A rigid body of mass m rotates with angular v | vedclass

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(C) The rotational kinetic energy of a body rotating about an axis is given by $E = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia about t

Vedclass · Accepted Answer

$\frac{1}{2} m (d^2 + K^2) \omega^2$

Vedclass · Answer

(C) The rotational kinetic energy of a body rotating about an axis is given by $E = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia about the axis of rotation.According to the parallel axis theorem,the moment of inertia $I$ about an axis at a distance $d$ from the center of mass is $I = I_{cm} + md^2$.Given that the radius of gyration about the axis passing through the center of mass is $K$,we have $I_{cm} = mK^2$.Substituting this into the parallel axis theorem,we get $I = mK^2 + md^2 = m(K^2 + d^2)$.Now,substituting $I$ into the kinetic energy formula: $E = \frac{1}{2} [m(K^2 + d^2)] \omega^2 = \frac{1}{2} m(d^2 + K^2) \omega^2$.

$A$ rigid body of mass $m$ rotates with angular velocity $\omega$ about an axis at a distance $d$ from its center of mass. The radius of gyration about an axis passing through the center of mass $G$ and parallel to the given axis is $K$. What is the rotational kinetic energy of the body?

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