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General: Calculating the Moment of Inertia

  • Thread starter ckelly94
  • Start date
  • #1
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Hey, sorry for not using the exact template; I just have a general question about how to calculate the moment of inertia that I will have to apply to a number of instances.

In this particular case, I need to calculate the moment of inertia for a rod pendulum. Of course, I could just use T=2∏√(l/g) but I want to show how accurate you can get for this lab I'm doing (i.e. calculate the period as a simple pendulum as opposed to a physical one). Furthermore, I will have to eventually calculate the expected period for a pendulum with two masses at different positions. The screw by which the rod is attached to the apparatus is not at the exact end of the rod and I want to account for the space between the screw and the end of the rod. Lastly, I want to include the width of the rod (as it is hollow). We're ignoring differences in the density of the rod material if there are any.

I know the basic equation for calculating the moment of inertia for any object is going to be
I=Ʃmr2 and I believe I will have to consider the rod as a sum of infinitesimally small parts and add the inertia of the masses, but I A) don't know to how logistically do that and B) don't know how to include the factor of space between the screw and the end of the rod (unless you just consider the screw to be the center of mass and subtract the moment of inertia of any of the rod above the screw).

Would someone be able to work me through coming to a general formula for finding the moment of inertia for this kind of pendulum, considering the following factors:
- L=total length of the rod
- LST=length from the screw to the end (shortest side)
- LSB=length from the screw to the other end (bottom; this is the longest side)
- M1=mass of first weight
- M2=mass of second weight
- Mr=mass of the rod and the screw combined
- Dr=width of the rod
- DL=width of the lip of the rod
- Lw=height of the weights used

Of course, you can pick any other letter or symbols, I just figured that would be easiest.

Thanks in advance :)
 
Last edited:

Answers and Replies

  • #2
rock.freak667
Homework Helper
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Since your screw (or pivot) is not at the direct end, you can use the parallel axis theorem to the moment of inertia about the screw.

I = Icenter + md^2 where d is the distance from the center of the rod to the screw and Icenter is the moment of inertia about the centroidal axis.

Also depending on the cross section of the rod, you might actually be dealing with a rectangular prism instead of a circular cross-section.
 
  • #3
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Sure, but that doesn't take into account the differing positions of the masses. I considered each mass to be a cylinder, along with the rod itself and used the parallel axis theorem to yield I=[1/12 m_1 [3(〖r_w〗^2+r_i^2 )+h^2 ]+mr_l1^2 ]+[1/12 m_2^2 [3(r_w^2+r_i^2 )+h^2 ]+mr_l2^2 ]+[1/12 m_r [3(r_r^2+r_rl^2 )+l^2 ]+mr_l3^2]
where r_w is the total radius of the masses
r_i is the inner radius of the masses
h is the height of the masses
r_ln is the distance from the screw to the center of the nth mass
l is the length from the pivot point to the center of mass.
 
  • #4
rock.freak667
Homework Helper
6,230
31
Oh well I thought the weights were bobs and thus would be considered point masses so I for each would be md^2
 

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