Spinning a bucket in a vertical circle

AI Thread Summary
To keep the water in a 6.5 kg bucket swung in a vertical circle of radius 1.2 m, the bucket must reach a specific speed at the top of the circle to maintain the necessary centripetal force. The critical condition is that the gravitational force must be equal to or greater than the required centripetal force at that point. The equation f = (mV^2)/r can be used to derive the speed needed, but it’s essential to recognize that this is a condition rather than a law. The discussion emphasizes understanding the forces acting on the bucket, particularly at the top of the circle, where the risk of spilling is highest. Proper analysis of these forces will lead to the correct speed calculation.
proace360
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Homework Statement


You swing a 6.5 kg bucket of water in a vertical circle of radius 1.2 m.
(a) What speed must the bucket have if it is to complete the circle without spilling any water?

m=6.5
r=1.2


Homework Equations


maybe f=ma
f=m(v^2/r)

The Attempt at a Solution


What I did was set fc=6.5(V^2)/1.2, and since f=ma, I had a (or v/t)=V^2/1.2. I end up with the equation 1.2vt, which just brings me in circles. What can I do?
 
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proace360 said:

Homework Statement


You swing a 6.5 kg bucket of water in a vertical circle of radius 1.2 m.
(a) What speed must the bucket have if it is to complete the circle without spilling any water?

m=6.5
r=1.2


Homework Equations


maybe f=ma
f=m(v^2/r)

The Attempt at a Solution


What I did was set fc=6.5(V^2)/1.2, and since f=ma, I had a (or v/t)=V^2/1.2. I end up with the equation 1.2vt, which just brings me in circles. What can I do?

First of all think about where the greatest challenge to keep the water in the bucket will be.

What condition must be met for the water to stay in the bucket and not come out?
 
Remember that the equation f = (mV^2)/r is a condition that must be satisfied for centrifugal force; it is not a law.
 
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