Re: What is relation between mass and period in inertial balance?

Hi Class, I think I found the site to which you are refering:

quest.arc.nasa.gov/smore/teachers/microgravity/6inertial.html

I will describe an inertial balance in a bit. Basically, an inertial balance is a type of Simple Harmonic Oscillator.

Simple Harmonic Oscillators are fascinating to physicists. We spend a significant amount of our training learning the ins and outs or back and forths of SHO's. We also learn how to tell really bad jokes, too.

For a simple pendulum you a period:

Period = 2 * PI * (LENGTH / g ) ^1/2

where g is the acceleration due to gravity.

From a spring system typically you get terms like

Period is proportional to (MASS / K) ^1/2

where K is the restoring force constant of Hooke's Law

FORCE = - K * DISPLACEMENT.

The inertial balance is more like a spring system. However instead of MASS being the term the moment of inertia is important. In fact, the moment of inertia is the real term for all SHO's.
But for pendulums with light strings and spring systems where the mass of the spring is not important compared to the mass being oscillated using the MASS term works out very well.

FORCE = - K' * ANGULAR DISPLACEMENT.

where K' is the angular restoring constant of the rod. It has the units of force per radian or just force.

The moment of inertia of the rod fixed at one end will be

I_rod = 1/3 * M * L ² where M is the mass of the rod and L is the length.

If we add a mass m to the end the moment of inertia will be the sum of the rod and the added mass:

I_total = 1/3 * M * L² + m * L².

The period of oscillation can be calculated:

Period = (2 * PI * L) / K'^1/2 * (1/3 * M + m)^1/2.

So the
period of oscillation is proportional to the square root of the total moment of inertia.

But let's look at the special case in which the mass m is much smaller than 1/3 * M. In this case we can do a bit of algebra to come up with

Period = (2 * PI * L) *(M / K')^1/2 * (1 + 3 * m / M)^1/2.

A favorite trick of physicists is to expand terms of an equation by their Taylor Series Approximation. The expansion for (1 + X)^1/2 to the linear term in X is:

(1 + X)^1/2 = (1 + 1/2 * X - ...)

Using this approximation for (1 + 3 * m / M)^1/2 we obtain

Period = (2 * PI * L) *(M / K')^1/2 * (1 + 3/2 * m / M).

So the Period is given by the equation of a straight line when the mass m of the object to be measured is much much smaller than 1/3 the mass M of the rod.

If you look at the NASA webpage you can see the graph is of a straight line.

3 * m /M .............. Square Root ............. Approximation

0.01                1.00499               1.00500
0.05                1.02470               1.02500
0.10                1.04881               1.05000
0.30                1.14018               1.15000
0.50                1.22474               1.25000
0.80                1.34164               1.40000

Sincerely,

Tom "Approximately" Cull