A barbell spins around a pivot at its center at A. The barbell consists of two s

Question

A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 450 grams (0.45 kg), at the ends of a very low mass rod of length d = 35 cm (0.35 m; the radius of rotation is 0.175 m). The barbell spins clockwise with angular speed 120 radians/s.


(a) What is the speed of ball 1?

(b) Calculate the translational angular momentum trans, 1, A of just one of the balls (ball 1).


(c) Calculate the translational angular momentum trans, 2, A of the other ball (ball 2).


(d) By adding the translational angular momentum of ball 1 and the translational angular momentum of ball 2, calculate the total angular momentum of the barbell, tot, A.

Explanation / Answer

(a) speed of ball 1 = rx = 0.175*120 = 21 m/s

(b) and (c) translational angular momentum of ball 1 = translational angular momentum of ball 2 =mvr

=0.45*21*0.175 = 1.65375 kg.m/s

(d) total translational angular momentum = 2*1.65375 = 3.3075 kg.m/s

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