total kinetic energy during pure rolling(v=ωrv=\omega r) will become,
KE=12mv2+12Iω2=12mv2+12mr22v2r2KE=\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}I{\omega }^{2} =\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}\dfrac{m{r}^{2}}{2}\dfrac{{v}^{2}}{{r}^{2}}
⇒KE=34mv2\Rightarrow KE=\dfrac{3}{4}m{v}^{2}
=3×18=54J=3\times 18=54J

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