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19 October, 06:13

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

+3
Answers (1)
  1. 19 October, 06:35
    0
    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
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