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Re: [Phys-l] "compound" pendulum



On 12/3/2010 9:14 PM, Stefan Jeglinski wrote:
I doubt "compound" is the correct description - if it has a proper
name I'd like to know what it is.

I'm picturing a pendulum that in its vertical position looks like this:

m2
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xx
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m1

The connecting rods are ideal (massless, infinitely stiff, no
friction, etc). The pivot axis (xx) is not in the center. Distance xx
to m2 is L2, distance xx to m1 is L1. I have derived the differential
equation describing its motion with a "torque" method and an "energy"
method and get the same result. In a nutshell, the "torque" method
sums the torques about the pivot axis and sets them equal to I*alpha.
The energy method is simply conservation of energy. In both cases the
answer I get is

theta'' + (C1/C2)*(g/L1)*sin(theta) = 0

where

C1 = 1 - m2*L2/m1*L1
C2 = 1 + m2*L2^2/m1*L1^2

This reduces correctly to the limiting cases of a simple pendulum
{m2=0,L2=0} or just a "balanced rod" that does not oscillate
{m1=m2,L1=L2}. Hopefully it is correct :-)

For pedagogical purposes, I'm trying to now solve the problem using a
sum of forces approach, but am having a problem conceptualizing and
also justifying it. For example, one approach is to consider the
free-body diagram for m1. The gravitational force in the
theta-direction on m1 is the usual: -m1*g*sin(theta).

After this I'm getting a bit sketchy.

I invoke the infinite stiffness of the rod to claim that the
gravitational force on m2 (not a restoring force as I've drawn it)
"transmits itself" to the other end of the rod so as to act against
the force of gravity on m1. In other words, because of m2, the net
force on m1 in the theta direction is really -m1*g*sin(theta) +
m2*(L2/L1)*g*sin(theta). This is really just invoking the torque
argument again and I think I'm pretty safe doing so in the limit of
an infinitely stiff rod. Then I set this net force equal to m1*a1
which is m1*L1*theta''. The result is missing the quantity C2:

theta'' + (C1)*(g/L1)*sin(theta) = 0

At this point I start arguing myself in circles. First I claim "wait
the forces have to be equal to something like (m1+m2)*Lcm*theta''"
then I start thinking that the sum of forces has to be a single one
acting through the center of mass but then why isn't a free-body
diagram for m1 not valid. And so on.

I can't see that I need 2 differential equations, 1 for m1 and 1 for
m2, because theta for both masses is the same, hence theta'' is the
same - besides, I could have done a free-body diagram for m2 instead
and tried to solve. I'm ignoring the forces along the length of the
pendulum for the same reasons as the simple pendulum (don't need to
find the tension in L1 and L2) - can't imagine I'm missing something
in that regard.

Am I missing the representation of some internal force? Was the
earlier torque or energy approach incorrect instead?


Stefan Jeglinski
Not sure how helpful it is to recite the canonical Starling & Woodall method, but here it is:

For the general case of a (compound) pendulum, the restoring force is equated to the angular acceleration times the moment of inertia. (in the Newtonian way)
Approximate the restoring force term in sin theta with theta for small deflections,
then theta = A sin 2 pi L/T
And so T = 2pi sqrt( I/mgL)

I is moment of inertia, A amplitude, L length, T periodic time.
For a simple case with all mass at length L from the pivot, the 'simple pendulum',
it is easy to find I which is m.L^2 leading to
T = 2.pi sqrt(L/g) in the usual way.

They then casually mention your case: where the moment of inertia I about the pivot is the sum of m.L^2 as before, PLUS the moment of inertia about the center of mass. This latter moment they call m.k^2 where k is the radius of gyration about
the center of mass, which leads to
T = 2.pi sqrt(( I^2 + k^2)/g.L)
They note this equation is quadratic in L, so there are two values of L which provide the same periodic time.
They confess that the terms L and k are not easily measured, and move quickly to note that when the pendulum is swung at this other L call it L2 which is close to the previous period, they show that
T = 2pi sqrt(( L1+ L2)/g)

Brian W