A disc of mass 3 , kg rolls down an inclined | vedclass

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(B) Using the principle of conservation of mechanical energy,the potential energy at the top is converted into translational and rotational kinetic en

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$100$

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(B) Using the principle of conservation of mechanical energy,the potential energy at the top is converted into translational and rotational kinetic energy at the bottom.$mgh = K_{trans} + K_{rot}$For a disc,the moment of inertia $I = \frac{1}{2} mR^2$ and the rolling condition is $v = R\omega$.$K_{rot} = \frac{1}{2} I \omega^2 = \frac{1}{2} (\frac{1}{2} mR^2) (\frac{v}{R})^2 = \frac{1}{4} mv^2$.Thus,$mgh = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2$.Given $m = 3 \, kg$,$h = 5 \, m$,and $g = 10 \, m/s^2$:$3 	imes 10 	imes 5 = \frac{3}{4} mv^2$$150 = \frac{3}{4} mv^2$$mv^2 = 200$Translational kinetic energy $K_{trans} = \frac{1}{2} mv^2 = \frac{1}{2} (200) = 100 \, J$.

$A$ disc of mass $3 \, kg$ rolls down an inclined plane of height $5 \, m$. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is .......... $J$. (Take $g = 10 \, m/s^2$)

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