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(A) The total kinetic energy $(K.E.)$ of a body rolling without slipping is given by $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Since the objects

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(A) The total kinetic energy $(K.E.)$ of a body rolling without slipping is given by $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Since the objects roll without slipping,$v = R\omega$,so $\omega^2 = \frac{v^2}{R^2}$.For a ring,the moment of inertia $I_{ring} = mR^2$. Thus,$K.E._{ring} = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.Given $K.E._{ring} = 4 \ J$,we have $mv^2 = 4 \ J$.For a disc,the moment of inertia $I_{disc} = \frac{1}{2}mR^2$. Thus,$K.E._{disc} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.Substituting $mv^2 = 4 \ J$,we get $K.E._{disc} = \frac{3}{4} 	imes 4 \ J = 3 \ J$.

$A$ ring and a disc roll on a horizontal surface without slipping with the same linear velocity. If both have the same mass and the total kinetic energy of the ring is $4 \ J$,then the total kinetic energy of the disc is: (in $J$)

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