source : yahoo.com
what is the magnitude of the net torque on the system about the axle of the pulley?
Torques on the axle
= 4 * 9.81 * 0.08 m-N
= 3.1392 m-N.
Angular momentum of the system
= angular momentum of the counterweight + angular momentum of the pulley
L = [4 * w * (0.08)^2] + [2 * (0.08)^2 w] = 0.0384 w kgm^2/s
Torque = dL/dt
=> 3.1392 = d/dt (0.0384 w)
=> dw/dt = 3.1392/0.0384 rad/s^2
=> angular acceleration, dw/dt = 81.75 rad/s^2
=> acceleration of the counterweight
= dw/dt * R
= 81.75 * 0.08 m/s^2
= 6.54 m/s^2.
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