A disc of mass $5kg$ and radius $50cm$ rolls on the ground at the rate of $10ms^{-1}$. Calculate the K.E. of the disc. $\Big(\text{Given}:\text{I}=\frac{1}{2}\text{MR}^2\Big)$
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Here, mass of the disc, M = 5kg Radius of the disc R = 50cm $=\frac{1}{2}\text{m}$ Linear velocity of the disc, $v = 10ms^{-1}$ $\text{As v}=\text{R}\omega$
$\therefore10=\frac{1}{2}\omega$ or $\omega=10\times2=20\text{ radian/sec.}$ Also, moment of intertia of disc about or axis through its centre. $\text{l}=\frac{1}{2}\text{MR}^2$ K.E. of the disc $=\frac{1}{2}\text{I}\omega^2+\frac{1}{2}\text{Mv}^2$
$=\frac{1}{2}\frac{\text{MR}^2}{2}\omega^2+\frac{1}{2}\text{Mv}^2$
$=\frac{1}{4}\times5\times\Big(\frac{1}{2}\Big)^2\times(20)^2+\frac{1}{2}\times5(10)^2$
$=375\text{J}.$
$\therefore10=\frac{1}{2}\omega$ or $\omega=10\times2=20\text{ radian/sec.}$ Also, moment of intertia of disc about or axis through its centre. $\text{l}=\frac{1}{2}\text{MR}^2$ K.E. of the disc $=\frac{1}{2}\text{I}\omega^2+\frac{1}{2}\text{Mv}^2$
$=\frac{1}{2}\frac{\text{MR}^2}{2}\omega^2+\frac{1}{2}\text{Mv}^2$
$=\frac{1}{4}\times5\times\Big(\frac{1}{2}\Big)^2\times(20)^2+\frac{1}{2}\times5(10)^2$
$=375\text{J}.$
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