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.


(h) Calculate the moment of inertia I of the barbell.


(j) Use the moment of inertia I and the angular speed || = 120 rad/s to calculate the rotational angular momentum of the barbell:


(k) How does this value, |rot|, compare to the angular momentum |tot, A| calculated earlier by adding the translational angular momenta of the two balls?


(l) Use the moment of inertia I and the angular speed || = 120 rad/s to calculate the rotational kinetic energy of the barbell:

Explanation / Answer

h) Calculate the moment of inertia I of the barbell.

=>2*(0.45)*(0.175)^2 = 0.0275625 kg-m^2


(j) Use the moment of inertia I and the angular speed || = 120 rad/s to calculate the rotational angular momentum of the barbell:

L = I* = 0.0275625*120 = 3.3075 N-m-s


(k) How does this value, |rot|, compare to the angular momentum |tot, A| calculated earlier by adding the translational angular momenta of the two balls?

both are be same

(l) Use the moment of inertia I and the angular speed || = 120 rad/s to calculate the rotational kinetic energy of the barbell:

=>1/2*I^2 = (1/2)*0.0275625*(120)^2 =198.45 J

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