An object with a mass of m = 4.6 kg is attached to the free end of a light strin
ID: 1266418 • Letter: A
Question
An object with a mass of m = 4.6 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 5.00 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
angular velocity of reel be w and velocity of the mass be v
then v = Rw
and also acceleration a = R*alpha
force balance equation along vertical direction gives us
mg - T = ma
and Torque = I*alpha
T*R = MR^2/2 *alpha
(mg-ma) = MR/2 * a/R = Ma/2
solving this we get a = 7.4 m/s^2
a)
Tension = mg - ma = 11.04 N
b)
a = 7.4 m/s^2
c)
v^2 - u^2 = 2as
v = sqrt(2as) = sqrt(2*7.4*5) = 8.6 m/s
d)
using conservation of energy model
loss in potential energy of m = gain in KE of (m+M)
mgh = .5mv^2 + .5Iw^2
mgh = .5mv^2 + MR^2 * v^2/4R^2
mgh = .5mv^2 + Mv^2 / 4
solving thie we get v = 8.6