Moment of inertia of hollow cylinder, axis orthogonal to lengthby jds17 Tags: classical mechanics, moment of inertia 

#1
Jan2211, 08:21 AM

P: 3

Hi, I am working through the Feynman lectures on physics and trying to calculate the moment of inertia stated in the title.
(the taxis of rotation going through c.m., orthogonal to length). My approach is to slice the cylinder into thin rods along the length, using the parallel taxis theorem and the result for a rod. Unfortunately, I get as result: I = M ( L^2 / 12 + r^2 / 2). I.e. the last numerator comes out as 2 instead of 4, as stated in section 192. The corresponding expression comes from summing up dm sum( z_i ^ 2), where dm is the mass of a single rod and z_i the height of the rod's center of inertia. Perhaps my mistake lies in handling the 2dim slices as 3dim rods? 



#2
Jan2211, 08:48 AM

Sci Advisor
P: 2,470

I would slice the cylinder into rings instead. It makes integration far easier.




#3
Jan2211, 09:14 AM

Mentor
P: 40,907





#4
Jan2211, 09:38 AM

P: 3

Moment of inertia of hollow cylinder, axis orthogonal to length
@Doc Al: Thank you for your reply, I took the cylinder as a hollow one, and this seems to be my mistake. I will try
to do the calculation again for the solid cylinder as soon as I get back home. @K^2: thank you, too, but I wanted to find out what was wrong with my thinking instead of doing a different calculation. I will try yours, too, although it seemed more complicated when I first considered it 



#5
Jan2211, 02:18 PM

P: 3

Hi, everything turned out nicely, considering a partition into concentric hollow cylinders, adding their M.I.s (calculated as before) up and going to the limit gives the answer in table 192!



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