Motion of a particle under a tangential force

[etotheipi]

Homework Statement

A particle in the x-y plane is subjected to a force $\vec{F}$ which is always perpendicular to the position vector $\vec{r}$. What's the equation of motion of the particle?

Relevant Equations

N/A

This is diverted from the Classical Physics forum. My first approach was this: the force $\vec{F}$ can be written in polar coordinates as $\vec{F} = F \hat{\theta}$. It follows that $F_{\theta} = ma_{\theta} \implies F = m(2\dot{r}\dot{\theta} + r\ddot{\theta})$.

This result also agrees with what we'd get if we analyse the motion in the rotating frame of the particle. The centrifugal force is $mr\dot{\theta}^2 \hat{r}$ away from the axis, the Coriolis force is $-2m\dot{r}\dot{\theta} \hat{\theta}$ and the Euler force is $-mr\ddot{\theta} \hat{\theta}$. So we again end up with $F = m(2\dot{r}\dot{\theta} + r\ddot{\theta})$ if we balance forces in the $\hat{\theta}$ direction.

However, now I try calculating the torque on the particle about the origin. This is $\tau_z = I_z \ddot{\theta} \implies Fr = mr^2 \ddot{\theta}$ or $F = mr\ddot{\theta}$. But this is different to the previous result!

I'm struggling to see why these two results differ. Any help would be appreciated!

 

[Lnewqban]

Why should the particle have some torque applied on it?

 

[etotheipi]

Why should the particle have some torque applied on it?

The force $\vec{F}$ constitutes a torque on the particle about the origin .

 

[etotheipi]

Actually, I realize my mistake. $\tau = I\alpha$ of course only holds for a fixed radial coordinate!

The full expression should be $\vec{\tau} = \frac{d\vec{L}}{dt} = m\vec{r} \times \ddot{\vec{r}} = mr\hat{r} \times (2\dot{r}\dot{\theta} + r\ddot{\theta})\hat{\theta} = mr(2\dot{r}\dot{\theta} + r\ddot{\theta})\hat{z} = Fr\hat{z}$, as expected! This reduces to the standard $I_z \alpha$ expression if $\dot{r}$ of all particles is zero.

I need to get more sleep