A string is attached to the rim of a small hoop of radius = 8.00?10?2 m and mass

Question

A string is attached to the rim of a small hoop of radius = 8.00?10?2 m and mass = 0.180 kg and then wrapped several times around the rim. If the free end of the string is held in place and the hoop is released from rest and allowed to drop, as shown in the figure (Figure 1) , calculate the angular speed and the translational speed of the rotating hoop after it has descended = 0.750 m . Use g = 9.80 m/s2 for the acceleration due to gravity. In order to determine the potential energy of the system, the zero point for gravitational potential energy must first be defined. This is the point where you set in your coordinate system. For the remainder of this problem, assume that the point is located at the bottom point of the hoop's motion, as shown in the figure. (Figure 2) What are the initial and final potential energies ( and ) of the hoop? What are the initial and final kinetic energies ( and ) of the hoop?

Explanation / Answer

In this case, the hoop is rotating about a fixed point along its rim (similar to a rotating tire). So the hoop is descending under the force of gravity, which accelerate it at g = 9.81 m/s². We an write down the equations of motion for constant acceleration: vf(t) = vi + gt = gt since vi = 0 d = ½gt² (0.75) = ½(9.81)t² => t = 0.39 sec The angular speed is given by: ? = v/(2r) where r is the radius of the hoop, and the reason we're using 2r here is because the hoop is not rotating about its center, but a fixed point along its rim. ? = v/(2r) = gt/2r = (9.81)(0.39)/2(0.08) = 23.98 radians/se

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