Centripetal motion - swinging a bucket full of water

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SUMMARY

The discussion focuses on centripetal motion involving a bucket with a mass of 1.00 kg containing 2.00 kg of water, swung in a vertical circular path with a radius of 1.00 m. The key problem is determining the minimum angular speed required to prevent the water from falling out at the top of the path and calculating the force exerted on the hand at both the highest and lowest points of the swing. Participants emphasized the importance of understanding forces and acceleration, particularly using free body diagrams to visualize the forces acting on the bucket.

PREREQUISITES
  • Understanding of centripetal acceleration
  • Familiarity with Newton's Second Law
  • Ability to draw and interpret free body diagrams
  • Basic knowledge of angular motion and forces
NEXT STEPS
  • Calculate minimum angular speed for centripetal motion in vertical circles
  • Explore the concept of centripetal force in practical applications
  • Learn about free body diagram techniques for complex systems
  • Investigate the relationship between angular speed and forces in circular motion
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Students and educators in physics, mechanical engineers, and anyone interested in understanding the dynamics of circular motion and forces acting on objects in motion.

SA32
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I'm having trouble coming up with a strategy for this problem. Any help is appreciated!

"You are swinging a bucket (mass 1.00 kg) containing 2.00 kg of water at a constant angular speed in a vertical circular path (the path lies in a vertical plane) of radius 1.00 m.

a.) What is the minimum angular speed that the bucket must have in order that the water does not fall out of the bucket at the top of its path?
b.) Assuming that you are swinging the bucket at the angular speed found in (a), what is the force on your hand due to the bucket handle (give magnitude and direction) at i.) its highest point and ii.) its lowest point?"

A
Okay, so, first let me see if I understand correctly why the water does not fall out of the bucket at the top of the path at a certain speed...

At the top of the circular path, there is a downward force of the object's weight and a downward force of my hand on the bucket. If I swing the bucket with enough force so that the magnitude of the acceleration of the bucket toward the center of rotation is greater than the magnitude of the acceleration due to gravity, the water cannot fall out because it is not accelerating as quickly as the bucket?

Then if the acceleration of the bucket is faster than the acceleration of the water, the water will not fall out. Which means what I'm looking for is the speed of the bucket so that it is accelerating just fast enough so that any slower and its acceleration would be less than the acceleration of water.

Does that mean I'm looking for when the acceleration of the bucket is equal to the acceleration of the water? If so I am not certain how to express this mathematically?

B
Really lost with this one... to increase the acceleration of the bucket I need to increase the force of my hand on the bucket. Does this mean the magnitude of the force of the bucket on my hand also increases? Again, I'm confused about how to express this mathematically.

Thanks ahead of time for any insight.
 
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Does that mean I'm looking for when the acceleration of the bucket is equal to the acceleration of the water? If so I am not certain how to express this mathematically?

You are doing fine. Think in terms of forces, and draw a free body diagram of the bucket when it is at the highest point of the curve. If you do that, you will see the forces that need to be balanced in order to keep the water from falling down.

Really lost with this one... to increase the acceleration of the bucket I need to increase the force of my hand on the bucket. Does this mean the magnitude of the force of the bucket on my hand also increases? Again, I'm confused about how to express this mathematically.

A free body diagram will help here too. Draw one for the position of the bucket at the top of the vertical circle, (that would be the same as you used in A), and a second one for the bucket at the bottom of the circle.

If you do this, you will see that there are more forces pulling on your hand at the bottom then there are at the top. You should also be able to express this as a sum of forces, based on your diagram and of course, Newton's Second Law.

Dorothy
 
Figured it out, thanks!
 

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