# Tension and centripital force HELP!

warmfire540

## Homework Statement

A 2.0 kg ball is attached to a vertical post with two strings, one 2.0 m long and the other 1.0 m long as shown in the figure. If the ball is set whirling in a horizontal circle, what is the minimum speed necessary for the lower string to be taught? If the ball has a constant speed of 6 ms-1, find the tension on each string.

|\
| \
| \ -2.0m
| \
| \
|----0 -2.0kg
|
| ^1m

## The Attempt at a Solution

I don't exactly understand what this means, I've tried tackling it several ways.
Only thing I can figure is that when the string is taught it is at maximum centripetal force. However I'm not sure how to find this.
I have so far
F(centripetal)=mv^2/r
F=2v^2/1
F=2v^2

lzkelley
your equations are correct, but there is no maximum centripetal force in this case.
To find out when the bottom string becomes taught:
image that the bottom string isn't there (when its not taught, it effectively isn't - it doesn't do anything). As the ball speeds up, it will move further out (increasing radius) from the pole (i.e. the angle between the pole (vertical) and the top string will increase). When the radius of rotation reaches the length of the bottom string (1m), the bottom string will be taught.
Its probably easier to find this in terms of the angle afore-mentioned.

Aryan Classes
Draw free body diagram
total 3 forces acting:
1. gravitational
2: Tension 1 (upper)
3 tension 2 (lower)

The resultant of the three forces is providing the necessary centripetal force.
Tension must be greater then zero.
solve accordingly.

warmfire540
your equations are correct, but there is no maximum centripetal force in this case.
To find out when the bottom string becomes taught:
image that the bottom string isn't there (when its not taught, it effectively isn't - it doesn't do anything). As the ball speeds up, it will move further out (increasing radius) from the pole (i.e. the angle between the pole (vertical) and the top string will increase). When the radius of rotation reaches the length of the bottom string (1m), the bottom string will be taught.
Its probably easier to find this in terms of the angle afore-mentioned.

For the first part, figured that the 2m string has two components of tension Tsin60 and Tcos60.
Tsin60=mg=19.6
Tcos60=mv^2/r=2v^2
tan60=19.6/2v^2
v=5.66 m/s

is that right? I calculated the angle that the 2m string made w/ the x-axis and went from there...

Now..for the second part..
i figured that the tension in the 1m string is the centripital force which is:
T=mv^2/r=2*36/1= 72N
Now for the second string..what would the tension be?
i'm guessing i have to use the Tsin60 and the Tcos60
Tsin60=19.6N
Tcos60=72N
T=sqrt(19.6^2+72^2)
T=74.6N