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6.System of Particles and Rotational Motion
medium
A uniform sphere of mass $500\; g$ rolls without slipping on a plane horizontal surface with its centre moving at a speed of $5.00\; \mathrm{cm} / \mathrm{s}$. Its kinetic energy is
A
$8.75 \times 10^{-4} \;\mathrm{J}$
B
$8.75 \times 10^{-3} \;\mathrm{J}$
C
$6.25 \times 10^{-4} \;\mathrm{J}$
D
$1.13 \times 10^{-} \;\mathrm{J}$
(JEE MAIN-2020)
Solution
$\mathrm{m}=0.5 \mathrm{kg}, \mathrm{v}=5 \mathrm{cm} / \mathrm{s}$
$\mathrm{KE}$ in rolling $=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$
$=\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)$
$=8.75 \times 10^{-4} \mathrm{J}$
Standard 11
Physics