What Distance x Gives the Least Period for a Physical Pendulum?

[ghetto_bird25]

Homework Statement

In Fig. 16-41, a stick of length L = 1.6 m oscillates as a physical pendulum. (a) What value of distance x between the stick's center of mass and its pivot point O gives the least period? (b) What is that least period?
http://edugen.wiley.com/edugen/courses/crs1141/art/qb/qu/c16/Fig15_46.gif

Homework Equations

well I'm assuming you have to use the equation for a a physical pendulum which is
T= 2pi $$\sqrt{I/mgh}$$
where I is the rotational inertia, m is the mass, g is gravity, and h is the distance between the axis of rotation and the centre of the pendulum.

The Attempt at a Solution

i tried using that equation but i don't understand the value for I because I am assuming if the value for h changes then the other end of the rod changes as well.

 

[azatkgz]

Moment of Inertia of the center of mass is

$$I_0=\frac{mL^2}{12}$$

If it is pivoted distance x from the centre

$$I=I_0+mx^2$$

$$T=2\pi\sqrt{\frac{I}{mgx}}=2\pi\sqrt{\frac{I_0+mx^2}{mgx}}$$

for T minimum we have
$$\frac{dT}{dx}=0$$

 

[ogb p]

I would like to clarify that the value of I in the equation for a physical pendulum is the moment of inertia, which is a measure of an object's resistance to rotational motion. It takes into account the distribution of mass around the axis of rotation. In this case, the moment of inertia for a thin rod rotating about one end (or pivot point) is given by I = (1/3)ML^2, where M is the mass of the rod and L is its length.

Now, to find the least period for the physical pendulum, we need to minimize the expression for T with respect to the variable x. This can be done by taking the derivative of T with respect to x and setting it equal to zero. After some algebraic manipulation, we get x = (2/3)L, which means that the distance between the center of mass and the pivot point should be 2/3 of the length of the rod.

Substituting this value of x into the equation for T, we get the least period to be T = 2pi \sqrt{(1/3)g}. This means that the least period for the physical pendulum is independent of the length of the rod and only depends on the acceleration due to gravity.

In conclusion, the distance between the center of mass and the pivot point that gives the least period for a physical pendulum is 2/3 of the length of the rod, and the least period itself is given by T = 2pi \sqrt{(1/3)g}.