Explain why when you have two physical pendula

[daveed]

can someone explain why when you have two physical pendula, one of which has a weight 2/3 of the way down falls at the same rate as one without a weight but is the same length?

 

[ceptimus]

The kinetic energy gained by each small section of the bar is proportional to its speed squared, and the speed varies linearly with the distance from the pivot point.

So to calculate the total kinetic energy of the whole bar, you integrate a function of x^2 between zero (at the pivot) and L, where L is the length of the bar.

Integral of x^2 is 1/3 x^3

Set L = 1 and the integral becomes 1/3

So the kinetic energy of the pivoting uniform bar is the same as if all its mass were concentrated at 1/3 of the distance from the pivot point.

 

[Doc Al]

can someone explain why when you have two physical pendula, one of which has a weight 2/3 of the way down falls at the same rate as one without a weight but is the same length?

The rod with the weight at 2/3 L can be approximated as a simple pendulum of length 2/3 L. The rod without the added weight cannot and must be treated as a physical pendulum (a uniform rod).

The equation governing the pendulum is this:
$$\tau = I \alpha$$
For a simple pendulum of length L, for small angles:
$$mg L \theta = m L^2 \alpha$$
or:
$$\alpha = g/L \theta$$
The solution to this differential equation gives the period of the simple pendulum as:
$$T = 2 \pi \sqrt{L/g}$$

For the uniform rod, the torque and rotational inertia differ:
$$\tau = I \alpha$$
$$mg L/2 \theta = 1/3 m L^2 \alpha$$
$$\alpha = g/(2/3L) \theta$$
Which gives a period of:
$$T = 2 \pi \sqrt{(2/3)L/g}$$
Which is equivalent to a simple pendulum of length 2/3L.