A counterweight of mass m = 5.20 kg is attached to a light cord that is wound ar

Question

A counterweight of mass m = 5.20 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 = 9.00 cm and mass M = 3.00 kg. The spokes have negligible mass. What is the net torque on the system about the axle of the pulley? 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. Using your result from (b) and T = dL/dt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.)

Explanation / Answer

a) Torque = weight*R = (5.2*9.81)*(9*10^-2) = 4.59108 Nm

Direction = counterclockwise

b) Total angular momentum = mv*R + I*(v/R) = mvR + (MR^2)*(v/R) = (mR + MR)v = (5.2*0.09 + 3*0.09) v

= 0.738*v

c) L = 0.738*v

= dL/dt = (dL/dv)(dv/dt) = a*(dL/dv) = a*0.738

4.59108 = a*0.738

a = 6.221 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.