An object with a mass of m = 5.3 kg is attached to the free end of a light strin

Question

An object with a mass of m = 5.3 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.255 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 5.4 m above the floor.

(a) Determine the tension in the string.

(b) Determine the magnitude of the acceleration of the object.

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

Explanation / Answer

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*0.255^2 = 0.097 Nm^2)

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

a = 5.3*9.81/(0.097/0.255^2 + 5.3)

a = 7.65 m/s^2 = acceleration of the object
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Tension = 5.3*(9.81 - 7.65) = 11.448 N
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v = sqrt(2sa) = sqrt(2*5.4*7.65)

v = 9.08 m/s = velocity when hitting the ground

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