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(D) When a body rolls without slipping,it possesses both translational kinetic energy and rotational kinetic energy.Translational kinetic energy is gi

Vedclass · Accepted Answer

$ \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2} $

Vedclass · Answer

(D) When a body rolls without slipping,it possesses both translational kinetic energy and rotational kinetic energy.Translational kinetic energy is given by $K_t = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity of the center of mass.Rotational kinetic energy is given by $K_r = \frac{1}{2}I\omega^2$,where $I$ is the moment of inertia about the center of mass and $\omega$ is the angular velocity.The total kinetic energy $K$ is the sum of these two energies:$K = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.Therefore,the correct option is $D$.

$A$ sphere of mass $m$ and moment of inertia $I$ about its center of mass is rolling without slipping down an inclined plane from rest. Which of the following represents its total kinetic energy?

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