What is the force exerted by the axle on the cylinder?

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Homework Help Overview

The problem involves a bucket of water suspended by a rope around a windlass, which is a solid cylinder. The scenario describes the dynamics of the falling bucket and the forces acting on the cylinder as it pivots on a frictionless axle. The objective is to determine the force exerted by the axle on the cylinder.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need for additional information to analyze the forces involved, such as the impact speed of the bucket or the time taken to fall. There are mentions of frictional torque and its effects on the system. Some participants suggest using the tension in the rope and gravitational forces to find the force on the axle.

Discussion Status

The discussion is ongoing with various interpretations of how to approach the problem. Some participants have offered guidance based on their understanding of the forces at play, while others express confusion about the instructions given. There is no explicit consensus on the method to find the force exerted by the axle.

Contextual Notes

Participants are working with the constraints of the problem, including the assumption of a frictionless axle and the need to consider both the gravitational force on the bucket and the tension in the rope. There is a focus on the relationship between linear and rotational dynamics.

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A bucket of water of mass 15.7kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.250m with mass 11.2kg. 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 10.1m to the water. You can ignore the weight of the rope.

I need to find the Force exerted on the cylinder by the axle.

I have found so far the tension in the rope while the bucket is falling which is 40.5 N, the speed on impact of the bucket, 12.1m/s, and the time it takes to fall as 1.67s.

But I have no clue how to proceed to find the force.
 
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to do this, you m ust have one more piece of information. What I'm seeing here is that there is a constant frictional torque applied by the axle on the cylinder. We could find this out if we had something like the time it took the bucket to fall, or the speed it hit the water with. Then we could see what the difference is between the real situation and the ideal conditions.

Regards,

Nenad
 
Nenad said:
to do this, you m ust have one more piece of information. What I'm seeing here is that there is a constant frictional torque applied by the axle on the cylinder. We could find this out if we had something like the time it took the bucket to fall, or the speed it hit the water with. Then we could see what the difference is between the real situation and the ideal conditions.

Regards,

Nenad

I don't think I understand what you are telling me to do. And all that information I have found. It is in the original post.
 
jaymode said:
A bucket of water of mass 15.7kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.250m with mass 11.2kg. 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 10.1m to the water. You can ignore the weight of the rope.

I need to find the Force exerted on the cylinder by the axle.

I have found so far the tension in the rope while the bucket is falling which is 40.5 N, the speed on impact of the bucket, 12.1m/s, and the time it takes to fall as 1.67s.

But I have no clue how to proceed to find the force.
The gravitational force on the dropping bucket provides all the accelerations: the bucket's downward acceleration and the cylinder's rotational acceleration (torque = tangential force x radius of cylinder)

[tex]mg - T= ma = F_{bucket}[/tex]

[tex]Tr = \tau = I_{cyl}\alpha[/tex]

[tex]\alpha = a_{bucket}r[/tex]

With that you should be able to work out the net force on the bucket and T. T is the force on the axle.

AM
 
Last edited:
ok, I see what you mean. I took the question the wrong way. Do what Andrew Mason said, it is the correct approach. Sorry about the confusion.
 
I took a different approach than what you were telling me to do. Since I had already found the tension on the string, I added that to the Force of gravity exerted on the cylinder and that turned out to be the force that axle was exerting.
 

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