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

Question

Physics

Prancer

13 February, 00:48

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

+4

Answers (1)

Know the Answer?

Anna Morrison · Answer 1

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