An object with a mass of m = 4.7 kg is attached to the free end of a light strin
ID: 1468044 • Letter: A
Question
An object with a mass of m = 4.7 kg is attached to the free end of a light string wrapped around a reel of radius R = 0.225 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 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
ANS for c)
The angular KE of solid disk:
KE angular = 1/2Iw² = (0.5)(1/2MR²)(V/R)² = 0.25MV² {where M = 3.00 kg and V = speed of object at floor}
KE angular = 0.25(3.00)V² = 0.75V²
KE linear = 1/2MV² {where M = 4.7 kg & V = speed of object at floor}
KE linear = (0.5)(4.7)V² = 2.35V²
GPE = mgh {where m = 4.7 kg & h = 5.10 m}
GPE = (4.7)(9.81)(5.10) = 235.14 J
GPE = total KE
235.14 = 0.75V² + 2.35V² = 3.10V²
V² = 235.14/3.10 = 75.85
V 8.709 m/s
Ans for b)
Vavg = V/2 = 8.709/2 = 4.35 m/s
Vavg = d/t = 5.10/t {where t = time for object to travel distance = d, to the floor}
t = 5.10/4.35 = 1.172 s
a = V/t = 8.709/1.172 = 7.430 m/s²
Ans for a)
T = tension in string = m(g - a) = 4.7(9.81 - 7.430) = 11.139 N