# Homework Help: Centripetal acceleration question

1. Oct 22, 2008

### physics_geek

1. The problem statement, all variables and given/known data
Young David who slew Goliath experimented with slings before tackling the giant. He found that he could revolve a sling of length 0.800 m at the rate of 6.00 rev/s. If he increased the length to 1.100 m, he could revolve the sling only 4.00 times per second.
(a) What is the speed of the stone for each rate of rotation?
at 6.00 rev/s
at 4.00 rev/s

(b) What is the centripetal acceleration of the stone at 6.00 rev/s?

(c) What is the centripetal acceleration at 4.00 rev/s?

2. Relevant equations
a = v^2/r
T = 2(pi)r/v

3. The attempt at a solution
a = (6.00 rev/s)^2/.800 m = 45m/s^2

im not sure if im even doing this rite..how do i get the speed?
do i have to convert rev/s to anythin?

2. Oct 22, 2008

### LowlyPion

You are on the right track but you didn't treat the 2π factor correctly.

a = v2/r

v = w * R = 2πf * R

a = (2πf* R)2/R = 4π2*f2*R

3. Oct 22, 2008

### physics_geek

what do w and n stand for?

4. Oct 22, 2008

### LowlyPion

w is angular velocity little omega

And that's not N. Let me copy it in a different font.

a = v2/r

v = w * R = 2πf * R

a = (2πf * R)2/R = 4π2 * f2 * R

5. Oct 22, 2008

cool
thanks a lot