A counterweight of mass m = 3.90 kg is attached to a light cord that is wound ar

Question

A counterweight of mass m = 3.90 kg is attached to a light cord that is wound around a pulley as shown in the figure below. The pulley is a thin hoop of radius R = 7.00 cm and mass M = 2.40 kg. The spokes have negligible mass.

(a) What is the net torque on the system about the axle of the pulley?

magnitude

(b) When the counterweight has a speed v, the pulley has an angular speed = v/R. Determine the magnitude of the total angular momentum of the system about the axle of the pulley.

magnitude

(b) When the counterweight has a speed v, the pulley has an angular speed = v/R. Determine the magnitude of the total angular momentum of the system about the axle of the pulley.

Explanation / Answer

A. The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain z, we should have:

tau = R x T = Rmg (z)

tau = 0.07*3.9*9.81 = 2.678 (z)

B. Taking into account rotation of the pulley and translation of the counterweight , the total angular momentum of the system is:

L = R x mv + Iw

L = mvR + MR^2*(v/R)

L = (m + M)Rv

L = 6.3*0.07*v = 0.441*v Kg.m

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