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Rotational and Translational Motion

  • Thread starter doopokko
  • Start date
  • #1
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Homework Statement



A bucket of water of mass 14.2 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.350 m with mass 12.1 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 11.0 m to the water. You can ignore the weight of the rope.

Solve for:
1) tension in the rope
2) time before the bucket hits the water
3) speed at which the bucket hits the water
4) force exerted on the cylinder by the axle during the fall

Homework Equations



Torque=R*F*sin(phi)
Torque=I*alpha
I=(M*R^2)/2

The Attempt at a Solution



M=12.1 kg (mass of cylinder)
R=0.175 (radius of cylinder)
m=14.2 (mass of bucket)

Torque = 0.175 * Tension = 1/2 * 12.1 * 0.175^2 * alpha

Torque = 0.175 * Tension = 0.18528125 * alpha

1.05875 * alpha = Tension

Weight = 14.2 * 9.8 = 139.16

ForceNet (on bucket) = Weight - Tension = m*a
ForceNet = 139.16 - Tension = 14.2 * a
139.6 - 1.05875*alpha = 14.2 * a

Tension = (1/2) * M * a

Tension = 41.57619

But I've been told that this isn't the correct answer. Can somebody set me straight, please?
 

Answers and Replies

  • #2
13
0
There shouldn't be problems, as long as the rope is assumed to be wrapped on the rim of the cylinder throughout the whole process.
 
  • #3
hage567
Homework Helper
1,509
2

Homework Statement



A bucket of water of mass 14.2 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.350 m with mass 12.1 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 11.0 m to the water. You can ignore the weight of the rope.

Solve for:
1) tension in the rope
2) time before the bucket hits the water
3) speed at which the bucket hits the water
4) force exerted on the cylinder by the axle during the fall

Homework Equations



Torque=R*F*sin(phi)
Torque=I*alpha
I=(M*R^2)/2

The Attempt at a Solution



M=12.1 kg (mass of cylinder)
R=0.175 (radius of cylinder)
m=14.2 (mass of bucket)

Torque = 0.175 * Tension = 1/2 * 12.1 * 0.175^2 * alpha

Torque = 0.175 * Tension = 0.18528125 * alpha

1.05875 * alpha = Tension

Weight = 14.2 * 9.8 = 139.16

ForceNet (on bucket) = Weight - Tension = m*a
ForceNet = 139.16 - Tension = 14.2 * a
139.6 - 1.05875*alpha = 14.2 * a

Tension = (1/2) * M * a

Tension = 41.57619

But I've been told that this isn't the correct answer. Can somebody set me straight, please?
It looks pretty good to me. I think you've got it right.

edit: Worked it out, got the same answer. Can't see why it would be wrong...:confused:
 
Last edited:
  • #4
4
0
Hahaha, I went through again and it turns out that that really was the right answer after all. I think there are just some wires crossed in my brain.

Thanks to everyone who looked this over, though.
 
  • #5
3
0
Hey guys, why is the tension 1/2 * M * a ??? Am a bit confused about that..
 
  • #6
Doc Al
Mentor
44,892
1,144
Hey guys, why is the tension 1/2 * M * a ??? Am a bit confused about that..
That fact can be deduced by combining these formulas (plus one other):
Torque=R*F*sin(phi)
Torque=I*alpha
I=(M*R^2)/2
 

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