Minimum Period of Oscillation Disk

In summary: I appreciate your help. In summary, the problem is asking for the minimum distance from the rim of a disk with radius R to the pivot point for the period of oscillation to be a minimum. The solution involves finding the derivative of the equation T = 2 pi sqrt((d/g) + (R^2/2gd)), setting it equal to zero, and solving for R - d.
  • #1
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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  • #2
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 [itex] d = \frac{R}{\sqrt{2}} [/itex].
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
[tex] T = 2 \pi \sqrt{\frac{d}{g} + \frac{R^2}{2gd}} [/tex]

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 [itex] 2 \pi [/itex] in my equation.]
 
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Related to Minimum Period of Oscillation Disk

What is the minimum period of oscillation disk?

The minimum period of oscillation disk is the shortest amount of time it takes for a disk to rotate or oscillate in a complete cycle. It is determined by the size and mass of the disk, as well as the forces acting upon it.

How is the minimum period of oscillation disk calculated?

The minimum period of oscillation disk can be calculated using the formula T = 2π√(I/mr²), where T is the period, I is the moment of inertia, m is the mass of the disk, and r is the radius of the disk.

Why is the minimum period of oscillation disk important?

The minimum period of oscillation disk is important because it can provide valuable information about the physical properties of the disk, such as its mass and moment of inertia. It is also important in understanding the behavior and stability of rotating systems.

How does the minimum period of oscillation disk differ from the natural frequency of a disk?

The minimum period of oscillation disk is the minimum time it takes for a disk to complete a full oscillation, while the natural frequency is the rate at which a disk naturally oscillates, without any external forces acting upon it. The minimum period is dependent on the mass and size of the disk, while the natural frequency is dependent on the material and shape of the disk.

Can the minimum period of oscillation disk be affected by external factors?

Yes, the minimum period of oscillation disk can be affected by external factors such as friction, air resistance, and external forces. These factors can change the moment of inertia and the forces acting on the disk, thus altering the minimum period of oscillation.

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