# Open Jar-Rotation Question

1. Feb 8, 2017

### Arman777

1. The problem statement, all variables and given/known data
An open jar of water moves in a vertical circle of radius $0.50m$ with a frequency that is small enough to put the water on the verge of falling out of the jar at the top of to the circle.If the same demonstration were repeated on Mars,where the $a_g$ is only $3.7 \frac {m} {s^2}$ ,what is the change in the circling frequency to again put the water on the verge of falling out at the top point.

2. Relevant equations
$a_r=\frac {v^2} {r}$
$v=2πfr$

3. The attempt at a solution
$a_r=\frac {v^2} {r}$
$v=2πfr$
$m\frac {v^2} {r}=mg$
from these
I found $0.27 Hz$, which its correct.But theres something that I confused.
$m\frac {v^2} {r}=mg$ how can these can be equal..weight must point downward and the other force is pointing inward...? I didnt quite understand the logic..I tried to understand but I couldnt

Thanks

2. Feb 8, 2017

### CWatters

The equation does not equate two different forces. The left hand side tells you the magnitude of the centripetal force required if the water is to move in a circle of radius r. The right hand side represents what actually provides that force at the top of the circle. If the right hand side of the equation was larger than the left the water would move in a circle that had a radius less than r (eg it would fall out).

3. Feb 8, 2017

### Arman777

Ohh,I see now Thanks..

4. Feb 8, 2017

### CWatters

So in order for the water to fall out...

mg > mv^2/r

In order for it to just stay in the jar..

mg = mv^2/r