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.

I worked out (a) and no matter what answer I get for (b) it's wrong. I can't do (c) without (b). Please help!!! ((b) is not 0.091)

(a) Magnitude = 2.6 Nm

Direction= to the right of 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.

Explanation / Answer

Forces on the block are tension up and weight down: m*a = m*g - T Torque on the pulley: tau = T*R Newton's second law for pulley tau = I*alpha Kinematics constraint (inextensible and no-slip rope): a = alpha*R Consolidate to two equations: T*R = I*a/R m*g - T = m*a Solve for both a and T: a = m*g*R^2/(m*R^2 + I) T = I*m*g/(m*R^2 + I) And since we can treat the pulley as a thin hoop: I = M*R^2 Thus, it simplifies to: a = m*g/(m + M) T = g*M*m/((m+M)) And the torque on the pulley: tau_pulley = g*M*m*R/((m+M)) Data: g:=9.8 N/kg; M:=2.9 kg; m:=3.7 kg; R:=0.06 m; Results: tau_pulley = 0.9559 Netwon-meters tau_totalsystem = 2.1756 Newton-meters T = 15.93 Newtons a = 5.494 meters/second^2 alpha = 91.57 radians/second^2

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