A doubt in the rotational analogue of F=ma

In summary, the conversation covers the derivation of the moment of inertia equation for a rigid body, as well as the significance of applying this equation about the centre of mass or the instantaneous axis of rotation. The conversation also touches on the importance of choosing a frame of reference and the relationship between the derivatives of a vector measured by two different frames.
  • #1
Hamiltonian
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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.
 
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  • #2
Hamiltonian299792458 said:
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.
 
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  • #3
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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  • #4
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.
 
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Related to A doubt in the rotational analogue of F=ma

1. What is the rotational analogue of F=ma?

The rotational analogue of F=ma is known as the angular analog, or the rotational equivalent, of Newton's second law of motion. It describes the relationship between torque, moment of inertia, and angular acceleration.

2. How is the rotational analogue of F=ma derived?

The rotational analogue of F=ma is derived using the same principles as Newton's second law of motion. It states that the net torque acting on an object is equal to the product of its moment of inertia and its angular acceleration.

3. What is the significance of the rotational analogue of F=ma?

The rotational analogue of F=ma is significant because it helps us understand the motion of rotating objects and how forces act on them. It is also essential in the study of rotational dynamics and engineering applications such as designing gears and motors.

4. Can the rotational analogue of F=ma be applied to all rotating objects?

Yes, the rotational analogue of F=ma can be applied to all rotating objects, regardless of their shape or size. However, it is important to note that the moment of inertia may vary depending on the object's mass distribution.

5. How does the rotational analogue of F=ma relate to the conservation of angular momentum?

The rotational analogue of F=ma is closely related to the conservation of angular momentum. According to this law, the total angular momentum of a system remains constant unless acted upon by an external torque. This means that the net torque acting on a system is equal to the change in its angular momentum over time.

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