A counterweight of mass m = 3.90 kg is attached to a light cord that is wound ar

Question

A counterweight of mass m = 3.90 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 = 1.30 kg. The spokes have negligible mass. What is the net torque on the system about the axle of the pulley? magnitude N middot m direction to the right along the axis of rotation When the counterweight has a speed v, the pulley has an angular speed omega = v/R. Determine the magnitude of the total angular momentum of the system about the axle of the pulley. Note that your numerical answer is multiplied by the symbol v in the answer box. kg middot m)v (c) Using your result from (b) and T = dL/dt, calculate the acceleration of the counterweight. (Enter the magnitude of the acceleration.) The expression for the angular momentum in part (b) has a factor of v. How do you determine dL/dt, from this expression? m/s2

Explanation / Answer

(b) total angular momentum =

= block + pulley =

= m v R + I w =

= 3.90 * v * 0.0700 + M R^2 * v/R =

= 0.273 v + 1.30 * 0.0700 * v =

= 0.364 v just enter 0.364 in the box

(c) acceleration = total force / total mass = Mg / (m+M) =

= 3.9 * 9.80 / (1.3 + 3.9) = 7.35 m/s^2

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