Mass Moment of Inertia - Thin - Cylinder

  • Thread starter Thread starter erobz
  • Start date Start date
erobz
Gold Member
Messages
4,442
Reaction score
1,839
Homework Statement
Apparently the Mass moment of inertia of a cylinder about the axis shown is ## I = \frac{1}{4}MR^2##
Relevant Equations
## \int_V \rho r^2 dm ##
1743632039598.png


The integral is a bit nasty.

$$ I = 4 w \rho \int_0^R x^2 \sqrt{R^2-x^2} dx $$

Just inquiring if I try it, I'm not chasing an approximation, or verify someone gets/has gotten the result.
 
Last edited:
Physics news on Phys.org
Never mind- I figured out how to confirm it with symbolic calculator. It does appear to be equivalent.

1743634759794.png
 
  • Like
Likes Lnewqban and berkeman
erobz said:
Homework Statement: Apparently the Mass moment of inertia of a cylinder about the axis shown is ## I = \frac{1}{4}MR^2##
Relevant Equations: ## \int_V \rho r^2 dm ##

View attachment 359392

The integral is a bit nasty.

$$ I = 4 w \rho \int_0^R x^2 \sqrt{R^2-x^2} dx $$

Just inquiring if I try it, I'm not chasing an approximation, or verify someone gets/has gotten the result.

Well you can substitute x for Rcos$ and it would become way easier , i got the final value after putting in the limits (πR^2)/16 . Thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top