Torque around the origin of a particle using moment of inertia (in 2D)

Question

(You can skip this derivation and go down to my final question if you already are familiar with the results $(1)$ and $(2)'$ from this derivation)

Suppose we are in the xy-plane:

In two dimensions, you can define the torque $\tau$ as $$\tau = I_O \ddot{\theta}$$ where $I_O$ is the moment of inertia of a body around the origin $O$ and $\ddot{\theta}$ is the angular acceleration of the bodys center of mass around the origin.

According to the parallel axis theorem, the moment of inertia of a particle around the origin is $$I_O = d^2M$$ where $d$ is the distance between the origin and the particle and $M$ is the mass of the particle.

Now let's suppose that the position in polar coordinates of a particle of mass $M$, referenced from the origin, is given by the vector $$\mathbf{r} = r\hat r+\theta\hat\theta$$

The distance to the origin is simply $r$ so we get $$I_O = r^2M$$

so the torque is simply $$ (1):= \tau = Mr^2\ddot{\theta} $$

We calculate the torque again by using the known fact that $$(2):=\tau = Force \cdot Distance = Fr$$ where $F$ is the net force on the particle in the $\hat \theta$ direction, since the force component in the $\hat r$ direction doesn't contribute to the torque.

According the newtons second law, we can write the force $F$ as the product of the particles mass $M$ and its acceleration $a_{\theta}$ in the $\hat \theta$ direction $$(3):= F=Ma_{\theta}$$

The acceleration of the particle in the $\hat \theta$ direction can be found by taking two time derivations of its position vector $\mathbf{r}$, and then taking the dot product with the unit vector $\hat \theta$. Doing that we get $$(4):= a_{\theta}=\ddot{\vec{r}} \cdot \hat \theta = r\ddot{\theta} + 2\dot{r}\dot{\theta} $$

With $(4)$ we can write $(3)$ as $$F=M( r\ddot{\theta} + 2\dot{r}\dot{\theta})$$

and $(2)$ as $$(2)':=\tau= Mr( r\ddot{\theta} + 2\dot{r}\dot{\theta}) = Mr^2\ddot{\theta} + 2Mr\dot{r}\dot{\theta} $$

Now, let us finally compare $(1)$ and $(2)'$: $$ (1)= \tau = Mr^2\ddot{\theta} $$ $$ (2)'= \tau = Mr^2\ddot{\theta} + 2Mr\dot{r}\dot{\theta} $$

We see that $(1)$ and $(2)'$ can only be equal when $\dot r = \dot \theta = 0$, which is only true when the particle has no velocity.

What is going on here? Is $I_O = d^2M$ only valid when the particle has no velocity? What about a real rigid body of multiple particles, can the parallel axis theorem not be applied when the velocity is non-zero?

Answer 1

I think you're mixing equations valid for rigid bodies with those that apply to non-rigid ones. Torque is the rate of change of angular momentum. For your particle example, the angular momentum is changing both because of the angular acceleration and because the moment of inertia is changing. So: \begin{align*} \text{angular momentum} = L &= Mr^2\dot{\theta} \\ \text{torque} = \dot{L} &= 2M r \dot{r}\dot{\theta} + Mr^2\ddot{\theta} \end{align*} If the particle were constrained at a fixed distance from the origin, of course there would be no extra term.