# Help me understand centripetal acceleration?

## Homework Statement

A child swings a tennis ball attached to a .750 m-string in a horizontal circle above his head at a rate of 5.00 rev/s What is the centripetal acceleration of the tennis ball?

## Homework Equations

angular speed $\omega$ = 2pi/T
speed = r$\omega$

$a_c = -v^2 /r$
$a_c = -\omega ^2 r$

## The Attempt at a Solution

So its 5 rev/s, or .2s in 1 revolution. Using angular speed I get 10pi radians per second
Plugging in,

$a_c = -(10pi)^2 (.750 m) = -740 m/s^2$

But my books answer is 740 m/s^2.

I don't understand why my answer is negative, even though I used the correct formulas. Does anyone know why?

## Answers and Replies

haruspex
Homework Helper
Gold Member
2020 Award
I don't understand why my answer is negative
The minus sign is in the formula because the acceleration vector is in the opposite direction to the radius vector. But you are working with scalars, not vectors, so you can drop the minus sign and just say in words which way the acceleration is (if you need to specify it, which you probably do not here).
By the way, the answer is not quite right. The given distance is the length of the string, not the radius of the circle. Gravity should be taken into account.

By the way, the answer is not quite right. The given distance is the length of the string, not the radius of the circle. Gravity should be taken into account.

What does gravity have to do with the length of the string??

haruspex
Homework Helper
Gold Member
2020 Award
What does gravity have to do with the length of the string??
It does not alter the length of the string, but it does alter the radius of the circle.

It does not alter the length of the string, but it does alter the radius of the circle.

Oh I see. I got you I think I understand why. I'm not sure why but the question didn't state assume no force of gravity, but I think it was suppose to. This is in the 2d motion chapter in my first physics class, so I don't think they consider stuff like that yet.

haruspex