I am pushing with a contact force of 40 N on the handle of the winch shown in th

Question

I am pushing with a contact force of 40 N on the handle of the winch shown in the diagram.

The handle of the winch is L = 0.750 m long, the radius of the winch R = 0.300 m. The moment of inertia (rotational inertia) of the winch is 24 kg m2 and friction in the bearings exerts a torque of 14.0 N m, which resists the winch turning.

The winch starts from rest. After one quarter of a rotation (pi/2 radians), if I exert a constant torque (by maintaining the 30 degree angle to the handle and the 40 N of force) what is the speed of the rope (2 s.f.)?

0.36 rad/s isn't correct

1.3 rad/s isn't correct

Explanation / Answer

torque due to force,

torque1 = L F cos30 = 0.750 x 40 x cos30 = 25.98 N m

friction torque = - 14 N m

Net torque = 25.98 - 14 =11.98 Nm

and Torque = I x alpha

alpha = 11.98 / 24 = 0.5 rad/s^2

USing wf^2 - wi^2 = 2(alpha)(theta)

wf^2 - 0 =2 (0.5)(pi / 2)

wf = 1.25 rad/s

Speed of rope , v = w R = 1.25 x 0.3 = 0.376 m / s    (not rad /s )

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