Minimum Period of Oscillation Disk

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SUMMARY

The discussion focuses on determining the optimal pivot point distance from the rim of a disk with radius R to minimize its period of oscillation. The correct relationship is established as d = R/sqrt(2), where d is the distance from the center of mass. Participants emphasize the importance of correctly applying derivatives to the equation T = 2π√(d/g + R²/2gd) and highlight the need to find R - d, as the problem specifically asks for the distance from the rim.

PREREQUISITES
  • Understanding of oscillation mechanics
  • Knowledge of calculus, specifically derivatives
  • Familiarity with the formula for the period of oscillation
  • Concept of center of mass in rigid body dynamics
NEXT STEPS
  • Study the derivation of the period of oscillation for rigid bodies
  • Learn about the application of derivatives in optimization problems
  • Explore the concept of center of mass and its implications in physics
  • Investigate the effects of varying pivot points on oscillation periods
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Physics students, mechanical engineers, and anyone interested in the dynamics of oscillating systems will benefit from this discussion.

Airp
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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
 

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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.]
 
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