Question

A bucket of water of mass 16.0kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.350m with mass 11.4kg . 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 10.1m to the water. You can ignore the weight of the rope. a) What is the tension in the rope while the bucket is falling? Take the free fall acceleration to be g = 9.80m/s^2 . b) With what speed does the bucket strike the water? Take the free fall acceleration to be g = 9.80m/s^2 . c) What is the time of fall? Take the free fall acceleration to be g = 9.80m/s^2 . 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 = 9.80m/s^2 .

Answer 1

Step-by-step explanation:

We shall consider tension to be T.

For motion of bucket in downward direction

mg - T = ma , m is mass of bucket , a is linear acceleration of bucket

for rotational motion of wheel

Tortque by force Tr = I x α I is moment of inertia , α is angular acceleration , r is radius of the wheel.

Putting the values in the equation above

mg - I x α / r = ma

mg - 1/2 M r ²x a / r² = ma

mg - 1/2 M a = ma

mg = 1/2 M a + ma

a = mg / (m + .5 M)

= 16 X 9.8 / (16 +.5 x 11.4)

= 7.22 m /s²

mg - T = ma

T = m( g - a )

= 16 x ( 9.8 - 7.22 )

= 41.28 N.

b )

v² = u² + 2 a h , v is velocity when bucket falls by height h

= 0 + 2 x 7.22 x 10.1

= 145.84

v = 12.07

c )

v = u + at , t is time of fall

12.07 = 0 + 7.22 t

t = 1.67 s

d )

Force by axle on cylinder = T

= 41.28 N .