Physics question

[alexjrgreen]

A point mass m is attached to one end of a rod of negligible mass and length l. The rod is suspended by its other end from a frictionless pivot, so that it can swing freely in a vertical plane, and is lifted to the horizontal and released. What is the average downwards force exerted by the rod on the pivot?

[Nik]

FWIW, the time-average as described depends on how long it was hung on-load prior to release, and how long off-load after release before you halt the experiment...

[chrismb]

A point mass m is attached to one end of a rod of negligible mass and length l. The rod is suspended by its other end from a frictionless pivot, so that it can swing freely in a vertical plane, and is lifted to the horizontal and released. What is the average downwards force exerted by the rod on the pivot?

Is this your homework?

Presuming you mean arithmetic average wrt time you have to integrate mg.sin(theta).[pi/omega] of the angle of the rod to its starting position between theta = 0 and 180. The last bit deals with how long it spends at any given point in space and I may not be right but have gone on dimensional analysis and it must be something like that.

Or I could make it easy - the modal average wrt time is 'zero' at it spends more time at its slowest speed at the end of the swings (as it 'floats' through 'zero gravity'), so I'll stick with that answer as I can work that out easy.

[alexjrgreen]

Is this your homework?

It's quite a while since I had homework to mark, and even longer since I had homework to do.

I'll give you some help:

Let theta be the angle between the rod and the vertical.

The tension in the rod due to the weight of the mass is then

mg cos(theta) [1]

The kinetic energy of the mass as it falls must equal the loss in potential energy

mgh = mv^2/2

where h is the distance the mass has fallen = l cos(theta)

so mv^2 = 2mgl cos(theta)

The tension in the rod needed to keep the mass moving in a circular path is

mv^2/l

= 2mg cos(theta) [2]

so the total tension in the rod is [1] + [2]

= 3mg cos(theta)

Now all you have to do is work out the average downwards force on the pivot.

[tomclarke]

A point mass m is attached to one end of a rod of negligible mass and length l. The rod is suspended by its other end from a frictionless pivot, so that it can swing freely in a vertical plane, and is lifted to the horizontal and released. What is the average downwards force exerted by the rod on the pivot?

Asymptotically gm.

The proof is trivial, the downwards force on the pivot from the rod must average the downwards force on the mass from gravitation, since the mass position stays bounded.

The rod does not execute SHM, so the instantaneous force is a bit more complex - will be gcos(theta)+lmcos(theta)d2theta/dt2, taking the vertical component of the gravitational and centripetal force, where theta is angle to vertical.

Best wishes, Tom

[chrismb]

I'll give you some help:

Sorry?! I thought I was giving you the help.

What has this got to do with Polywell theory?

[chrismb]

= 3mg cos(theta)
Now all you have to do is work out the average downwards force on the pivot.

But where is the essence of how long it spends at any given point? You are averaging by angle, not by time, so this won't be the average force. Or, otherwise, as I said you can play tunes on what 'average' means as it is a very loose mathematical definition (one that the "there-is-no-GW sceptics" like to exploit).

[alexjrgreen]

What has this got to do with Polywell theory?

What has an apparently simple but actually hard to solve ordinary differential equation got to do with Polywell theory?

[alexjrgreen]

A point mass m is attached to one end of a rod of negligible mass and length l. The rod is suspended by its other end from a frictionless pivot, so that it can swing freely in a vertical plane, and is lifted to the horizontal and released. What is the average downwards force exerted by the rod on the pivot?

Asymptotically gm.

The proof is trivial, the downwards force on the pivot from the rod must average the downwards force on the mass from gravitation, since the mass position stays bounded.

Nice. The force varies from 0 to 3mg, but the average is mg.



(y axis units: mg Newtons, x axis units: 1/2 T seconds)

[chrismb]

What has the question "What has an apparently simple but actually hard to solve ordinary differential equation got to do with Polywell theory?" got to do with Polywell theory?

[Betruger]

It was asked on a Polywell board.

[chrismb]

It was asked on a Polywell board.

eh!?!!?!!?!!?!

Am I going daft? Did I just read that right?

So you're saying that anything mentioned on the Polywell forum has something to do with Polywell because it is mentioned on a Polywell forum?

[blaisepascal]

It was asked on a Polywell board.

eh!?!!?!!?!!?!

Am I going daft? Did I just read that right?

So you're saying that anything mentioned on the Polywell forum has something to do with Polywell because it is mentioned on a Polywell forum?

Well, that would explain what global climate change, anti-Obama politics, and any of another half-dozen issues which have taken over the General section of this forum have to do with polywell.

[KitemanSA]

It was asked on a Polywell board.

eh!?!!?!!?!!?!
Am I going daft? Did I just read that right?
So you're saying that anything mentioned on the Polywell forum has something to do with Polywell because it is mentioned on a Polywell forum?

Well, that would explain what global climate change, anti-Obama politics, and any of another half-dozen issues which have taken over the General section of this forum have to do with polywell.

Sorry, bad logic. The "General Forum" guidelines state:

This is the place for any "off topic" posts — discussion you want to have with fellow Talk-Polywell members, but which does not relate directly to polywell fusion.

Please feel free to introduce yourself, tell us a bit about your background, and share your hopes and dreams in this forum.

Such topics are appropriate for the General Forum.
The "Theory" forum is supposed to pertain to Polywell. But the question does indeed pertain to Polywell along the vein of: "How much AVERAGE force will the oscillating electrons place on the MaGrid?" Ok, it is a bit ofa stretch, but... there is nothing else to talk about theory wise right now!

[Aero]

But a Polywell can't be designed for average force. Rather, it must be designed for peak forces including a safety factor. Designing for average forces is like saying that you should be safe jumping out of a third story window. After all, your average deceleration would be less than one g, and that's very survivable.
But this belongs in the Design thread ...