A solid sphere of mass $1 \mathrm{~kg}$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3} \mathrm{~J}$. The speed of the centre of mass of the sphere is __________ $\operatorname{cm~s}^{-1}$
Answer (integer)
10
Solution
$\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}=7 \times 10^{-3}$
<br/><br/>$\Rightarrow$ $\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2}\left(\frac{2}{5} \mathrm{MR}^{2}\right)\left(\frac{\mathrm{V}}{\mathrm{R}}\right)^{2}=7 \times 10^{-3}$
<br/><br/>$\Rightarrow$ $\frac{1}{2} \mathrm{MV}^{2}\left[1+\frac{2}{5}\right]=7 \times 10^{-3}$
<br/><br/>$\Rightarrow$ $\frac{1}{2}(1)\left(\mathrm{V}^{2}\right)\left(\frac{7}{5}\right)=7 \times 10^{-3}$
<br/><br/>$\Rightarrow$ $\mathrm{V}^{2}=10^{-2}$
<br/><br/>$\Rightarrow$ $\mathrm{V}=10^{-1}=0.1 \mathrm{~m} / \mathrm{s}=10 \mathrm{~cm} / \mathrm{s}$
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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