0

I am required to find the moment of inertia of the lever for a project in physics. This is my attempt:

Please note that we have not been taught this yet in class so i have not been taught this officially yet.

The seperate radiuses are the distance from the fulcrum to the end of each side of the lever.

$L = 1.57m$

$r_1 = .97m$

$r_2 = .6m$

$M_{total} = 2.3 kg$

$$ I = \frac{M_{total}}{L}\int_{0}^{.97} x^2 dx + \frac{M_{total}}{L} \int_{0}^{.6} x^2 dx $$

$$ I = \frac{M_{total}}{L}[\frac{.97^3+.6^3}{3}] $$

$$ I = \frac{2.3(.97^3+.6^3)}{4.71} = \frac{165347}{300000} $$

But this seems way too easy? Am i doing something wrong?

Qmechanic
  • 220,844
Eric L
  • 127
  • 1
  • 7

1 Answers1

2

Yes, you did it good, and it is that easy. The expression is correct because you have: $I = \int x^2 dm = \int x^2 \rho dx = \rho \int x^2 dx= \frac{M_{total}}{L} \int x^2 dx $

You can always check if your calculation is correct by using a different method: by finding the moment of inertia for a standar shaped object form a table, and using the parallel axis theorem:

the parallel axis theorem can be used to determine the mass moment of inertia of a body about any axis, given the body's moment of inertia about a parallel axis through the object's center of mass and the perpendicular distance between the axes