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Question

(A) The total kinetic energy $(KE_{total})$ of a rolling body is given by the sum of translational and rotational kinetic energy: $KE_{total} = KE_{tr

Vedclass · Accepted Answer

$200$

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(A) The total kinetic energy $(KE_{total})$ of a rolling body is given by the sum of translational and rotational kinetic energy: $KE_{total} = KE_{trans} + KE_{rot} = \frac{1}{2}mv^2 (1 + \frac{K^2}{R^2})$.For a circular disc,the radius of gyration $K$ is given by $K^2 = \frac{1}{2}R^2$,so $\frac{K^2}{R^2} = \frac{1}{2}$.Substituting this into the formula: $KE_{total} = \frac{1}{2}mv^2 (1 + \frac{1}{2}) = \frac{1}{2}mv^2 (\frac{3}{2}) = \frac{3}{4}mv^2$.We know that translational kinetic energy is $KE_{trans} = \frac{1}{2}mv^2$.Therefore,$KE_{trans} = \frac{2}{3} KE_{total}$.Given $KE_{total} = 300 \, J$,we get $KE_{trans} = \frac{2}{3} 	imes 300 \, J = 200 \, J$.

$A$ circular disc is rolling on a horizontal plane. Its total kinetic energy is $300 \, J$. The translational kinetic energy of the disc is:

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