Translational kinetic energy:

$\overline{){\mathbf{K}}{{\mathbf{E}}}_{{\mathbf{T}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{\mathbf{v}}}^{{\mathbf{2}}}}$

Rotational kinetic energy:

$\overline{){{\mathbf{KE}}}_{{\mathbf{R}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{{\mathit{I}}}_{\mathbf{c}\mathbf{m}}{{\mathit{\omega}}}^{{\mathbf{2}}}}$

Total kinetic energy:

$\overline{){\mathbf{KE}}{\mathbf{=}}{\mathit{K}}{{\mathit{E}}}_{{\mathbf{T}}}{\mathbf{+}}{\mathit{K}}{{\mathit{E}}}_{{\mathbf{R}}}}$

The total kinetic energy:

This problem illustrates the two contributions to the kinetic energy of an extended object: rotational kinetic energy and translational kinetic energy. You are to find the total kinetic energy of a dumbbell of mass when it is rotating with angular speed and its center of mass is moving translationally with speed .

Denote the dumbbell's moment of inertia about its center of mass by . Note that if you approximate the spheres as point masses of mass each located a distance from the center and ignore the moment of inertia of the connecting rod, then the moment of inertia of the dumbbell is given by , but this fact will not be necessary for this problem. Find the total kinetic energy of the dumbbell. Express your answer in terms of m, v, , and .

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