The diagram shows a mass hanging from a string that is wrapped around a flywheel

Question

The diagram shows a mass hanging from a string that is wrapped around a flywheel. The flywheel is free to rotate about an axel through its center (frictionless). The string is wrapped around a portion of the flywheel that has a radius of 0.0440 m. The hanging mass is 7.18 kg and, when released, is observed to accelerate downward at a rate of 2.75 m/s2. What is the rotational inertia of the flywheel?

Explanation / Answer

First do a free body diagram showing the forces acting on the mass and another showing the forces acting on the pulley. Them mass has only two forces acting on it - gravity pulling down and the tension on the rope pulling up. Assuming the mass accelerates as it falls, the sum of teh forces on the mass are: ma = T - mg where T = tension from the rope and a = acceleration. Now the pulley sees only the tension from the rope acting in a downward direction. Since the pulley does not move up or down there is no linear accleration. But there is a torque. If the pully has a radius R, the torque, H, is: H = T*R = I q where I = moment of inertial and q = angular acceleration. So we can write T in terms of R, I, and q ---> T = I q/R Hence ma = I q/R - mg Now q is related to a by: q = a/R so that ma = I a/R^2 - mg --> a = I a/(mR^2) - g ---> a (1 - I/(mR^2)) = g ---> a = g/[1 - I/(mR^2)] Note I/(mR^2) > 1 so |a|

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