What's the angular acceleration?

[kuruman]

@kuruman, I would like to know your point of view of whether we can calculate moment of inertia in the case of the original question, i.e. when the particle is moving in straight line, but the axis about which we need to calculate torque is some distance apart from particle's line of motion.

Why can't we have $I=mr^2$ in this particular case? It's just that the moment of inertia about the origin is not constant in time but there is nothing untoward about that. Watch this.

Start with the angular moment about the origin when the particle is at some arbitrary position $\vec r$ moving with arbitrary constant velocity $\vec v$. Then its angular momentum momentum about the origin is given by $\vec L= \vec r\times(m\vec v).$ Using results posted in #27, $\dfrac{d\vec r}{dt}=\vec v=\dot r~\hat r+r \dot \theta ~\hat \theta.$ Substitute,$$\vec L= r~\hat r\times[m(\dot r~\hat r+r \dot \theta ~\hat \theta)]=m[(r\dot r~(\hat r\times\hat r)+mr^2\dot \theta~(\hat r\times \hat \theta)]=mr^2 \dot\theta~\hat z.$$With the definitions $I=mr^2$ and $\dot \theta\hat z=\vec \omega$, you have the well known result, $\vec L = I\vec \omega$.

Note, that both $I$ and $\vec \omega$ change with respect to time, nevertheless their product is constant because angular momentum about the origin is conserved since the torque about the origin is zero.

On edit:
If there is a torque about the origin, then you can find an expression for it simply by using $$\vec \tau=\frac{d\vec L}{dt}=\frac{d}{dt}\left(mr^2 \dot\theta~\hat z\right).$$ The Cartesian unit vector is constant in time, so you only have to use the product rule on $r^2$ and $\dot\theta$.

 

[NTesla]

@kuruman , @PeroK, @haruspex,
As there is a general form of $\alpha =\ddot{\theta }+\frac{2\dot{r}\dot{\theta }}{r}$, Is there any general form for $\omega$ ? Or is it just $\frac{d\theta}{dt}$ ?

 

[PeroK]

@kuruman , @PeroK, @haruspex,
As there is a general form of $\alpha =\ddot{\theta }+\frac{2\dot{r}\dot{\theta }}{r}$, Is there any general form for $\omega$ ? Or is it just $\frac{d\theta}{dt}$ ?

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

 

[kuruman]

@kuruman , @PeroK, @haruspex,
As there is a general form of $\alpha =\ddot{\theta }+\frac{2\dot{r}\dot{\theta }}{r}$, Is there any general form for $\omega$ ? Or is it just $\frac{d\theta}{dt}$ ?

You can easily derive the expression. From what has been said above, $$\vec \omega=\frac{\vec L }{I}=\frac{\vec r \times (m \vec v)}{mr^2}=\frac{\vec r\times \vec v}{r^2}.$$I don't know if it qualifies as a "general" form, but the same expression is given in the Wikipedia link provided by @PeroK in post #33.