A disc of radius 1,m and mass 4,kg rolls on a | vedclass

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(A) Given:Radius $R = 1\,m$Mass $M = 4\,kg$Linear velocity $v = 10\,cm/s = 0.1\,m/s$For a disc rolling without slipping,the moment of inertia about th

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$0.01$

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(A) Given:Radius $R = 1\,m$Mass $M = 4\,kg$Linear velocity $v = 10\,cm/s = 0.1\,m/s$For a disc rolling without slipping,the moment of inertia about the central axis is $I = \frac{1}{2}MR^2$.The angular velocity is $\omega = \frac{v}{R}$.Rotational kinetic energy $K_{rot} = \frac{1}{2}I\omega^2$.Substituting $I = \frac{1}{2}MR^2$ and $\omega = \frac{v}{R}$:$K_{rot} = \frac{1}{2} \left( \frac{1}{2}MR^2 ight) \left( \frac{v}{R} ight)^2 = \frac{1}{4}Mv^2$.$K_{rot} = \frac{1}{4} 	imes 4\,kg 	imes (0.1\,m/s)^2 = 1 	imes 0.01 = 0.01\,J$.

$A$ disc of radius $1\,m$ and mass $4\,kg$ rolls on a horizontal plane without slipping in such a way that its centre of mass moves with a speed of $10\,cm/s$. Its rotational kinetic energy is (in $,J$)

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