A solid spherical ball of mass 1 , kg and rad | vedclass

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(D) The rotational kinetic energy $K_R$ is given by the formula $K_R = \frac{1}{2} I \omega^2$. For a solid sphere,the moment of inertia $I$ about its

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$9/20 \, J$

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(D) The rotational kinetic energy $K_R$ is given by the formula $K_R = \frac{1}{2} I \omega^2$. For a solid sphere,the moment of inertia $I$ about its center is $I = \frac{2}{5} M R^2$. Given: $M = 1 \, kg$,$R = 3 \, cm = 0.03 \, m$,and $\omega = 50 \, rad/s$. Substituting the values: $I = \frac{2}{5} 	imes 1 	imes (0.03)^2 = \frac{2}{5} 	imes 0.0009 = 0.00036 \, kg \cdot m^2$. Now,$K_R = \frac{1}{2} 	imes 0.00036 	imes (50)^2$. $K_R = \frac{1}{2} 	imes 0.00036 	imes 2500$. $K_R = 0.00018 	imes 2500 = 0.45 \, J$. Since $0.45 = 45/100 = 9/20$,the rotational kinetic energy is $9/20 \, J$.

$A$ solid spherical ball of mass $1 \, kg$ and radius $3 \, cm$ is rotating with an angular velocity of $50 \, rad/s$ about an axis passing through its center. The rotational kinetic energy is:

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