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 = 1.30 kg. The spokes have negligible mass.

(a) What is the net torque on the system about the axle of the pulley?

magnitude _______ N 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 = 1.30 kg. The spokes have negligible mass. L with arrow/dt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.) _________ m/s2 tau (a) What is the net torque on the system about the axle of the pulley? magnitude _______ N A m direction (to the right along the axis of rotation) or (to the left along the axis of rotation) (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 A m)v (c) Using your result from (b) and

Explanation / Answer

Here ,

M = 1.30 Kg

R = 0.09 m

m = 5.20 Kg

a)

Here , Let the tension in the string is T

Now, acceleration of the block ,a = net force/effiective mass

a = (5.20 *9.8)/(5.20 + 0.5 * 1.3 R^2/R^2)

a = 8.71 m/s^2

Now , mg - T = ma

5.2 *9.8 - T = 5.2 * 8.71

T = 5.66 N

Torque = 5.66 * 0.09

TOrque = 0.51 N.m

the direction of torque is left of axle

B)

Angular momemtum = 0.5 * 1.3 * 0.09^2 * v^2 /.09^2 + 5.2 * .09 * v

Angular momemtum = (1.118 kg.m)v

C)

the acceleration of wounterweight is 8.71 m/s^2

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