A circular disc is rolling on a horizontal pl | vedclass

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(A) The total kinetic energy $(KE_{total})$ of a rolling body is the sum of its translational and rotational kinetic energies: $KE_{total} = KE_{trans

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) The total kinetic energy $(KE_{total})$ of a rolling body is the sum of its translational and rotational kinetic energies: $KE_{total} = KE_{trans} + KE_{rot}$.For a circular disc,the moment of inertia is $I = \frac{1}{2} mR^2$,where $K^2 = \frac{I}{m} = \frac{1}{2} R^2$,so $\frac{K^2}{R^2} = \frac{1}{2}$.The total kinetic energy is given by $KE_{total} = \frac{1}{2} mv^2 (1 + \frac{K^2}{R^2})$.Since $KE_{trans} = \frac{1}{2} mv^2$,we have $KE_{total} = KE_{trans} (1 + \frac{1}{2}) = KE_{trans} (\frac{3}{2})$.Given $KE_{total} = 300\,J$,we find $KE_{trans} = \frac{2}{3} 	imes 300\,J = 200\,J$.

$A$ circular disc is rolling on a horizontal plane. Its total kinetic energy is $300\,J$. What is its translational kinetic energy in $J$?

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