A counterweight of mass m 360 kg is attached to a light cor

A counterweight of mass m = 3.60 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 = 1.10 kg. The spokes have negligible mass. What is the net torque on the system about the axle of the pulley? Magnitude Note that the block will be accelerating downward. What does this mean about the relative magnitudes of mg and 77 N m direction When the counterweight has a speed v, the pulley has an angular speed omaga = v/R. Determine the magnitude of the total angular momentum of the system about the axle of the pulley. (kg m)v Using your result from (b) and r rightarrow = dL/dt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.) m/s^2

Solution

a)

Net Torque on the system about the axle of the pulley

T_net=Rmg =0.07*3.6*9.8

T_net = 2.47 N-m

The direction is along the rotation axis towards to the right.

b)

The total angular momentum of the system is

L=mvR+IW =mvR +(MR²)(V/R)

L=(m+M)vR =(3.6+1.1)*0.07v

L=(0.329 Kg-m)v

c)

T=dL/dt

mgR =(M+m)R(dV/dt)=(M+m)Ra

=>a=mg/(m+M) =3.6*9.8/(3.6+1.1)

a=7.51 m/s²