Question

A bucket of water of mass 14.7 is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.280m with mass 11.6 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 10.3m 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/s squared

Part B
with what speed does the bucket strike the water?
-Take the free fall acceleration to be g=9.80 m/s^2

Part C
What is the time of fall
-Take the free fall acceleration to be g=9.80m/s^2

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/s^2

Answer 1

Answer:

Explanation:

Tension T in the rope will create torque in solid cylinder ( axle ). If α be angular acceleration

T R = 1/2 M R²α ( M is mass and R is radius of cylinder )

= 1/2 M R² x a / R ( a is linear acceleration )

T = Ma / 2

For downward motion of the bucket

mg - T = m a ( m is mass and a is linear acceleration of bucket downwards )

mg - Ma / 2  = ma

a = mg / ( M /2 + m )

Substituting the values

a = 14.7 x 9.8 / ( 5.8+ 14.7 )

= 7 m / s²

A )

T = Ma / 2

= 5.8 x 7

= 40.6 N

B ) v² = u² + 2 a h

= 2 x 7 x 10.3

v = 12 m /s

C )

v = u + a t

12 = 0 + 7 t

t = 1.7 s