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?
top: Ftension= mv2/r - mg
bottom: Ftension= mv2/r + mg
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!