Circular Motion and centripetal acceleration

Click For Summary
SUMMARY

The discussion centers on the dynamics of a bucket of water swung vertically on a string, specifically analyzing the tension changes and centripetal acceleration. The minimum centripetal acceleration required to prevent the water from falling out is established as 9.81 m/s², with the tension being highest at the bottom of the swing, calculated as 2mg. Participants debate the feasibility of maintaining uniform circular motion, concluding that the tension and centripetal force must vary due to changes in speed and gravitational potential energy as the bucket moves through its path.

PREREQUISITES
  • Understanding of centripetal acceleration and its formula, ac = v²/r
  • Knowledge of forces acting on objects in circular motion
  • Familiarity with gravitational force and its impact on tension in strings
  • Basic principles of energy conservation, particularly potential and kinetic energy
NEXT STEPS
  • Study the relationship between centripetal force and gravitational force in vertical circular motion
  • Explore the concept of non-uniform circular motion and its implications on tension
  • Investigate the equations governing energy conservation in circular motion scenarios
  • Learn about the dynamics of pendulums and their comparison to circular motion
USEFUL FOR

Physics students, educators, and anyone interested in understanding the mechanics of circular motion and forces acting on objects in motion.

Jimmy87
Messages
692
Reaction score
19

Homework Statement


Q.1. Explain the changes in the tension of a piece of string which is being swung vertically with a bucket of water at the end. What would the minimum centripetal acceleration need to be for water to not fall out?

Homework Equations


ac = v^2/r

The Attempt at a Solution


The tension in any vertical path will always be greatest at the bottom as the weight force needs to be overcome as well as providing a net force towards the center which is required for circular motion. The centripetal acceleration would need to be 9.81 m/s^2, therefore, at the top gravity will provide all the centripetal force and no tension is required from the string.

From researching the bucket problem online, it seems to be that when the bucket is weightless at the top then the tension at the bottom would be equal to twice the weight of the bucket. However, this assumes uniform circular motion. Can this bucket problem ever be uniform circular motion? Can you ever have water in a bucket having a centripetal acceleration of 9.81 all the time? Surely as it comes down towards the ground the kinetic energy is greater therefore v^2/r cannot be constant?
 
Last edited:
Physics news on Phys.org
It would be better to separate the two problems in separate threads.

What is the equation for the centripetal force opposing the gravitational force mg?
 
Astronuc said:
It would be better to separate the two problems in separate threads.

What is the equation for the centripetal force opposing the gravitational force mg?

Wouldn't it be: Fc = 2mg since the centripetal force at the top is mg and the only way to get that at the bottom (when gravity is opposing the bucket) is to have 2mg.

Can a bucket be swung vertically so that the centripetal force is always mg? Is that physically possible?
 
Since the questions ask you to "explain" and "discuss", I'd say you're right on the money.
I think you've thought about the main issues.

One comment I'm not sure about: "this assumes uniform circular motion" - I don't see why?
At first I thought you were referring to the fact that when you swing the bucket in practice, your hand has to move and the string does not go to a fixed centre of a circular orbit, so maybe the orbit is not a circle? Then I decided you did not mean that and you were thinking the angular speed could not be constant.
As you suggest, I doubt that any such motion could have constant angular speed. I just think about the point when the string is horizontal.
 
What is the equation for static equilibrium? Think the condition of weightlessness?

Think about the forces on the mass (m) of water.
 
Merlin3189 said:
Since the questions ask you to "explain" and "discuss", I'd say you're right on the money.
I think you've thought about the main issues.

One comment I'm not sure about: "this assumes uniform circular motion" - I don't see why?
At first I thought you were referring to the fact that when you swing the bucket in practice, your hand has to move and the string does not go to a fixed centre of a circular orbit, so maybe the orbit is not a circle? Then I decided you did not mean that and you were thinking the angular speed could not be constant.
As you suggest, I doubt that any such motion could have constant angular speed. I just think about the point when the string is horizontal.

Thanks for your answer. What I mean by "this assumes uniform circular motion" is that when you read about the minimum speed to keep the water to stay in the bucket they show that the tension in the string is zero at the top and 2mg at the bottom. This therefore gives a centripetal force of mg at the top and bottom and hence uniform circular motion. They say that "if you swing the bucket such that ac is always 9.81 m/s^2..." but I don't see how you can keep a constant centripetal acceleration as surely it inevitably goes slower as it reaches the top. Therefore, surely it is wrong to say that the tension in the string is 2mg at the bottom and mg at the top because this implies a constant centripetal force and hence a uniform circular motion?
 
Ok. I think you are right. I had not followed through, from agreeing with you that v and ω are changing, to realising that as well as the tension in the string changing, the centripetal force must also change in the way you describe and so, again as you say, the tension at the bottom must be greater than 2mg.
And it is easily calculable since the only source of changed speed is the change in PE.
 

Similar threads

Replies
55
Views
3K
Replies
1
Views
812
Replies
19
Views
3K
  • · Replies 12 ·
Replies
12
Views
990
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K