Not an actual hw problem, but it might as well be

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AI Thread Summary
The discussion revolves around calculating the centripetal acceleration experienced at the edge of a 7200 rpm 2.5-inch hard drive. The user assumes a radius of 1.25 inches (0.03175 m) and converts the rotational speed into radians per second. Using the formula for centripetal acceleration, they calculate an acceleration of approximately 18049.53 m/s², which translates to about 1841.79 g. There is a sense of uncertainty about the accuracy of the calculations, but the user believes the approach is correct. The conversation highlights the application of physics principles to understand the forces acting on spinning hard drive components.
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Homework Statement


Basically I bought a new 7200 rpm hard drive and I was curious about it since it is a 2.5 inch notebook hard drive. What I was wondering was how many g's were pulled at the edge of the disk while it spun at 7200 rpm

Homework Equations



\frac{v^{2}}{r}

I'm just going to make up a number and say the radius is 1.25 inches on a 2.5 inch drive so let's say the radius of the spinning disk then is 0.03175 m

The Attempt at a Solution



so basically what I then did was
7200 rpm where 1 revolution is 2pi rad
7200*2pi rad is about 45238.9342 rad/min
divide by 60 to convert to seconds
753.982237 s^-1
then \omega^{2}*r becomes my centripetal acceleration, right?

Therefore assuming a radius of .03175 the edge of the disk is feeling an acceleration of 753.982237^2 * 0.03175 = 18049.5325 m/s^2

which would be roughly 1841.78903 g?

ugh I'm embarrassed that I"m not confident about this answer. Does everything look good?
 
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Looks good to me.
 
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