Question
A solid sphere of mass $0.1kg$ and radius $2.5cm$ rolls without sliding with a uniform velocity of $0.1ms^{-1}$ along a straight line on a smooth horizontal table. Find its total energy.
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Answer
Energy of a body rolling on the table is E = Translational K.E. + Rotational K.E. $=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{l}\omega^2$ Now, $\text{l}=\frac{2}{5}\text{mr}^2,\omega=\frac{\text{v}}{\text{r}}$
$\therefore\ \text{E}=\frac{1}{2}\text{mv}^2+\frac{1}{2}\times\frac{2}{5}\text{mr}^2\times\frac{\text{v}^2}{\text{r}^2}$
$=\frac{1}{2}\text{mv}^2+\frac{1}{5}\text{mv}^2$
$=\frac{7}{10}\text{mv}^2=\frac{7}{10}\times0.01\times(0.1)^2$
$=7\times10^{-4}\text{J.}$
$\therefore\ \text{E}=\frac{1}{2}\text{mv}^2+\frac{1}{2}\times\frac{2}{5}\text{mr}^2\times\frac{\text{v}^2}{\text{r}^2}$
$=\frac{1}{2}\text{mv}^2+\frac{1}{5}\text{mv}^2$
$=\frac{7}{10}\text{mv}^2=\frac{7}{10}\times0.01\times(0.1)^2$
$=7\times10^{-4}\text{J.}$
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