In which situations does the equation for torque t=Iα hold?

[Leo Liu]

The equation $\|\vec\tau\|=I\ddot\theta$ offers a way to calculate the torque acting upon a body which rotates about a fixed axle by using inertia and angular acceleration. This equation can be derived from $\|\vec L\|=I\dot\theta$.

What I would like to know are the requirements for applying this equation. Thank you.

 

[PeroK]

Try reading:

https://en.wikipedia.org/wiki/Angular_acceleration

 

[etotheipi]

I don't think your equation as written is correct, since $\ddot{\theta}$ can certainly be negative. Better to write something like $\tau_z = I \ddot{\theta}_z$ or simply $\tau = I \ddot{\theta}$.

Anyway, suppose a body is undergoing rotation about some fixed point $\mathcal{O}$ (this causes no loss in generality, since any infinitesimal motion can be described as the composition of a translation and a rotation, and by extension any motion can be described by the position of some reference point and the rotation around that reference point).

The moment of inertia tensor at the point $\mathcal{O}$, which we might call $I_{\mathcal{O}}$, is defined by an integral $(I_{\mathcal{O}})_{ij} = \int dV \rho(x_k x_k \delta _{ij} - x_i x_j)$, N.B. summation over repeated indices, over the volume $\Omega$ which comprises the body. The $\{ x_i \}$ coordinates are relative to the origin $\mathcal{O}$. Then$$L_i = (I_{\mathcal{O}})_{ij} \omega_j$$Now if the moment of inertia is constant during the motion,$$\tau_i = (I_{\mathcal{O}})_{ij} \dot{\omega}_j \equiv (I_{\mathcal{O}})_{ij} \alpha_j$$Furthermore, if the angular acceleration is constrained to only one direction, e.g. perhaps $\boldsymbol{\alpha} = \alpha \boldsymbol{z} = \ddot{\theta}_z \boldsymbol{z}$, then all the terms in the sum on the RHS drop out except for one, in which case$$\tau_i = (I_{\mathcal{O}})_{iz} \alpha_z$$If you chose your coordinate system to coincide with the principal axes, then $a\neq b \implies I_{ab} = 0$ (i.e. the moment of inertia tensor has diagonal matrix representation), and simply you have$$\tau_z = (I_{\mathcal{O}})_{zz} \alpha_z, \quad \quad \tau_x = \tau_y = 0$$In cases like this it is conventional to rewrite something like $(I_{\mathcal{O}})_{zz} \equiv (I_{\mathcal{O}})_{z}$.

 

[kuruman]

To answer the question posed in the title of the thread, it holds in situations where the moment of inertia does not have a time dependence. More generally, $$\vec \tau=\frac{d\vec L}{dt}.$$

 

[vanhees71]

If $I$ in the OP is meant to be just a scalar this holds for the rotation of a rigid body around a fixed axis. $I$ is then the moment of inertia of the body around this fixed axis.

In general for a rigid body you have a tensor of inertia, explained nicely in ##3, describing the rotation of the body around a point fixed in the body.