An object with a mass of m = 4.95 kg is attached to the free end of a light stri
ID: 1449749 • Letter: A
Question
An object with a mass of m = 4.95 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 5.20 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
(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
a) on block ( Applying Fnet = ma )
mg - T = ma
4.95g - T = 4.95a
on reel :
Applying torque = I x alpha
R (T) = ( M R^2 / 2) (a/R)
T = Ma /2 = 1.5a
putting value of T in first equation,
4.95g - 1.5a = 4.95a
a = 7.52 m/s^2
and T = 1.5a = 1.5 x 7.52 = 11.28 N .........Ans
b) a = 7.52 m/s^2 (calculated inpart A)
c) using vf^2 - vi^2 = 2 ad
v^2 - 0^2 = 2( 7.52)(5.20)
v = 8.84 m/s
d) On mass:
Work done by tension force + work done by gravity = change in KE
- 11.28 x 5.20 + (4.95 x 9.8 x 5.20) = 4.95v^2 /2 - 0
v = 8.84 m/s