A counterweight of mass m = 5.00 kg is attached to a light cord that is wound ar

Question

A counterweight of mass m = 5.00 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-1.00 kg. The spokes have negligible mass. (a) What is the net torque on the system about the axle of the pulley? magnitude direction | Select N·m (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. kg·m)v (c) Using your result from (b) and = dL/dt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.) m/s

Explanation / Answer

given m = 5kg

R = 7 cm = 0.07 m

M = 1 kg

a. net torque on the axle of the system = mg*R = T

T = 5*9.81*0.07 = 3.4335 Nm

direction - counterclockwise

b. moment of inertia of the whole system = MR^2 + mR^2 = 6*0.07^2 = 0.0294 kg m^2

hence

angular momentum = Iw

but w = v/R

hennce

angular momentum = Iv/R = 0.42v kg m

c. now, T = dL/dt

so, 3.4335 = 0.42*a

a = 8.175 m/s/s

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