An object with a mass of m = 4.65 kg is attached to the free end of a light stri
ID: 1449980 • Letter: A
Question
An object with a mass of m = 4.65 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.270 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 5.10 m above the floor.
a)Determine the tension
b)Determine the magnitude of the acceleration of the object.
c)Determine the speed in which the object hits the floor
Explanation / Answer
Ans:-
A]Tension T = mg-ma (m = mass of object) ....................1
b]Torque = T*R = (mg - ma)*R = I* = I*(a/R) --> solve for a
(I = 1/2 mR^2 = 1/2*3*0.27^2 = 0.10935 Nm^2)
a = mg/(I/R^2 + m)
a = 4.65*9.81/(0.10935/0.27^2 + 4.65)
a = 7.42 m/s^2 = acceleration of the object
From equation 1
a] Tension = 4.65*(9.81 – 7.42) = 11.13 N
C]By kinematics eqn
v = sqrt(2sa) = sqrt(2*5.1*7.42)
v = 8.7 m/s = velocity when hitting the ground