Question

A dumbbell consists of two point-like masses M connected by a massless rod of length 2L. There are no external forces. If the dumbbell rotates with angular velocity w while its center of mass moves linearly (translates) with velocity v such that v = ωL, the ratio of the rotational kinetic energy to the translational (linear) kinetic energy is:

Answer 1

Answer:

2

Explanation:

The rotational kinetic energy of the dumbbell is:

$E_r = \frac{1}{2}I\omega^2$

where I is the moments of inertia of the two point-like system

$I = 2Mr^2 = 2ML^2$

$E_r = ML^2\omega^2$

The translation kinetic energy is:

$E_t = \frac{1}{2}Mv^2 = \frac{1}{2}M\omega^2L^2$

Therefore the ratio of the rotational kinetic energy to the translational (linear) kinetic energy is:

$\frac{E_r}{E_t} = \frac{ML^2\omega^2}{\frac{1}{2}M\omega^2L^2} = 2$