Minimum Period of Oscillation Disk

AI Thread Summary
To find the minimum period of oscillation for a disk, the pivot point must be located at a distance d from the center of mass, where d = R/sqrt(2). The discussion highlights confusion regarding the derivative and the treatment of constants like R^2. It is emphasized that the derivative should be set to zero to find critical points, and simplification is not necessary during this process. Additionally, the final answer should reflect the distance from the rim, which is R - d. Clarification on these points helps guide the solution effectively.
Airp
Messages
23
Reaction score
0

Homework Statement


how far from the rim of a disk of Radius R must the pivot point be located in order for its period of oscillation to be a minmum where R is the distance from the point to the centre of mass?

I'm stuck at the derivative because I saw a similar problem where the answer is T=R/sqrt(2) but if you look at where I am right now the derivative of the whole thing is a bit too big. So I think I might have made a mistake or I have no idea how to do a derivative at all (also I'm wondering what to do with R^2 since it is a constant) and I would like to have some input from other people as to if I'm on the right track, because I feel lost.

Any help would be greatly appreciated! :)

Homework Equations


2+d^{2}}{dg}}.gif

3. The Attempt at a Solution [/B]

IMG_20160209_160016.jpg
 

Attachments

  • IMG_20160209_160016.jpg
    IMG_20160209_160016.jpg
    29.9 KB · Views: 614
Physics news on Phys.org
Airp said:

Homework Statement


how far from the rim of a disk of Radius R must the pivot point be located in order for its period of oscillation to be a minmum where R is the distance from the point to the centre of mass?
I think you mean to say d is the distance from the point to the centre of mass (not R; that's the radius).
I'm stuck at the derivative because I saw a similar problem where the answer is T=R/sqrt(2)
I think you mean to say d = \frac{R}{\sqrt{2}}.
but if you look at where I am right now the derivative of the whole thing is a bit too big. So I think I might have made a mistake or I have no idea how to do a derivative at all (also I'm wondering what to do with R^2 since it is a constant) and I would like to have some input from other people as to if I'm on the right track, because I feel lost.

Any help would be greatly appreciated! :)

Homework Equations


2+d^{2}}{dg}}.gif

3. The Attempt at a Solution [/B]

View attachment 95586
You seem to be on the right track so far, except somewhere along the way you lost a value of "2" in one of your terms.

Try that again starting with
T = 2 \pi \sqrt{\frac{d}{g} + \frac{R^2}{2gd}}

and keep plugging away with the derivative of that.

There's no need to simplify too much along the way. Since you are setting the derivative equal to zero, some of the more scary parts go way quite quickly. What's left over is pretty manageable.

[Edit: Oh, and one last thing: don't forget that the problem isn't asking for d itself, but rather R - d, since it says "how far from the rim." But you can make that as your final step.]

[Another edit: I corrected an unintentional omission of 2 \pi in my equation.]
 
Last edited:
  • Like
Likes Airp
Thank You!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top