An object with a mass of m = 5.2 kg is attached to the free end of a light strin
ID: 1459743 • Letter: A
Question
An object with a mass of m = 5.2 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 6.70 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
(d) 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
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)
a = mg/(I/R^2 + m)
a = 5.2*9.81/(3/2 + 5.2)
a = 7.6 m/s^2 = acceleration of the object
t
-------------
Tension = 5.2*(9.81 - 7.6) = 11.4 N
------------
v = sqrt(2sa) = sqrt(2*6.70*7.60)
v = 10 m/s = velocity when hitting the ground
--------
Apply law of conservation of energy:
0.5Iw^2+0.5mv^2=mgh
0.25Mr^2(v/r)^2+0.5mv^2=mgh
0.25Mv^2+0.5mv^2=mgh
V=sqrt(mgh/0.25M+0.5m)=10 m/s