# Physical pendulum oscillation

## Homework Statement

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50 cm mark. The period of oscillation is 2.5 s. Find d

## Homework Equations

T=2PI*sqrt(I/mgh)
I(com)=(1/2)mL^2
parallel axis theorem
I=I(com)+m(L/2-d)^2
L=length of stick
h=L/2=center of mass

## The Attempt at a Solution

so T=2.5 s
center of mass = .50 m
I=(1/2)mL^2 + m*(L/2-d)^2
then plug T, I, and h in for the equation
T=2PI*sqrt(I/mgh)
2.5 s=2PI*sqrt((3L^2/4-2Ld+d^2)/(gL/2))
however when I solve for d I do not get 5.6 cm..anyone see my mistake?

I just spent another 30 minutes working on the problem...and no luck :(

mezarashi
Homework Helper
I think your original 'relevant equations' are wrong.
The moment of inertia of a stick through its center of mass of total length L is: (mL^2)/12
Combining this with the parallel axis theorem, you should get through the point d cm from its center of mass:

I = (mL^2)/12 + md^2

Note that h in the period equation is actually d. It represents the distance from the center of mass. So if you actually swing it through the center of mass, guess what? It swings forever and if you divide by h approaching 0, the answer is a period approaching infinity.

Hope that helps.

ah! that helps a lot. Thank you so much. When I changed it to (1/12)mL^2 and used d instead it worked out great. I was forgetting about where it was rotating. Once again, thanks so much!

mezarashi
Homework Helper
You're welcome ^_^

Ya u got it, but in case u wonder how 2 get the equation for the period, here is my quick derivation:

$$\Sigma\tau=I*\ddot{\theta}=r\timesF=$$
$$-d*m*g*sin(\theta)=-d*m*g*\theta$$ we can approximate sin(theta) ~ theta for small theta

=>$$-d*m*g*\theta - I*\ddot{\theta} = 0$$

Solving Differential Equations

w = sqrt(d*m*g / I)

T = 2*pi/lamba =2*pi*sqrt(I / d*m*g)