The diagram shows a rope passing around a circular wheel of radius 0.500 meters.

Question

The diagram shows a rope passing around a circular wheel of radius 0.500 meters. At t = 0s, a torque is applied to the wheel which is initially at rest. This torque is applied with two horizontal tensions, 90.0 Newtons (lower rope) and F Newtons (upper rope), where F is unknown. Assume the rope does not slip relative to the wheel as the wheel rotates and that the two forces are constant in magnitude and direction after t = 0s . The moment of inertia of the wheel about its axis of rotation is I = 100kg.m^2

If the angular momentum of the wheel is L = ?500kg.m.s-2 at t=10 seconds, find the value of the unknown force F

Explanation / Answer

torque T = I * Alpha

where I is moment of inertia

Alpha = angular accleration = (Tf - Ti)/I

Alpha = ( F-90)*0.5/(I)

here W = 0 + Alpha dT = (F-90)* 0.5/I * 10

Angular MOmentum L = I W = ( I *(F-90)* 5 /I = 500

F = 190 N (ANSWER)

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