A counterweight of mass m = 3.80 kg is attached to a light cord that is wound ar

Question

A counterweight of mass m = 3.80 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 = 1.90 kg. The spokes have negligible mass.

magnitude N m direction ---Select---to the right along the axis of rotationto the left along the axis of rotation A counterweight of mass m = 3.80 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 = 1.90 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 ? = 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 pi = ldt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.)

Explanation / Answer

(a)   the net torque on the system about the axle of the pulley
= 3.8 * 9.81 * 0.07
=2.609Nm
answer

(b) counterweight has a speed v, the pulley has an angular speed = v/R.
Moment of inertia of system
= mr^2
= 1.9 * 0.07^2
= 0.0093
magnitude of the total angular momentum of the system about the axle of the pulley
= MoI * ?
= 0.0093 * v/ 0.07
= 0.133 v
answer

(c) net torque = MoI * angular acc
2.609 = 0.0093 * angular acc
angular acc = 280.53 rad/s^2
acceleration of the counterweight
= r * angular acc
= 0.07 * 280.53 m/s^2

= 19.63m/s^2
answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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