- #1

Rockdog

- 23

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A physics student whirls a small stone on a string in a horizontal circle. The circle has a radius of 1.4 m and a height of 2.2 m above the ground. Unfortunately, after she has whirled it for 33 seconds, the string breaks and the stone flys off, landing 12 from her (and just barely missing her physics teacher taking a nap near by).

question is:

What was the magnitude of the centripetal acceleration on the stone just before the string broke?

Well, I know that for centripetal acceleration, a=v^2/r

where a=acceleration

v=velocity

r= radius.

The other equation I know is T= (2(pi)r)/v

where T=period

r=radius

v=velocity

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At first I thought this was easy, but after half an hour of frustration, computer won't take any of my answers.

I know that one thing is find what is the velocity for the stone to land 12meters away, but I don't know how to do it. I tried making a triangle where the sides are 1.4m and 2.2m, and solve for the hypotheneuse, which gives me 2.6088 meters, but I don't know what that means really, if anything.

I tried manipulating the two forumulas to no avail, but no luck whatsoever. What am I not getting this time?