Rotational Motion

JEE Main level, numerical answer, medium

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Question 273 of 280

A solid sphere of mass 2kg2kg is making pure rolling on a horizontal surface with kinetic energy 2240J2240J. The velocity of centre of mass of the sphere will be ______ ms−1m{s}^{-1}.

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Answer: 40

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,
2240=122(v)2+12[25(2)R2](vR)22240=\dfrac{1}{2}2(v{)}^{2}+\dfrac{1}{2}\left[\dfrac{2}{5}(2){R}^{2}\right]{\left(\dfrac{v}{R}\right)}^{2}
⇒2240=v2+25v2\Rightarrow 2240={v}^{2}+\dfrac{2}{5}{v}^{2}
⇒v=40ms−1\Rightarrow v=40m{s}^{-1}.

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