If I was to swing a bucket full of water (in a vertical plane) with a circle radius of 0.5m. What would be the smallest velocity that the bucket should have at the top of the circle if no water is to be split?
And @Pete W, I've replied to your comment on the previous question under comments :)
And @Pete W, I've replied to your comment on the previous question under comments :)
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This is a neat experiment and terrifying the first time you try it. You almost think that you better wear the best rain gear you can find. Actually it really doesn't take that much. You have two forces that must be in equilibrium. The first is the circular force (F = m*v^2/r) and the second is the gravitational pull F=mg
These two forces must be equal so
g = 9.81
r = 0.5 m
m is the mass of the bucket. M
Mg = M v^2/r The M's cancel
9.81 = v^2 / 0.5
v^2 = 4.905
v = 2.21 m/s which is not overly much.
These two forces must be equal so
g = 9.81
r = 0.5 m
m is the mass of the bucket. M
Mg = M v^2/r The M's cancel
9.81 = v^2 / 0.5
v^2 = 4.905
v = 2.21 m/s which is not overly much.