# Physical pendulum frequency

1. Nov 5, 2009

### ehrenfest

1. The problem statement, all variables and given/known data
A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of configuration I?
(B) (3/2)^1/2
(C) 6/5
(D) 3/2
(E) 5/3

2. Relevant equations

Angular Frequency of a physical pendulum of length L with a mass at its endpoint: sqrt(g/L)

"The center of mass of a system moves as if it were a single particle of mass equal to the total mass of the system, acted on by the total external force, and independent of the nature of the internal forces" (Marion and Thornton page 333)

3. The attempt at a solution

Why is it wrong to apply the quote from Marion and Thornton and say that this situation is equivalent to the situation of having a physical pendulum of length 3/4 L with both of the masses at the endpoint?

2. Nov 5, 2009

### clamtrox

The mass switch affects potential and kinetic energies in a different way. I guess M&T are describing a system that is in uniform motion (every particle moving with the same velocity).

3. Nov 5, 2009

### clamtrox

Also, if you know Lagrangian mechanics, deriving the correct answer is really straightforward: you just sum over two oscillators with different lengths.

4. Nov 5, 2009

### ehild

The formula you cited refers to the "mathematical pendulum". For a physical pendulum with mass m and moment of inertia I with respect to the axis of rotation, the angular frequency is

$$\omega = \sqrt{\frac{mgs} {I}}$$

where s is the distance of CM from the axis.

ehild

5. Nov 5, 2009

### ehrenfest

Marion and Thornton derive that principle for any system of particles that obeys the weak form of Newton's third law (that forces are equal and opposite). Is the weak form of Newton's third law violated here?