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Take a uniform rectangle with base $b$ and height $h$ and let it rotate about $b$. The kinetic energy is:

$$ T = \dfrac{1}{2} I_b \omega^2 = \dfrac{1}{2} \dfrac{bh^3}{3} \omega^2 $$ (with $I_b$ being the moment of inertia about the axis formed by its base)

Here is the problem:

When I try to calculate the kinetic energy via the center of mass I get a different result:

$$ T = \dfrac{1}{2}m v^2_{cm} + \dfrac{1}{2} I_{cm} \omega^2 = \dfrac{1}{2} ( \dfrac{h^2}{4} + \dfrac{bh^3}{12}) \omega^2 $$

(The CM being at a height of $\dfrac{h}{2}$)

What is the mistake here ? It seems that I'm missing an factor of $bh$ in the $v_{cm}$ term?

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  • $\begingroup$ yes I see the difference, but the first one is purely rotational since the pivot is the base, the second one needs to have an additional $v_{cm}$ term since the pivot is about the CM. $\endgroup$
    – lohey
    Commented Dec 14, 2022 at 7:50
  • $\begingroup$ @josephh see physics.stackexchange.com/questions/599461/… $\endgroup$
    – lohey
    Commented Dec 14, 2022 at 7:53
  • $\begingroup$ Okay. So the second equation you have used the parallel axis theorem, right? $\endgroup$
    – joseph h
    Commented Dec 14, 2022 at 7:57
  • $\begingroup$ I've taken $I_b = \dfrac{bh^3}{3}$ and $I_{cm} = \dfrac{bh^3}{12}$ as given. You could get $I_b$ from $I_{cm}$ via the parallel axis theorem . $\endgroup$
    – lohey
    Commented Dec 14, 2022 at 8:01
  • $\begingroup$ Right. That's your answer. $\endgroup$
    – joseph h
    Commented Dec 14, 2022 at 8:04

2 Answers 2

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The moment of inertia of an uniformly dense rectangle ($ \rho(x,y) = \rho $) about its base is in fact:

$$ I_b = \rho \dfrac{bh^3}{3} = \underbrace{\rho bh}_\text{ $= \rho A = m$ } \dfrac{h^2}{3} $$

So the $I_b$ in my original question assumes $\rho = 1$ !

(my bad here ,I blindly copied the $I_b$ online without thinking)

Also there is the error in the second $T$ (a missing $m$) which with $ m = \rho bh $ would make the two kinetic energies the same.

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The mistake is your dropping m in your second equation and you are evidently using

$m=b h$ in your first equation, so you had better use the same value of m in your second equation.

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