An object of mass 10 kg and radius 0.5 m is | vedclass

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(D) The total kinetic energy $(K.E.)$ of a body rolling without slipping is given by the sum of its translational and rotational kinetic energies:$K.E

Vedclass · Accepted Answer

$0.4$

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(D) The total kinetic energy $(K.E.)$ of a body rolling without slipping is given by the sum of its translational and rotational kinetic energies:$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since $I = mk^2$ and $\omega = v/R$,where $k$ is the radius of gyration and $R$ is the radius of the object:$K.E. = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(v/R)^2 = \frac{1}{2}mv^2 \left(1 + \frac{k^2}{R^2}ight)$Given: $m = 10 \ kg$,$R = 0.5 \ m$,$v = 2 \ m/s$,$K.E. = 32.8 \ J$.Substituting the values:$32.8 = \frac{1}{2} 	imes 10 	imes (2)^2 \left(1 + \frac{k^2}{(0.5)^2}ight)$$32.8 = 5 	imes 4 \left(1 + \frac{k^2}{0.25}ight)$$32.8 = 20 \left(1 + 4k^2ight)$$1.64 = 1 + 4k^2$$0.64 = 4k^2$$k^2 = 0.16$$k = 0.4 \ m$.

An object of mass $10 \ kg$ and radius $0.5 \ m$ is rolling without slipping with a velocity of $2 \ m/s$. Its total kinetic energy is $32.8 \ J$. The radius of gyration of the object is .......... $m$.

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