Question

A physics teacher swings a pail of water in a vertical circle 5.0 ft. in a radius at constant speed. (a) What is the speed of the pail of water? (b) What is the maximum time per revolution if the water is not to spill?

Answer 1

hey.. i know a physics teacher who did this

here is the link
http://www.youtube.com/watch?v=hMDiyAEG7L4

oh wait.. thats me :D :D :D!!

Answer 2

@Mashy I totally thought you were going to clip that little ledge to your right at first! X.x



1) The forces on the bucket at the top of the loop (if it's going *just fast enough to keep the water in the bucket are
\[\sum F_y = mg=ma_{centripetal}\]
\[g=a_{centripetal}\]
\[g=\frac{v^2}{r}\]
With that you can solve for the velocity

2) The time per revolution has a name, called its period , T. In terms of frequency, T can be expressed as

\[ T=\frac{1}{f}\]

Which just says that if something is rotating at some frequency (revolutions per second), then to find how long it takes to make 1 whole revolution (its period, T) is just the inverse of that (T is in units of seconds per revolution).

The frequency can be found from its angular velocity

\[f=\frac{\omega}{2 \pi}\]

Saying that to find how many whole revolutions per second something is going is how many radians per second it's going divided by the number of radians in a revolution (there are 2 pi radians in one revolution).

So lastly, its angular velocity is given by

\[ \omega = \frac{v}{r}\]

Saying that the number of radians per second something is traveling is its tangential velocity divided by the distance from the axis of rotation (so, since the bucket is spinning in a circle, the bucket's distance from the axis of rotation is the radius of the circle [the circle's center is the axis of rotation]).


The last caveat of this problem is saying "what's the *maximum time per revolution." Looking at our original expression for velocity,
\[g=\frac{v^2}{r}\]
In order for the water not to fall out of the bucket, its centripetal acceleration has to be *greater than or equal to* the gravity
\[g≤\frac{v^2}{r}\]
Following the inequality up through its period shows us that the maximum period is given by the minimum velocity.

\[v≥\sqrt{ \ gr}\]
\[ \frac{v}{r}≥\frac{\sqrt{ \ gr}}{r}\]
\[ \omega ≥ \sqrt{ \frac{gr}{r^2}} \quad \quad \longrightarrow \quad \omega ≥ \sqrt{ \frac{g}{r}}\]
\[\frac{\omega}{2\pi} ≥\frac{1}{2\pi} \sqrt{ \frac{g}{r}}\]
\[f≥\frac{1}{2\pi} \sqrt{ \frac{g}{r}}\]
\[ \frac{1}{f} ≤2\pi \sqrt{\frac{r}{g}}\]
notice the flip in the inequality here because we're taking the reciprocal - this leaves us with

\[T≤2\pi \sqrt{\frac{r}{g}}\]
Showing that it's the maximum value of T we're looking for when we find the minimum velocity!