Circular motion of a bucket filled with water

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SUMMARY

The discussion centers on the dynamics of a bucket filled with water undergoing vertical circular motion. It establishes that a contact force exists between the water and the bucket at the top of the rotation due to the bucket providing centripetal force, which must exceed gravitational acceleration (g) for the water to remain inside. If the bottom of the bucket is removed while at the top, the water will exit tangentially, following the principles of centripetal acceleration. The critical equation derived is R = mv²/r - mg, indicating that a reaction force exists as long as the centripetal acceleration (v²/r) is greater than g.

PREREQUISITES
  • Understanding of centripetal force and acceleration
  • Familiarity with Newton's laws of motion
  • Basic knowledge of fluid dynamics
  • Ability to analyze forces in circular motion
NEXT STEPS
  • Study the principles of centripetal acceleration in detail
  • Learn about fluid pressure and its relation to forces in motion
  • Explore the dynamics of objects in vertical circular motion
  • Investigate the effects of varying tension in extensible strings on motion
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Physics students, educators, and anyone interested in the mechanics of circular motion and fluid dynamics will benefit from this discussion.

Faiq
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Consider a bucket filled with water in vertical circular motion.
Why does there exist a contact force between water and the bucket when the bucket is at the top of center of rotation?
What will happen if I remove the bottom part of the bucket when it is on the top?
 
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Faiq said:
Why does there exist a contact force between water and the bucket when the bucket is at the top of center of rotation?
Because the bucket accelerates downwards at more than 1g.
Faiq said:
What will happen if I remove the bottom part of the bucket when it is on the top?
What do you think?
 
The water will leave through the bottom and travel (as the theory always tells us) on a tangential path. This is exactly the same as if any mass were being swung on a string in vertical plane and going fast enough to be kept in a circle. Naturally, there are practicalities involved with actually getting all the water to leave at once and, in reality, there would be a 'fan' of water out of the bottom, each bit would be tangential to the path of the bucket.
There exists a contact force between bucket and water because the bucket is providing a centripetal force to constrain the water to move in a circular path. It has to be going fast enough to have a centripetal acceleration of at least g, for the bucket to 'overtake' the Earth's gravitational acceleration of g downwards and still be 'pressing inwards' on the water. At the bottom, the contact force will be at least twice the weight of the water, if the rotation is uniform.
 
A.T. said:
Because the bucket accelerates downwards at more than 1g.

What do you think?
How do you make up the assumption that acceleration is greater than g. Isn't it possible that the reaction force cancels some of the weight leaving the centripetal acceleration smaller than g?
 
Faiq said:
How do you make up the assumption that acceleration is greater than g.
Well, try to rotate it slower and report back.
 
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Faiq said:
Consider a bucket filled with water in vertical circular motion.
Why does there exist a contact force between water and the bucket when the bucket is at the top of center of rotation?
What will happen if I remove the bottom part of the bucket when it is on the top?
What do you get if you do a force balance on the water in the bucket?
 
Chestermiller said:
What do you get if you do a force balance on the water in the bucket?
R+mg = mv^2/r
 
Faiq said:
R+mg = mv^2/r
Excellent. So this is the equation that applies at the moment that the bucket is directly overhead. What can you conclude from this equation?
 
Many things can be concluded. What are we emphasizing on?
 
  • #10
Faiq said:
Many things can be concluded. What are we emphasizing on?
Conclusions related to the questions you asked.
 
  • #11
Well if you're hinting about the reaction force then we can conclude that
R = mv^2/r - mg
 
  • #12
Faiq said:
Well if you're hinting about the reaction force then we can conclude that
R = mv^2/r - mg
So what does that tell you?
 
  • #13
Well it tells me that as long as v^2/r > g, there will be a reaction force. If v^2/r = g then reaction force will be 0
 
  • #14
Faiq said:
Well it tells me that as long as v^2/r > g, there will be a reaction force. If v^2/r = g then reaction force will be 0
Excellent!

Now, if there were a small hole in the bottom, would water spray out if the inequality applied?
 
  • #15
No because centripetal acceleration is moving the water towards itself more as compared to gravity.
 
  • #16
Faiq said:
No because centripetal acceleration is moving the water towards itself more as compared to gravity.
But the reaction force tells you that there is fluid pressure at the bottom of the bucket. The pressure is equal to the reaction force divided by the cross sectional area of the bucket.
 
  • #17
How about when v^2/r = g. What will happen to reaction force then?
 
  • #18
Faiq said:
How about when v^2/r = g. What will happen to reaction force then?
It would be zero(as you said), and the fluid pressure at the bottom would be zero(gauge). So the water would be in free fall, and none would spray out the hole.
 
  • #19
Okay one more thing, if the tangential velocity is towards the right at top. The water must be pushed from the left side of the bucket. Wouldn't that give rise to a reaction from the left side?
 
  • #20
Faiq said:
Okay one more thing, if the tangential velocity is towards the right at top. The water must be pushed from the left side of the bucket. Wouldn't that give rise to a reaction from the left side?
Is the tangential velocity constant?
 
  • #21
Direction is changing so no
 
  • #22
Faiq said:
Direction is changing so no
Direction is changed by the centripetal force, not by tangential forces.
 
  • #23
Faiq said:
Direction is changing so no
So, if it is not accelerating tangentially, the sides of the bucket do not have to apply sideways force as the bucket passes overhead.
 
  • #24
Chestermiller said:
So, if it is not accelerating tangentially, the sides of the bucket do not have to apply sideways force as the bucket passes overhead.
Oh okay, got it thank you very much. If the bucket was connect through an extensible string, would there be an angular acceleration, if the motion was vertical? If no, can you give me a case where there is an angular acceleration?
 
  • #25
Faiq said:
If no, can you give me a case where there is an angular acceleration?
It's your scenario. You simply say that you accelerate it angularly with your arm.
 
  • #26
Faiq said:
Oh okay, got it thank you very much. If the bucket was connect through an extensible string, would there be an angular acceleration, if the motion was vertical? If no, can you give me a case where there is an angular acceleration?
It is not very clear what you mean by this.

Possibly you imagine a bucket in motion in a not-quite-circular path in a vertical plane. It is restrained by a bungee cord rather than by a string. The bungee stretches at the bottom portion of the path and retracts on the upper portion. But at the top and bottom of the arc, there is still no tangential acceleration -- the bungee is at right angles to the trajectory and is neither stretching further nor retracting further.

Possibly you imagine that a string is being reeled in (or payed out). Because of this, the path of the bucket as it passes overhead is not horizontal. There is an angular acceleration as a result.

Or you can use an inextensible string. Just move your hand in a circle that "leads" the bucket by 90 degrees so that the string tension has both a component radial to the bucket's path and a non-zero component that is tangential. (This hand motion is very natural - you do it automatically without even thinking about it). Now, as the bucket passes directly overhead, the string is still accelerating it.
 
  • #27
jbriggs444 said:
It is not very clear what you mean by this.

Possibly you imagine a bucket in motion in a not-quite-circular path in a vertical plane. It is restrained by a bungee cord rather than by a string. The bungee stretches at the bottom portion of the path and retracts on the upper portion. But at the top and bottom of the arc, there is still no tangential acceleration -- the bungee is at right angles to the trajectory and is neither stretching further nor retracting further.

Possibly you imagine that a string is being reeled in (or payed out). Because of this, the path of the bucket as it passes overhead is not horizontal. There is an angular acceleration as a result.

Or you can use an inextensible string. Just move your hand in a circle that "leads" the bucket by 90 degrees so that the string tension has both a component radial to the bucket's path and a non-zero component that is tangential. (This hand motion is very natural - you do it automatically without even thinking about it). Now, as the bucket passes directly overhead, the string is still accelerating it.
I was concerned with the angular acceleration. Not the tangential acceleration.
 
  • #28
When there is a vertical motion with an object connected to the centre by an extensible string, there are varying magnitudes of tension at each point because of the change in centripetal force. What I am asking is whether this situation could give rise to an angular acceleration?
 
  • #29
Faiq said:
I was concerned with the angular acceleration. Not the tangential acceleration.
Can you explain the distinction you see between the two? There is a distinction that can be made, but the relevance escapes me.
 
  • #30
Angular acceleration is change in angular velocity with respect to time. Tangential acceleration is change in linear velocity with respect to time.
 

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