# A doubt in the rotational analogue of F=ma

• Hamiltonian

#### 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.

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.

• Hamiltonian
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.

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• • Hamiltonian, berkeman, vanhees71 and 1 other person
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.

• Hamiltonian