Acceleration of a speck around a circle

[Lori]

Homework Statement

a grinding wheel 0.5 m in diameter roates at a rate of 8.00 x 10^2 revolutions per minute. find the magnitude of the acceleration of a speck of metal cuaght in the outer edge of the wheel

Homework Equations

a=4pi^2r/T^2

[/B]

The Attempt at a Solution

I was wondering if i used the right equation?
To get T (time for one revolution) i divided 60 seconds by 800 revolutions .I plugged that into my equation for acceleration along with .25 m radius. The answer i get is 1750 m/s^2

 

[DaveC426913]

Can you show your work?
I think your T may be off.

 

[NFuller]

Yes the formula is correct.

 

[DaveC426913]

Yes the formula is correct.

OK, but is the answer?

 

[NFuller]

I think your T may be off. If 800 revs take a minute, how many revs in a second? It should be larger than one.

The period should be smaller than one.
$$T=\frac{60\text{s}}{800\text{rev}}=0.075\text{s}$$
Using this period I get the same answer as Lori.

 

[DaveC426913]

The period should be smaller than one.
$$T=\frac{60\text{s}}{800\text{rev}}=0.075\text{s}$$
Using this period I get the same answer as Lori.

Yeah. I mucked it up first time. Didn't edit it fast enough.

 

[Lori]

Yeah. I mucked it up first time. Didn't edit it fast enough.

i always mess up with the period too! I have to remember it's the time for one revolution in seconds