An object with a mass of m = 4.9 kg is attached to the free end of a light strin

Question

An object with a mass of m = 4.9 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.245 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 5.20 m above the floor.

(a) Determine the tension in the string.
N

(b) Determine the magnitude of the acceleration of the object.
m/s2

(c) Determine the speed with which the object hits the floor.
m/s

Explanation / Answer

Hello

Tension T = mg-ma (m = mass of object)

Torque ? = T*R = (mg - ma)*R = I*? = I*(a/R) --> solve for a

(I = 1/2 mR^2 = 1/2*3*0.245^2 = 0.0900375 Nm^2)

a = mg/(I/R^2 + m)

a = 4.9*9.81/(0.09/0.245^2 + 4.9)

b) a = 7.51 m/s^2 = acceleration of the object
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a) Tension = 4.9*(9.81 - 7.51) = 11.266 N
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v = sqrt(2sa) = sqrt(2*5.2*7.5)

c) v = 10.8244 m/s = velocity when hitting the ground

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