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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2023
  • Medium
  • Numerical

A solid sphere of mass \(2\,kg\) is making pure rolling on a horizontal surface with kinetic energy \(2240\,J\). The velocity of centre of mass of the sphere will be \(..........ms ^{-1}\).

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Answer & Solution

Correct Answer

(A) \(40\)

Step-by-step Solution

Detailed explanation

\(KE =\frac{1}{2} mv ^2+\frac{1}{2} I \omega^2\)

\(2240=\frac{1}{2} 2( v )^2+\frac{1}{2} \frac{2}{5}(2) R ^2 \cdot\left(\frac{ v }{ R }\right)^2\)

\(2240= v ^2+\frac{2}{5} v ^2\)

\(\Rightarrow v =40\,m / s\)

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