Question

A bucket of water of mass 15.9 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.

Part A

What is the tension in the rope while the bucket is falling?

Take the free fall acceleration to be g = 9.80 m/s2 .

Part B

With what speed does the bucket strike the water?

Take the free fall acceleration to be g = 9.80 m/s2 .

Part C

What is the time of fall?

Take the free fall acceleration to be g = 9.80 m/s2 .

Part D

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/s2 .

Answer 1

Given that

M = 15.9 kg ; m = 12.4 kg ; d = 10.2 m

diameter = 0.350 m so radius = 0.175 m

(a)Let T be the tension in the rope.

T = m (g - a)

we need to know "a"

we know that moment of inetria of the drum is :

I = mr²/2

Torque = = I = T r

I = T r

mr²/2 = M( g - a) r ; But = a/r

solving for a we get:

a = 2Mg/(m+2M) = 2 x 15.9 x 9.8 / (12.4 + 2 x15.9) = 7.05 m/s2

T = M (g - a) = 15.9 x (9.8 - 7.05) = 43.7 N

Hence, T = 43.7 N

(b)from equation of motion

v² = u² + 2 a s

in our case, v =? u = 0 ; a = 7.05 and s = h = 10.2

v = sqrt (2 a h) = sqrt (2 x 7.05 x 10.2) = 12 m/s (11.99 = 12 aprox)

Hence, v = 12 m/s

(c)t will be given by:

t = sqrt(2 h /a) [derived using H = ut + 1/2 a t² ; where u = 0 ]

t = sqrt (2 x 10.2 / 7.05) = 1.7 sec

Hence, t = 1.7 sec

(d)The force on the cylinder axle would be the sum of weight of cylinder and Tension.

F = mg + T = 12.4 x 9.8 + 43.7 = 165.22 N

Hence, F = 165.22 N.