Question

a popular trick of many physics teachers is to swing a pail of water around in a vertical circle fast enough so that the water doesn't spill out when the pail is upside down. If Mr. Lowell's arm is .60 meters long, what is the minimum speed with which he can swing the pail so that the water doesn't spill out at the top of the path?

Answer 1

Need centripetal acceleration v^2/r at least as great as gravity, g=9.8m/s^2. Set them equal. Then,

v^2 = r g = (0.6 m) (9.8 m/s^2) = 5.88 m^2/s^2

v = sqrt(5.88) = 2.4 m/s

Extra: v = 2 pi r / T, the circumference of the circle divided by the time for one revolution.

T = 2 pi r / 2.4 = 1.6 s per revolution. Move quickly, Mr. Lowell.

Answer 2

Thank you! but the answer that my physics teacher gave us was 2.6 m/s

Answer 3

You should challenge his answer! (because the above answer is correct ^_^)

Answer 4

v=2.4m/s is the correct answer that is.

Answer 5

but i don't know how to get that 2.4.

Answer 6

@AllTehMaffs how did you get 2.4?

Answer 7

So at the stop of the swing, where the bucket is completely upside down, there's only 1 force acting on the water - gravity!


And you also know that since it isn't falling out of the bucket, that it's experiencing a centripetal acceleration that's greater than or equal to the force of gravity (if it were less than gravity, the water would fall out). So

\[F_g≤ma_{centripetal}\]
\[\qquad F_g=mg\]
\[\qquad a_{centripetal}=\frac{v^2}{r}\]
\[mg≤m\frac{v^2}{r}\]
\[g≤\frac{v^2}{r}\]
Showing
\[v≥\sqrt{ \ gr}\]
The velocity has to be greater than or equal to the square root of gravity times the radius - in this case, the radius is the length of the arm (.6m), and we're looking for the lower bound of values (the smallest velocity), so
\[v=\sqrt{9.81m/s^2)(.6m)}\approx 2.4m/s\]

^_^