# Force Exerted Non-Uniform Circular Motion

• akgtdoskce
If down is positive and you asked for the force on the bucket exerted by the tension then it will be positive.

## Homework Statement

A bucket of water (10kg) is swung vertically. Radius is 1 m. It takes 1 second to spin the bucket in a full circle.
a. How much force has to be exerted at the top of its motion?
b. How much force must be exerted at the bottom?

## Homework Equations

vtop=sqrt(gr)
top: Ftension= mv2/r - mg
bottom: Ftension= mv2/r + mg
ac=v2/r

## The Attempt at a Solution

This is a fairly straightforward question, but I'm getting different answers. I'm not sure if this exerted force means the same thing as force of tension. If so, then I have force at the top = 0 N, force at the bottom is 588 N.
If v=sqrt (9.8) as the minimum velocity at the top, then conservation of energy gives 1/2mv2top+mg2r=1/2mv2bottom.
Thus, velocity at the bottom of the bucket's motion is 7 m/s. Using the equation F=mv2/r+mg, I get the 588 N answer.
However, when I initially completed this in class, I was told that 490 N was the correct answer for part b. Does this somehow have to do with the 1 second time provided in the original problem, such as assuming constant velocity? I've been thinking over this for quite a while and I just keep getting more confused.
Thanks for any help in advance!

akgtdoskce said:
I'm not sure if this exerted force means the same thing as force of tension.
Yes.

akgtdoskce said:
I have force at the top = 0 N
You are not told it is the minimum speed to maintain circular motion, so you've no basis for saying the tension is 0 there.

akgtdoskce said:
Does this somehow have to do with the 1 second time provided
Yes.
akgtdoskce said:
such as assuming constant velocity?
No, it won't be constant velocity.
Start again, but taking the tension at the top to be some unknown value. Derive everything else in terms of that, including the time to complete a swing (that's the hard bit). Then set that time equal to 1.

haruspex said:
Start again, but taking the tension at the top to be some unknown value. Derive everything else in terms of that, including the time to complete a swing (that's the hard bit). Then set that time equal to 1.
Okay, so if I use 1/2mv2top+mg2r=1/2mv^2bottom, and top: Ftension= mv2/r - mg bottom: Ftension= mv2/r + mg I get Tension at bottom = mv^2(top)/r+4mg/r+mg, which simplifies to Tbottom=Ttop + 6mg.

As for the time, I'm not really sure what to do with that. T=2*pi*r/v, but that is constant velocity.

I used a vertical circular motion simulator to help me visualize the situation, but it returns tension at the top as -296.78 N. Is this even possible?

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akgtdoskce said:
Tbottom=Ttop + 6mg.
Correct.
akgtdoskce said:
As for the time, I'm not really sure what to do with that.
Hint: SHM.
akgtdoskce said:
tension at the top as -296.78 N. Is this even possible?
Depends what sign convention was used. If up is positive and you asked for the force on the bucket exerted by the tension then it will be negative.