A counterweight of mass m = 3.50 kg is attached to a light cord that is wound ar

Question

A counterweight of mass m = 3.50 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 = 6.00 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 N · m direction (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 tau with arrow = dL with arrow/dt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.) m/s2

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:

net = R × T~= Rmg~z= 0.06 × 3.50 × 9.8~z= 2.058 The net torque has a magnitude of = 2.058 N.m and its direction is along the rotation axis towards the right

b)Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:L~ = R× mv + IL = mRv + MR vR= (m + M)Rv= (3.50 + 2.40) × 0.06= 0.354 Kg.m

c)

=dL/dt= mgR = (M + m)Rdv/dt= (M + m)R

a =mg/(m + M)=3.50 × 9.8/(5.90)= 5.81 m/s2

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