Rotational and Translational Motion

AI Thread Summary
The discussion revolves around a physics problem involving a bucket of water and a windlass. The main calculations include determining the tension in the rope, the time before the bucket hits the water, the speed at impact, and the force on the cylinder by the axle. Initial calculations for tension yielded 41.57619 N, which was initially questioned but later confirmed as correct. The confusion about the relationship between tension and acceleration was clarified through the application of torque and moment of inertia equations. This problem illustrates the interplay between rotational and translational motion in a practical scenario.
doopokko
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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?
 
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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.
 
doopokko said:

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:
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.
 
Hey guys, why is the tension 1/2 * M * a ? Am a bit confused about that..
 
Dupain said:
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):
doopokko said:
Torque=R*F*sin(phi)
Torque=I*alpha
I=(M*R^2)/2
 
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