A bucket of water of mass 14.6 kg is suspended by a rope wrapped around a windla
ID: 1461386 • Letter: A
Question
A bucket of water of mass 14.6 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.280 m with mass 11.3 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 10.9 m to the water.
What is the tension in the rope while the bucket is falling? Take the free fall acceleration to be g = 9.80 m/s2 .
With what speed does the bucket strike the water? Take the free fall acceleration to be g = 9.80 m/s2 . m/s
What is the time of fall?
While the bucket is falling, what is the force exerted on the cylinder by the axle?
Explanation / Answer
R = d/2
= 0.28/2
= 0.14 m
moment of Inertia of cyliinder, I = (1/2)*M*R^2
= (1/2)*11.3*0.14^2
= 0.11074 kg.m^2
let a is the acceleration of bucket and alfa is the angulara cceleration of cyllinder.
Let T is the tension in the string.
Let m = 14.6 kg
Apply, net force acting on bucket, Fnet = m*g - T
m*a = m*g - T
T = m*g - m*a
net torque actung on cyllinder = I*alfa
T*R = I*a/R
(m*g - m*a)*R = (1/2)*M*R^2*a/R
m*g - m*a = (1/2)*M*a
m*g = a*(M/2 + m)
a = m*g/(M/2 + m)
= 14.6*9.8/(11.3/2 + 14.6)
= 7.07 m/s^2
so, T = m*(g - a)
= 14.6*(9.8 - 7.07)
= 39.8 N <<<<<<<<<<<<<<<------------Answer
v = sqrt(2*a*h)
= sqrt(2*7.07*10.9)
= 8.8 m/s <<<<<<<<<<<<<<<------------Answer
time taken
t = (v - u)/a
= (8.8 - 0 ) /7.07
= 1.24 s <<<<<<<<<<<<<<<------------Answer
force exerted on the cylinder by the axle = T
= 39.8 N <<<<<<<<<<<<<<<------------Answer
because net force on the cyllinder = 0
F - T = 0
F = T
= 39.8 N
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