****Please indicate final answers clearly ****** A bucket of water of mass 14.8
ID: 1489420 • Letter: #
Question
****Please indicate final answers clearly ******
A bucket of water of mass 14.8 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.350 m with mass 12.4 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 10.2 m to the water. You can ignore the weight of the rope. What is the tension in the rope while the bucket is falling? Take the free fall acceleration to be g = 9.80 m/s^2. With what speed does the bucket strike the water? Take the free fall acceleration to be g = 9.80 m/s^2. What is the time of fall? Take the free fall acceleration to be g = 9.80 m/s^2. While the bucket is falling, what is the force exerted on the cylinder by the axle? Take the free fall acceleration to be g = 9.80 m/s^2.Explanation / Answer
On bucket of water applying Fnet = ma
14.8g - T = 14.8a .... (i)
on cylinder using torque = I x alpha
(r x T) = (m r^2 / 2) (a/r)
T = 12.4a/2 = 6.2a
putting T in equation (i)
14.8g - 6.2a = 14.8a
a = 6.91 m/s^2
a) T = 6.2a = 6.2 x 6.91 =42.84 N
b) using v^2 - u^2 = 2ad
v^2 - 0 = 2 * 6.91 * 10.2
v = 11.87 m/s
c) for time using v = u + at
11.87 = 0 + 6.91t
t = 1.72 sec
d) on cylinder
F = T + mg = 42.84 + (12.4 * 9.80) = 164.36 N
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