How do I find the Moment of Inertia?

user_name12

1. The problem statement, all variables and given/known data

In this experiment a physical pendulum consisting of a rod and two large rubber stoppers (on top of each other) is swung from a point 1 mm from the end point. Calculate the moment of inertia of the pendulum about the point of suspension?

Data:
m=mass of rod=97.1 g
M=mass of the stoppers and the bolt=200 g
L=1 meter=length of rod
d=distance between the center of mass (CM) of the rod and the point of suspension=499 mm
D=distance between the center of mass of the stoppers and the Point of suspension=879 mm
R=distance between the Center of Mass of the rod-stopper system and the point of suspension=689 mm
theta (max)=the max angular displacement of the system from equilibrium (i.e. the max angular amplitude). Remember, theta (max) should be less than about pi/12 radian for the motion to be simple harmonic)=90 degrees
T(sub e)=the experimental period of oscillation of the pendulum. (measure and record at least three sets of 10 complete oscillations)=17.39 seconds/10 oscillations.
r=the distance from the center of the ruler to the point of rotation (.499 m)





2. Relevant equations
Note: I was sick when my teacher taught this so I learned the following from a friend.
use parallel axis theorem+MOI of point masses


(I)totalrod = (I)ruler + (I)mass

(I)mass= MR^2 where R = the distance from the center of the point mass to the rotation

and M is the mass for the point mass (200 g)

(I)ruler is the parralel axis theorem which is like

(I)ruler= (I)centerofmass + Md^2

where (I)center of mass is equal to 1/3Mr^2 where r is the distance from the center of the ruler to the point of rotation (.5 meters) + Md^2...

3. The attempt at a solution
I was sick...I dunno Sorry!
Help..eh :)
Thanks so much! This is due tomorrow so Urgency is somewhat required LOL...I wish I was smart

rock.freak667

Well you know that I_{point mass}=mr², so you can find the I for the masses about the point of suspension (POS).

For a rod, about its center, the moment of inertia is 1/12ML².

Now the parallel axis theorem states that

I_POS=I_C+mR_POS²

You should be able to get the I for the rod about the POS now.

Now that both of the 'I's are about the same axis, the total I about that axis, is just the sum of the individual 'I' values.