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

[lampshade]

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?

 

[Doc Al]

Looks good to me.

 

[m2003]

Hello! First of all, it's great that you're curious about your new hard drive and wanting to understand the physics behind it. Your calculations and approach seem to be correct, and your final answer of roughly 1842 g does seem to be in the range of what is expected for a 7200 rpm hard drive. However, it's important to note that this is just an estimate and the actual g-force experienced by the edge of the disk could vary depending on factors such as the design and materials of the drive. Additionally, g-force is not the only factor that determines the performance of a hard drive, as there are other factors such as data transfer rates and seek times that also play a role. Overall, it's great that you're thinking critically about your new hard drive and its capabilities. Keep exploring and learning!