# Why does plate need an external torque to continue its rotational motion with constant angular velocity?

Consider the following question:

a-) By using the principal axis frame, calculate the torque needed to rotate a rectangular plate with sides a and b about its diagonal with a constant angular velocity $$ω_0$$.

(b) For the set-up in part (a), calculate the torque by using the Lagrangian formulation. Compare it with what you found in part (a).

I've solved the part a-) by directly using Euler's equation(s) $$\vec \tau = I \dot w + w \times (Iw);$$ since $$\dot w = 0$$ is given, we can direct find the torque needed, which turns out a nonzero in the $$\hat z$$ direction.

However, physically speaking, I don't understand why the plate needs an external torque to continue rotation with a constant angular velocity; the rotation axis passes through the CM, and once we provided the necessary torque to set $$\vec w = \vec w_0$$, the plate should not need any torque.

Question:

Why does plate need an external torque to continue its rotational motion with constant angular velocity ?

• I would suggest you to check what the angular momentum does when the body rotates as stated in the problem's text. I also have no intuition on why the angular momentum is not oriented like $\omega$ (important feature!) but perhaps with a bit of patience and looking at the trajectories of the particles which compose the object you can find it out. Note that for Euler's equation to give "counterintuitive" results $I$ must be a tensor. – JTS Apr 12 '19 at 16:40

## 1 Answer

It's simply because angular momentum is conserved, not angular velocity. In components, we have $$L_i = \sum_j I_{ij} \omega_j.$$ Since the plate is rotating, the $$I_{ij}$$ change over time; in the absence of external torques the $$L_i$$ are constant, but that doesn't mean the $$\omega_i$$ are.

I'm not sure what would make this intuitive to you, but a simpler case might be a barbell with masses on the ends. Picture it oriented diagonally but rotating about the vertical axis -- this should look quite unnatural, and impossible if the barbell is isolated.