Re: An IITJEE problem
- From: mL <mL.beyond@xxxxxxxxxxxxx>
- Date: Sat, 29 Apr 2006 10:02:54 GMT
Hi again,
N:dlzc D:aol T:com (dlzc):
Dear mL:..
"mL" <mL.beyond@xxxxxxxxxxxxx> wrote in message news:33t4g.54794$d5.209097@xxxxxxxxxxxxxxxxxxHi David,
Dear mL:To satisfy the given answer, L has to be the distance
"mL" <mL.beyond@xxxxxxxxxxxxx> wrote in message news:Uvl4g.54699$d5.209196@xxxxxxxxxxxxxxxxxxN:dlzc D:aol T:com (dlzc):Thanks mL.. Counter rotating with A rotating clockwise,Dear Nishu:Try this figure:
"Nishu" <amitk_dni@xxxxxxxxxxx> wrote in message news:1146152710.843745.302370@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxI have a SHM problem."Simple Harmonic Motion"
A rod of length L is placed on two circular dics.The problem setup is lacking much.
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.
================================
(o) L (o)
A B
I assume the disks are acting like wheels, soIf the wheels, A and B, are *counter rotating*
there would be no SHM.
(driven at the same angular speed) the rod will
move (slide) to and fro.
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.
between the wheel axes (as shown in my figure).
L is given in the problem statement as the length of the rod. The wheel spacing is inconsequential, as long as d + x < L.
Nishu made a correction downthreads:
"Let me again remember you that the L is distance between
centre of wheels and may not be length of rod itself."
..As it turns out, the period doesn't depend on the angularThe period of oscillation is a function of wheel speed.Your formula gets trashed if you do not use ASCII.Seems to be correct.
Is this t = 2 . pi . sqrt( L / (2 . k . g) ) ?
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).
wheel speed.
It *must*. Lets say I set the wheels rotating at 1 revolution per day. What is the period of the oscillation? Physically, the period will be on the order of a day, and likely the rod will dampen to a stop without ever swinging to significant "negative" values of x, with those wheel speeds.
The rod is slipping against the wheels during the
entire cycle of motion - and, the velocity of the rod
varies continuously as it should do in every SHM.
Think about it like this, what are the units of sqrt( L / g ) (since 2 . k is dimensionless)?
To solve the problem ...
- draw a free body diagram of the *rod* to identify
the forces acting on it:
weight mg, normal forces N_A and N_B,
and friction forces kN_A and kN_B
- set up the required equations (e.g. two force eqns
and one torque eqn)
- eliminate redundant quantities (m, N_A, N_B), and
simplify to get the wanted eqn of motion:
x" + w^2 x = 0, with w^2 = 2k g/L
Thus, the period of motion is
T = 2pi/w = 2pi sqrt(L/(2k g))
Mel
.
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