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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
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Eric L
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1 Answers1

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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

  • So its multiplied by the ratio of the mass divided by the length of the whole board? That was the big part that confused me. – Eric L Dec 04 '14 at 03:58
  • @EricLawson yes, I updated the answer to show why this is correct, hope it helps –  Dec 04 '14 at 04:04
  • Also, if i were to use

    $$ \tau_{net} = I \alpha $$

    Since

    $$ \omega_{f}^2 = \omega_i^2 + 2 \alpha \Delta \Theta $$

    $$ \tau_{net} = I \frac{\omega_f^2 - \omega_i^2}{2 \Delta \Theta} $$

    Would this be correct?

     – Eric L Dec 04 '14 at 04:07
  • it is correct assuming $\alpha$ is a constant (because the second equation assumes that) –  Dec 04 '14 at 04:12