# Centripetal acceleration at the ends of a rotating rod?

## Homework Statement

Sorry, I don't have a picture, so I can only relate this in words, so I have to indirectly translate it to drawing:

So you have a 0.15 m rod titled at a 45 degree angle above a horizontal , like this: \

Now rotate it some distance away from a vertical axis. The bottom of the rod rotates around a large horizontal rotation axis, and the top rotates around a smaller rotation axis. The MIDDLE of the rod is rotating at 58.33 rev/sec and has a centripetal acceleration of 19600 m/s^2, and it is rotating 0.15 m away from the vertical axis. Now you want to find the acceleration at the top axis, and the acceleration at the bottom axis. The answers: it accelerates between, though I believe not at, 12740 and 26460 m/s^2 (i think for top&bottom respectively, but it could be switched, i forgot). I want to try to find how to GET to these answers.

## Homework Equations

a = (2*3.14*f)^2 * r = v^2/r, and f stands for frequency
angular velocity = 2(3.14)f
velocity = 2(3.14)f * r

## The Attempt at a Solution

So I assumed that the angular velocity would remain constant, while velocity will change. So the a for the top&bottom ends differ only by radius. To find their radii, I assumed that since the rod is titled upwards at 45 degrees, it is in a square, and 0.15 * sin (45) showed that the distance between the top and bottom of the rod was 0.1061m. The distance of the middle of the rod to the vertical axis was .15 - x. I found x to be half of 0.1061m when I drew it out, so 0.09695 is the radius of the top (smaller) horizontal rotation axis, and 0.09695 + 0.1061 = 0.20307 was the radius of the bottom axis.

However, when I plug these values in I get 13026 (top) and 27279 (bottom), so I made an error somewhere, possibly at the very beginning, though definitely in the trigonometric process.

I also assumed frequency was the same, which seems right but that could be the error. Also apparently, if frequency is the same, the radius of the top is 0.0948m, and the bottom is 0.19697m. How do I get to these values?!?!?!

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Delphi51
Homework Helper
Welcome to PF.
I worked it out using ma = F = 4π²mR/T², T = 1/58.33.
All very close to what you found.
Unless we are both misunderstanding the description, it looks like the given answer is wrong.

tiny-tim
Homework Helper
welcome to pf!

hi graphsome! welcome to pf! (have an omega: ω and try using the X2 tag just above the Reply box )
The MIDDLE of the rod is rotating at 58.33 rev/sec and has a centripetal acceleration of 19600 m/s^2, and it is rotating 0.15 m away from the vertical axis.
no, you must have misunderstood the diagram with that frequency and acceleration, r is less than 0.15 …

find it using a = ω2r (which is easier than a = v2/r) Delphi51
Homework Helper
r is less than 0.15
I used R = .09697; graphsome used R = .09695. What did you get?

tiny-tim
Homework Helper
Hi Delphi51! I got 0.146. Ah... that would make sense. 0.15m is indeed just an estimate of 0.14605, which is the true answer (which I did not write down as it was already estimated to be 0.15). This gives 0.0944144m, which still doesn't really makes sense, but it's much closer.

It's too late anyways though, I already submitted the problem online and the values I got were wrong. But i'd still like to know how to get the right answer.

Delphi51
Homework Helper
Very confusing!
has a centripetal acceleration of 19600 m/s^2, and it is rotating 0.15 m away from the vertical axis. Now you want to find the acceleration at the top axis, and the acceleration at the bottom axis.
So was the 0.15 m part of the question or not?