A barbell consists of two massive balls connected by a low-mass rod. The barbell

Question

A barbell consists of two massive balls connected by a low-mass rod. The barbell slides across a low-friction icy surface, spinning as it moves, as shown in the diagram. The mass m of each ball is 0.4 kg. The distance d between the centers of the balls is 0.27 m. The speed v of the center of mass of the barbell is 0.65 m/s, and the barbell makes one complete revolution in 3 seconds.

What is the translational kinetic energy of the barbell?

Considering only the motion of the barbell relative to its center of mass, what is the speed of one ball as it rotates around the center of the barbell?

What is the rotational kinetic energy of the barbell? (This is the kinetic energy associated with motion relative to the center of mass.)

What is the total kinetic energy of the barbell?

Explanation / Answer

So each barbell ball has a mass of 0.6 kg, and the whole thing moves at 0.46 m/s (the speed of the center of mass). Since we're concerned only with the translational motion, you can treat the system (for now) as a point particle of 0.6 kg mass and 0.46 m/s speed (haha, simplification much?) The translational kinetic energy is therefore

K_trans = .5mv²
= .5 (0.4 kg) (0.65 m/s)²
K_trans = 0.0845 J (to two significant digits)

"Considering only the motion of the barbell relative to its center of mass, what is the speed of one ball as it rotates around the center of the barbell?"

Since the barbell makes a complete revolution in 3 seconds, and the distance between the centers of the balls is 0.27 m, you can reason out that the ball travels the circumference of this circle in 3 seconds. Since you know the diameter, the circumference is simply that diameter (the distance between the balls) multiplied by . Then the rotational speed must be that circumference divided by the time it takes to move a whole circle, namely 3 seconds. So

v_rot = *d/t
= 3.14*(0.27 m)/(3 seconds)
v_rot = .2826 m/s
K_rot = .5I²

where I is the inertial moment of the object and is the angular velocity. This, however, is not normally since for the purpose of the problem we will assume that all the mass of the whole system is represented by the two balls. This means that you can just use

K_rot = .5 m(v_rot)²

since both masses are moving at the speed of v_rot relative to the center of mass (whereas in the angular kinetic energy equation, the equation accounts for the fact that not all parts of a rotating system would be moving at the same speed relative to the center of mass). Substitution then yields

K_rot = .5 (2*0.4 kg)(.2826 m/s)²
K_rot = .03194 J

"What is the total kinetic energy of the barbell?"

Add together the translational and rotational energies. You'll find

K_tot = K_trans + K_rot
= 0.0845 J + .03194 J
K_tot = .1164 J

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