Re: An IITJEE problem

Hi David,

Dear mL:

"mL" <mL.beyond@xxxxxxxxxxxxx> wrote in message news:Uvl4g.54699$d5.209196@xxxxxxxxxxxxxxxxxx
N:dlzc D:aol T:com (dlzc):
Dear Nishu:

"Nishu" <amitk_dni@xxxxxxxxxxx> wrote in message news:1146152710.843745.302370@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
I have a SHM problem.
"Simple Harmonic Motion"

A rod of length L is placed on two circular dics.
The co-efficient of friction between rod and disks
is k. The rod is displaced by a small distance x.
Determine the time period of the oscillations.

_______________________
O O <----disks
<-------------- L ----------------->


The answer is t = 2??(L/2kg) [g- acc due to gravity)
But I want the solution.
The problem setup is lacking much.
Try this figure:

================================
(o) L (o)
A B

I assume the disks are acting like wheels, so there would be no SHM.
If the wheels, A and B, are *counter rotating*
(driven at the same angular speed) the rod will
move (slide) to and fro.

Thanks mL.. Counter rotating with A rotating clockwise, and B rotating counterclockwise. L is given as the length of the rod. The distance between A and B needs to be greater than x (for any x) and less than L.

To satisfy the given answer, L has to be the distance
between the wheel axes (as shown in my figure).

Your formula gets trashed if you do not use ASCII.
Is this t = 2 . pi . sqrt( L / (2 . k . g) ) ?
Seems to be correct.

The period of oscillation is a function of wheel speed. So any result will have to be expressed in terms of that. From the result given above, it looks like it might be 1 revolution per second (2 . pi radians).

As it turns out, the period doesn't depend on the angular
wheel speed.

[...]

Mel
.

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