(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?


1 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.

  • $\begingroup$ When we talk about angular acceleration, do we mean the angle between the axis of rotation and the body's center of mass? Or do we mean the angle which the body is rotated around its center of mass? Because the angular acceleration around an arbitrary fixed point outside the body is different than the angular acceleration around the body's center of mass. $\endgroup$ Commented Jul 16, 2018 at 14:40
  • $\begingroup$ Depending on the context it could mean either. The system described in your question is a point mass rotating about the origin, so the angle $\theta$ conforms to the first definition. Alternatively, one can consider an extended rigid body, not a point mass, and sensibly define an angle of rotation about its centre of mass. If you want to describe the rotation of such a body about an axis outside the body, you'll need more than one angle, in general. In all cases, however, the total angular momentum $L$ can be defined, about any axis, and will obey $\dot{L}=\tau$, the externally applied torque. $\endgroup$
    – user197851
    Commented Jul 16, 2018 at 16:02
  • $\begingroup$ Actually, I need to clarify the point about a body rotating about a fixed point outside the body. If it rotates "rigidly", all the vectors in the body, as well as the vector from the axis to the centre of mass, rotate at the same angular velocity $\dot{\theta}$ and with the same angular acceleration $\ddot{\theta}$. Is this what you meant? If it may rotate independently about its centre of mass as well as rotating about the outside fixed point, then you need two angles to describe the motion. $\endgroup$
    – user197851
    Commented Jul 16, 2018 at 17:32

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