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 = 0.230 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

Tension T = mg-ma (m = mass of object) Torque t = T*R = (mg - ma)*R = I*a = I*(a/R) --> solve for a (I = 1/2 mR^2 = 1/2*3*0,24^2 = 0,0864 Nm^2) a = mg/(I/R^2 + m) a = 5,5*9,81/(0,0864/0,24^2 + 5,5) a = 7,70 m/s^2 = acceleration of the object ------------- Tension = 5,5*9,81 - 7,7) = 11,605 N ------------ v = sqrt(2sa) = sqrt(2*5,2*7,7) v = 8,948 m/s = velocity when hitting the ground

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