# Homework Help: Tension at the Top of the String

1. Jan 13, 2010

1. The problem statement, all variables and given/known data

If a ball is being swung in a vertical circle at the end of a light cord at a constant speed of 3.35m/s when the the radius of the vertical circle is 0.70m and the mass of the ball is 0.160kg calculate the tension in the cord at the top of the swing.

2. Relevant equations
fc=T+fg

3. The attempt at a solution
m=0.160kg v=3.26m/s r=0.7m
T=fc-fg fc=T+fg T= ( mv2/r) - mg
now what confuses me is that this equation is used for tension in the cord at the bottom of the circle and it gives me the same answer what can i do to get a differant answer for tension in the cord at the top of the cord

2. Jan 13, 2010

### tiny-tim

Welcome to PF!

(try using the X2 tag just above the Reply box )
I think you're confused because you're using fc, by which I assume you mean centripetal (or centrifugal?) force.

This is an F = ma equation, with all the forces on one side, and the acceleration on the other side.

What you are calling "fc" is not a force, it's the centripetal acceleration …

just remember that both T and centripetal acceleration are always inward, but fg can be either inward or outward.

3. Jan 13, 2010

Re: Welcome to PF!

fc is centripetal force and is given by centripetal force= mv2/r this question is supposed to be answered using T= mv2/r + mg but i am confused on how to execute it

4. Jan 13, 2010

### tiny-tim

No no no

v2/r is centripetal acceleration

multiply it by m, put it in F = ma, and it should be equal to the total forces (in the vertical direction), in this case T and ±mg.

(if your book says that it is always +mg, then your book is wrong)

You are confused because you are tyring to treat this as a problem with three forces … when you accept that there are only two forces, everything will become clearer.

5. Jan 13, 2010

there are example problems for this but they are on the downswing of the string they use the equation that I am using but its supposed to be used differantly for the upswinng I am still very confused:shy:

6. Jan 13, 2010

### tiny-tim

On the downswing, T and mg are in opposite directions, but on the upswing, T and mg are in the same direction.

So the first has ma = T - mg, the second has ma = T + mg.

7. Jan 13, 2010

so for my question it would be ma=T+mg and the ma is mv2/r so T=mv2/r -mg so (0.160kg)(3.26m/s)2/(0.7m) - (0.160kg)(-9.81m/s2) or do i use positive 9.81 for this example and do (0.160kg)(3.26m/s)2/(0.7m) - (0.160kg)(+9.81m/s2)

8. Jan 13, 2010

### tiny-tim

We always say g = 9.71 (positive), so the downswing has ma = T - mg, and the upswing has ma = T + mg.

9. Jan 13, 2010