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6.System of Particles and Rotational Motion
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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 $..........ms ^{-1}$.
A
$40$
B
$38$
C
$37$
D
$36$
(JEE MAIN-2023)
Solution
$KE =\frac{1}{2} mv ^2+\frac{1}{2} I \omega^2$
$2240=\frac{1}{2} 2( v )^2+\frac{1}{2} \frac{2}{5}(2) R ^2 \cdot\left(\frac{ v }{ R }\right)^2$
$2240= v ^2+\frac{2}{5} v ^2$
$\Rightarrow v =40\,m / s$
Standard 11
Physics
