An object with a mass of m = 4.50 kg is attached to the free end of a light stri

Question

An object with a mass of m = 4.50 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.225 m and mass of M = 3.00 kg. The reel is a solid disk, free to rotate in a vertical plane about the horizontal axis passing through its center as shown in the figure below. The suspended object is released from rest 6.60 m above the floor. Determine the tension in the string. Determine the magnitude of the acceleration of the object. Determine the speed with which the object hits the floor. Verify your answer to part (c) by using the isolated system (energy) model. (Do this on paper. Your Instructor may ask you to turn in this work)

Explanation / Answer

The net force on the mass is

Fnet = ma = 4.5kg * a = mg - T = 4.5kg * 9.8m/s² - T = 44.1N - T
so T = 44.1 - 4.5a
torque = F*r = (44.1 - 4.5a) * 0.255m = 11.245 - 1.3147a
but also = I = ½mr² * a/r = ½ * 4.5kg * 0.255m * a = 0.573a
Since = , 11.245 - 1.317a = 0.573a
11.245 = 1.89a

a = 5.95m/s² acceleration

tension T =44.1 - 4.5 * 5.95 = 17.325 N tension

v² = u² + 2as = 0 + 2 * 5.95m/s² * 6.60m = 65.3 m²/s²
v = 8.86 m/s velocity

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