A doubt in the rotational analogue of F=ma

[Hamiltonian]

If we have a cylinder rolling on the ground $\tau = I\alpha$ can only be applied about the point in contact with the surface(Instantaneous axis of rotation) and its CoM I don't see why this should be the case.

why can the equation $\tau = I\alpha$ only be applied about the axis passing through CoM or the IAOR?

I thought the answer to this can be found by seeing how the equation is derived I faced an issue here, I am able to derive for a point mass $m$ moving in a circular path of radii R under the application of a force F
$$\vec{R}\times \vec{F} = \vec{\tau} = \vec{R}\times M(\vec{a_t}) = \vec{R}\times M(\vec{R} \times \vec{\alpha}) = MR^2\vec{\alpha}$$
in this case $MR^2$ is the moment of inertia of the body about its CoM.

if say there is a disc fixed at its CoM and a Force F acts on its circumference we know that $\vec{\tau} = \vec{R}\times\vec{F}$
but how exactly do we get $\vec{\tau} = \frac{(MR^2)}{2}\vec{\alpha}$ I am unable to derive $\tau = I\alpha$ in this case I don't see why we can just replace the point particle with a rigid body and just directly account for it by replacing $MR^2$ with the I of the rigid body.

 

[A.T.]

Summary:: -derivation of $\tau = I\alpha$ for a rigid body.
-why can the equation $\tau = I\alpha$ only be applied about the axis passing through CoM or the IAOR.

You can compute different $I$ around different points. But the $\alpha$ around a point other than CoM includes the linear $a$ of the CoM (from $F=ma$). Thus the CoM is often chosen as the reference point in order to separate linear and angular dynamics.

 

[etotheipi]

I think this can be a confusing topic, so it might help to start building up a better framework for tackling rotational dynamics problems. A rigid body just consists of a system of particles $\mathcal{S}$ which satisfy$$\frac{d}{dt} |\boldsymbol{x}_i - \boldsymbol{x}_j| = 0 \quad \forall \quad \mathcal{P}_i, \mathcal{P}_j \in \mathcal{S}$$You can analyse the motion using a coordinate system of arbitrary origin and basis, $\mathcal{F} = \{\mathcal{O}; \{ \boldsymbol{e}_i \} \}$. You can write the angular momentum of $\mathcal{S}$ with respect to $\mathcal{F}$ as$$\boldsymbol{L} = \sum_{i \in \mathcal{S}} m_i \boldsymbol{x}_i \times \dot{\boldsymbol{x}}_i$$The time derivative is$$\begin{align*}
\frac{d\boldsymbol{L}}{dt} = \sum_{i \in \mathcal{S}} m_i \left[ \dot{\boldsymbol{x}_i} \times \dot{\boldsymbol{x}_i} + \boldsymbol{x}_i \times \ddot{\boldsymbol{x}_i} \right] = \sum_{i \in \mathcal{S}} \boldsymbol{x}_i \times m_i \ddot{\boldsymbol{x}}_i &= \sum_{i \in \mathcal{S}} \boldsymbol{x}_i \times \boldsymbol{F}_i \\

&\equiv \sum_{i \in \mathcal{S}} \boldsymbol{\tau}_i
\end{align*}$$where we defined $\boldsymbol{\tau}_i = \boldsymbol{x}_i \times \boldsymbol{F}_i$, with $\boldsymbol{F}_i$ being the net force on the $i$^th particle. In the limit of a continuous rigid body, you can just exchange $\sum m_i \rightarrow \int d^3 \boldsymbol{x} \rho(\boldsymbol{x})$, i.e.$$\boldsymbol{L} = \int_V d^3 \boldsymbol{x} \rho(\boldsymbol{x}) \boldsymbol{x} \times \dot{\boldsymbol{x}}$$It is also standard to consider a second body-fixed frame $\overline{\mathcal{F}} = \{\overline{\mathcal{O}}; \{ \overline{\boldsymbol{e}}_i \} \}$ which rotates with the body, whose basis vectors are time dependent rotations of the basis vectors of $\mathcal{F}$. The point $\overline{\mathcal{O}}$ can be any point on the rigid body, although it's often convenient to choose it to be the centre of mass. There exists a unique orthogonal matrix $R(t)$ with components $R_{ij}(t)$ such that $$\overline{\boldsymbol{e}_i} = R_{ij}(t) \boldsymbol{e}_j$$The body frame basis vectors can then be shown to have derivatives$$ \left( \frac{d\overline{\boldsymbol{e}}_i}{dt} \right)_{\mathcal{F}} = \dot{R}_{ij}(t) R_{kj}(t) \overline{\boldsymbol{e}}_k \equiv \boldsymbol{\omega} \times \overline{\boldsymbol{e}}_i$$where we now introduced the angular velocity, $\boldsymbol{\omega}$, of the frame $\overline{\mathcal{F}}$ with respect to $\mathcal{F}$ [N.B. here I used the $()_{\mathcal{F}}$ notation to emphasise that the derivative is measured with respect to the frame $\mathcal{F}$, i.e. treating the $\{ \boldsymbol{e}_i \}$ as constant]. We can now consider the the difference in the time derivative of an arbitrary vector$$\boldsymbol{a} = a_i \boldsymbol{e}_i = \overline{a}_i \overline{\boldsymbol{e}}_i$$as measured by the frames $\mathcal{F}$ and $\overline{\mathcal{F}}$. We can write$$\left(\frac{d\boldsymbol{a}}{dt} \right)_{\mathcal{F}} = \frac{da_i}{dt} \boldsymbol{e}_i = \frac{d\overline{a}_i}{dt} \overline{\boldsymbol{e}}_i + \overline{a}_i \frac{d\overline{\boldsymbol{e}_i}}{dt} =
\frac{d\overline{a}_i}{dt} \overline{\boldsymbol{e}}_i + \boldsymbol{\omega} \times \boldsymbol{a}$$However, the derivative with respect to the $\overline{\mathcal{F}}$ frame (where the $\{\overline{\boldsymbol{e}}_i \}$ are instead constant) is$$\left( \frac{d\boldsymbol{a}}{dt} \right)_{\bar{\mathcal{F}}} = \frac{d\overline{a}_i}{dt} \overline{\boldsymbol{e}}_i$$and hence we obtain the relationship between the derivatives of $\boldsymbol{a}$ measured by the two frames,$$\left(\frac{d\boldsymbol{a}}{dt} \right)_{\mathcal{F}} = \left( \frac{d\boldsymbol{a}}{dt} \right)_{\bar{\mathcal{F}}} + \boldsymbol{\omega} \times \boldsymbol{a}$$Armed with this, we can derive some equations pertaining to rigid body rotation. The position vectors of general point $\mathcal{P} \in \mathcal{S}$ in the rigid body with respect to the $\mathcal{F}$ and $\overline{\mathcal{F}}$ are related by$$\boldsymbol{x} = \overline{\boldsymbol{x}} + \boldsymbol{\xi}, \quad \boldsymbol{\xi} = \mathcal{O}' - \mathcal{O}$$and thus$$\left( \frac{d\boldsymbol{x}}{dt} \right)_{\mathcal{F}} = \left( \frac{d\overline{\boldsymbol{x}}}{dt} \right)_{\mathcal{F}} + \left( \frac{d\boldsymbol{\xi}}{dt} \right)_{\mathcal{F}} = \left( \frac{d\overline{\boldsymbol{x}}}{dt} \right)_{\bar{\mathcal{F}}} + \boldsymbol{\omega} \times
\overline{\boldsymbol{x}} + \left( \frac{d\boldsymbol{\xi}}{dt} \right)_{\mathcal{F}}$$We can define $\boldsymbol{v} \equiv \left( \frac{d{\boldsymbol{x}}}{dt} \right)_{\mathcal{F}}$ as the velocity of $\mathcal{P}$ with respect to $\mathcal{F}$, and $\overline{\boldsymbol{v}} \equiv \left( \frac{d\bar{\boldsymbol{x}}}{dt} \right)_{\bar{\mathcal{F}}}$ as the velocity of $\mathcal{P}$ with respect to $\overline{\mathcal{F}}$, and we can also define $\boldsymbol{\mathcal{V}} \equiv \left( \frac{d\boldsymbol{\xi}}{dt} \right)_{\mathcal{F}}$ as the velocity of $\mathcal{O}$ with respect to $\overline{\mathcal{O}}$. So, in more familiar notation,$$\boldsymbol{v} = \overline{\boldsymbol{v}} + \boldsymbol{\mathcal{V}} + \boldsymbol{\omega} \times \overline{\boldsymbol{x}}$$If the body is rigid, then by definition the velocity of $\mathcal{P} \in \mathcal{S}$ with respect to $\overline{\mathcal{F}}$ is $\overline{\boldsymbol{v}} = \boldsymbol{0}$, so we just get $\boldsymbol{v} = \boldsymbol{\mathcal{V}} + \boldsymbol{\omega} \times \overline{\boldsymbol{x}}$.

Now, we can define a rank 2 tensor $I$ (the moment of inertia at the point $\overline{\mathcal{O}}$), very often used with components written with respect to the body-fixed $\{ \overline{\boldsymbol{e}}_i \}$ basis, defined by$$\overline{I}_{ij} = \int_V d^3 \overline{\boldsymbol{x}} \rho(\overline{\boldsymbol{x}})((\overline{\boldsymbol{x}} \cdot \overline{\boldsymbol{x}}) \delta_{ij} - (\overline{\boldsymbol{x}})_i (\overline{\boldsymbol{x}})_j)$$You will often choose the body-fixed basis vectors to align with the principal axes of the rigid body in order to simplify this calculation (the MoI matrix becomes diagonal). Anyway, let's consider the expression $\overline{I}_{ij} \overline{\omega}_j$ for a rigid body undergoing rotation about a fixed point $\overline{\mathcal{O}}$,$$\overline{I}_{ij} \overline{\omega}_j = \int_V d^3 \overline{\boldsymbol{x}} \rho(\overline{\boldsymbol{x}})((\overline{\boldsymbol{x}} \cdot \overline{\boldsymbol{x}}) \overline{\omega}_i - (\overline{\boldsymbol{x}})_i (\overline{\boldsymbol{x}})_j \overline{\omega}_j) = \int_V d^3 \overline{\boldsymbol{x}} \rho(\overline{\boldsymbol{x}}) \left[ \overline{\boldsymbol{x}} \times (\boldsymbol{\omega} \times \overline{\boldsymbol{x}}) \right]_i = \overline{L}_i$$since we have $\dot{\overline{\boldsymbol{x}}} =\boldsymbol{\omega} \times \overline{\boldsymbol{x}}$. Be careful to remember here that the $\overline{\omega}_i$ and $\overline{L}_i$ are the components of angular velocity and angular momentum written with respect to the body fixed $\{ \overline{\boldsymbol{e}}_i \}$ basis, but measured with respect to the lab frame! In coordinate free notation, the angular momentum about $\overline{\mathcal{O}}$ of a rigid body undergoing rotation about this fixed point ($\boldsymbol{\mathcal{V}} = \boldsymbol{0}$) thus satisfies $\boldsymbol{L} = I\boldsymbol{\omega}$.

We need to prove one more theorem to answer your question, that the angular momentum of a rigid body undergoing general motion is the sum of the angular momentum in the centre of mass frame and the angular momentum of the centre of mass with respect to $\mathcal{F}$. This is not too difficult; let $\boldsymbol{\xi}$ be the position vector of the centre of mass, then$$\begin{align*}

\boldsymbol{L} &= \int_V d^3 \boldsymbol{x} \rho(\boldsymbol{x}) (\boldsymbol{\xi} + \overline{\boldsymbol{x}}) \times \frac{d}{dt} \left( \boldsymbol{\xi} + \overline{\boldsymbol{x}} \right) \\

&= \int_V d^3 \boldsymbol{x} \rho(\boldsymbol{x}) \left[ \boldsymbol{\xi} \times \frac{d}{dt} \boldsymbol{\xi} + \overline{\boldsymbol{x}} \times \frac{d}{dt} \overline{\boldsymbol{x}}\right] = \boldsymbol{L}_{\text{CM}} + \overline{\boldsymbol{L}}

\end{align*}$$where we used that $$\int_V d^3 \boldsymbol{x} \rho(\boldsymbol{x}) \overline{\boldsymbol{x}} = \boldsymbol{0}$$from the definition of the centre of mass. Hence, it's common to write something like, if $I$ is the moment of inertia tensor at the centre of mass, $$\boldsymbol{\tau} = \frac{d}{dt} \boldsymbol{L} = \frac{d}{dt} \left( I \boldsymbol{\omega} + M \boldsymbol{\xi} \times \boldsymbol{\mathcal{V}} \right)$$And you can, for instance, decompose $I\boldsymbol{\omega}$ with respect to the body-fixed basis$$\frac{d}{dt} (I \boldsymbol{\omega}) = \frac{d}{dt}(\overline{I}_{ij} \overline{\omega}_j \overline{\boldsymbol{e}}_i) = \frac{d}{dt} (\overline{L}_i) \overline{\boldsymbol{e}}_i + \overline{L}_i \boldsymbol{\omega} \times \overline{\boldsymbol{e}}_i$$With this theory now you can analyse whatever rigid body you like.

 

[kuruman]

In short, as long as you write $\tau=I~\alpha$, with the torque $\tau$ and the moment of inertia $I$ of the rigid body expressed about the same reference point, Newton's 2nd law for rotation will be valid and $\alpha$ will be about the same point.

This post is a repeat of what @A.T. said except that the importance of matching torque and moment of inertia reference points was not spelled out in post #2.