Rotational Motion

JEE Main level, numerical answer, medium

252 of 280

Question 252 of 280

A solid sphere of mass 1kg1kg rolls without slipping on a plane surface. Its kinetic energy is 7×10−3J7\times {10}^{-3}J. The speed of the centre of mass of the sphere is ______ cms−1cm{s}^{-1}.

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Answer: 10

Let velocity of centre of mass of sphere be vv.
Total kinetic energy of solid sphere is KE=12mv2+12Iω2KE=\dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}I{\omega }^{2}, where, II is moment of inertia of the solid sphere about the axis passing through it's centre of mass and ω=vR(pure rolling condition)\omega =\dfrac{v}{R}\left(\text{pure rolling condition}\right).
Therefore,
7×10−3=12×1×(v)2+12[25(1)R2](vR)27\times {10}^{-3}=\dfrac{1}{2}\times 1\times (v{)}^{2}+\dfrac{1}{2}\left[\dfrac{2}{5}(1){R}^{2}\right]{\left(\dfrac{v}{R}\right)}^{2}
⇒12v2(75)=7×10−3\Rightarrow \dfrac{1}{2}{v}^{2}\left(\dfrac{7}{5}\right)=7\times {10}^{-3}
⇒v2=10−2\Rightarrow {v}^{2}={10}^{-2}
⇒v=10−1=0.1ms−1=10cms−1\Rightarrow v={10}^{-1}=0.1m{s}^{-1}=10cm{s}^{-1}

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