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 = 3.00 kg. The spokes have negligible mass.(a) What is the net torque on the system about the axle of the pulley?V N.mto the right along the axis of rotation vmagnitude 2.4695direction(b) When the counterweight has a speed v, the pulley has an angular speed w = v/R. Determine the magnitude of the total angular momentum of the system about the axle of the pulley.(3.705Note that your numerical answer is multiplied by the symbol v in the answer box. kg • m)v(c) Using your result from (b) and 7 = L/dt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.)6.925The expression for the angular momentum in part (b) has a factor of v. How do you determine dL/dt, from this expression? m/s2

Given: M=3.6 kg R=0.07 m g=9.8 m/s m=3 kgFormula used: τ=R⇀×T→ L→=R→×mv→+Iω→ τ=dldt r=radius M,m=masses g=gravitational constant v=velocity a=accelerationa)Torque τ=R⇀×T→ τ=R⇀mg z→ Where M=3.6 kg and R=0.07 m τ=3.6×0.07×9.81 z→ τ=2.469z→ N.m The direction is along the axis of rotation towards rightb) To find angular momentum L→=R→×mv→+Iω→ L→=R→×mv→+MR2VR L→=m+MRV Where M=3.6 kg, R=0.07 m, m=3 kg L→=3.6+30.07V On solving, L→=0.462 kg.mVc)We know, τ=dldt mgR=m+MRdvdt a=mgm+M Where M=3.6 kg, R=0.07 m, g=9.8 m/s and m=3 kg a=3.6×9.83.6+3 a=5.345 m/s2.

Answered: A counterweight of mass m = 3.60 kg is…