A counterweight of mass m = 5.30 kg is attached to a light cord that is wound ar

Question

A counterweight of mass m = 5.30 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 = 7.00cm and mass M = 2.20 kg. The spokes have negligible mass.

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

magnitude: ?

(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*m)v

? m/s^2

Explanation / Answer

  a)
Torques on the axle
= 5.30 * 9.81 * 0.07 m-N
= 3.639m-N.

b)
Angular momentum of the system
= angular momentum of the counterweight + angular momentum of the pulley
L = [ 5.30* w * (0.07)^2] + [2 * (0.07)^2 w] = 0.0357 w kgm^2/s

c)
Torque = dL/dt
=> 3.639 = d/dt ( 0.0357 w)
=> dw/dt =3.639/0.0384 rad/s^2
=> angular acceleration, dw/dt = 94.76 rad/s^2
=> acceleration of the counterweight
= dw/dt * R
=94.76 * 0.07 m/s^2
= 6.63 m/s^2.

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