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

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The force exerted by the axle on the cylinder can be determined using the principles of rotational dynamics and Newton's second law. The mass of the bucket is 15.7 kg, and the mass of the cylinder is 11.2 kg with a diameter of 0.250 m. The tension in the rope while the bucket falls is calculated to be 40.5 N, with the bucket impacting the water at a speed of 12.1 m/s after falling 10.1 m in 1.67 seconds. The net force on the bucket is derived from the gravitational force minus the tension, and the tension itself represents the force exerted on the axle.

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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)

mg - T= ma = F_{bucket}

Tr = \tau = I_{cyl}\alpha

\alpha = a_{bucket}r

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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