Find the minimum velocity for bucket in circular motion

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SUMMARY

The discussion focuses on calculating the minimum velocity required for a bucket of water to complete a vertical circular motion without spilling, specifically at the top of the circle with a radius of 1.00m. The key equation involved is the balance of forces, where the gravitational force must equal the centripetal force required to keep the water in the bucket. The participant suggests using the equation for centripetal acceleration, a = v²/r, and recognizes the importance of equating gravitational force (mg) with centripetal force (mv²/r) to find the minimum velocity.

PREREQUISITES
  • Understanding of centripetal force and acceleration
  • Familiarity with gravitational force equations
  • Basic knowledge of circular motion dynamics
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of centripetal acceleration formulas
  • Learn how to apply Newton's laws to circular motion problems
  • Explore examples of forces in vertical circular motion
  • Investigate the role of gravitational force in different scenarios
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and circular motion, as well as educators looking for practical examples to illustrate these concepts.

Tui
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Hey, first of all I'm not looking for the answer I just want a nudge in the right direction :)

Homework Statement


"a bucket of water is rotated in a vertical circle of radius 1.00m. What is the buckets minimum speed at the top of the circle if no water is to spill out?


Homework Equations


I'm not really sure where to start so.. :|


The Attempt at a Solution



I figure I must find the size of the force pulling the bucket downwards and equate that with the one pushing the water against the bucket? (I realize I've probably stuffed some force up here. this is confusing for me :))

If anyone could help with the equations I need then I can hopefully work this out on my own. Thanks a lot in advance
 
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Hey could it be that a(centrefugal or whatever :P) = v^2 / r

So I put -9.8 in for acceleration and work out v?
 
We need to pull the water to make it rotate or centripetal force
Since at the top the gravity is pulling the water down.

If mg=mv^/2, the resultant force is just to keep it rotating and not going anywhere else.
 

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