Re: Big Ben...

["Edwin R. Schweber" <edschweb@ix.netcom.com>]

Delissa Souder wrote (in part):
>        I know that the period of a pendulum is dependent upon the length
> of the pendulum and gravity. (T=2 x pi x (sqr rt. (length/gravity)))  I
> was watching the Discovery channel and they were talking
> about some of the clocks and they showed the inside of Big Ben in
> London.  They showed the pendulum in Big Ben with a small stack of old
> pennies on it.  By adding or subtracting pennies on the bob of the
> pendulum, they could adjust the period of the pendulum.  How is this so?
> I've always thought the period of the pendulum is independent of its mass.
Dwight,

     The formula 2(pi) sq rt (l/g) is an idealization that assumes 
that the bob is a geometric point. When the formula is applied to 
a real pendulum we take l as the distance from the support to the 
center of mass of the bob. By adding pennies to the pendulum, Big Ben's
keepers are changing the location of the center of mass.

    If we look at the situation more rigorously, taking into account 
the actual dimensions of the pendulum, the formula for the period
 becomes

     T = 2 (pi) sq rt (I/mgh)

where I is the moment of inertia of the pendulum about the pivot and 
h is the distance of the center of mass of the pendulum from the
pivot. I has a factor of mass disguised inside of it, so the mass
will still cancel out. But since I also depends on the dimensions
of the pendulum, adding the pennies also changes I. Of course,
adding the pennies also changes h, but it has a greater effect on I, 
so this change , unlike the case for the mass, does not cancel.

    But having said this, it seems that you can only shorten the
period by adding the pennies. What would they do if the penniless
period was too short to begin with?

Ed Schweber (email: edschweb@ix.netcom.com)
Physics Teacher
Solomon Schechter Day School
West Orange, NJ