Bucket Problem. Energy Method?

In summary, the bucket swings around in a vertical plane and passes through the top of the circle with a speed v1. At point 1 (top of the loop), the centripetal acceleration is mg and the tension in the rope is T1. At point 2 (bucket at same height as the center of the circle), the speed of the bucket is v2 and the centripetal acceleration is ac,2. At point 3 (bottom of the swing), the speed of the bucket is v3 and the centripetal acceleration is ac,3.f
  • #1
2
0

Homework Statement


Bob picks up a bucket of water by a rope and swings it around in a vertical plane. The bucket passes through the top of the circle with a speed v1.
(Note: You may neglect air resistance.)

a.)At point 1 (top of the loop):
How large is centripetal acceleration of the bucket? (Express as a multiple of g.)
ac,1 = · g
How much is the tension in the rope? (Express as a multiple of the bucket's weight.)
T1 = · mg

b.) At point 2 (bucket at same height as the center of the circle):
What is the speed of the bucket? (Express as a multiple of v1.)
v2 = · v1
How large is centripetal acceleration of the bucket?
ac,2 = · g
How much is the tension in the rope?
T2 = · mg

c.) Repeat when the bucket is at point 3 (bottom of the swing):
v3 = · v1
ac,3 = · g
T3 = · mg

Homework Equations



mg=(mv^2)/r


The Attempt at a Solution



I figured out the solutions to part a which is 1 and 0. But I'm having a hard time with part b and c. Top of the swing T=mv^2/r + mg. Left of swing T = mv^2/r. Bottom of swing T= mv^2/r -mg. Right?
 
  • #2
Welcome to PF, roarono.
I don't see how you got numerical answers for (a) when you don't know the radius or speed.
Top of the swing T=mv^2/r + mg. Left of swing T = mv^2/r. Bottom of swing T= mv^2/r -mg.
At the top, Fc = mv²/r and mg is part of this. The tension must provide the rest, so T = mv²/r - mg. At the bottom I would say
T = mv²/r + mg because the tension must provide the Fc as well as hold up the weight.
 
  • #3
Well, the problem is from webassign. And the way it wants me to put answers in is by multiples of the unit or multiples of gravity, mass, velocity, etc. The reason why I got 1 and 0 is because at the top of the swing, the centripetal acceleration is mg=(mv^2)/r which comes out to be a = g. So the answer is 1. Which will be 1 * g = g. That's how the website wants me to input the answers.
 
  • #4
at the top of the swing, the centripetal acceleration is mg=(mv^2)/r
I don't see how you get that. It will be equal to g for particular values of v and r, but not for all values of v and r. Does the question go on to say something about it being the minimum speed that avoids spilling the water? I do seem to be having trouble seeing things this week; I'm probably missing something!
 

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