A bucket of mass 1.60 kg is whirled in a vertical circle of radius 1.00 m.

AI Thread Summary
The discussion revolves around a physics problem involving a bucket of mass 1.60 kg being whirled in a vertical circle with a radius of 1.00 m. The tension in the rope at the lowest point is 25.0 N, and the first part of the problem requires finding the speed of the bucket, which is determined to be 2.42 m/s. For the second part, participants clarify that to prevent the rope from going slack at the top of the circle, the speed must be calculated using Newton's second law, leading to a required speed of 3.13 m/s. The conversation emphasizes understanding the forces acting on the bucket at different points in its circular motion. Overall, the key takeaway is the application of Newton's laws to solve for speed in circular motion scenarios.
fineztpaki
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Homework Statement


A bucket of mass 1.60 kg is whirled in a vertical circle of radius 1.00 m. At the lowest point of its motion the tension in the rope supporting the bucket is 25.0 N.
(a) Find the speed of the bucket.
(b) How fast must the bucket move at the top of the circle so that the rope does not go slack?

The Attempt at a Solution


I figured out A, the speed of the bucket to be 2.42 m/s but i don't understand how to find B.. can someone help please?
 
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When the rope just goes slack what happens to its tension?
 
the tension decreases?
 
fineztpaki said:
the tension decreases?
To what value?
 
less than 25?
 
Is there any formula i would be able to use to solve this?
 
If there's any tension in the rope, it's not slack. The only formula you need is Newton's 2nd law.
 
How do I get speed, or velocity (m/s) from that?
The answer is supposed to be in m/s ... I'm still confused
 
You solve part (b) the same basic way you solved part (a). What forces act on the bucket? Apply Newton's 2nd law. The differences: The bucket is at the top instead of the bottom. (What does that change?) And the tension is different. (What must it be just as the rope goes slack?)

Once you set up your equation you solve for v just like you did in part (a).
 
  • #10
3.13
 
  • #11
alright got it! thanks!
 
  • #12
fineztpaki said:
3.13
Yep. In m/s.
 
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