# Kinetic energy of a rectangle rotating about its base [closed]

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?

• 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. Commented Dec 14, 2022 at 7:50
• @josephh see physics.stackexchange.com/questions/599461/… Commented Dec 14, 2022 at 7:53
• Okay. So the second equation you have used the parallel axis theorem, right? Commented Dec 14, 2022 at 7:57
• 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 . Commented Dec 14, 2022 at 8:01
• Right. That's your answer. Commented Dec 14, 2022 at 8:04

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.

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.