Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Physics
Classical Physics
Mechanics
A doubt in the rotational analogue of F=ma
Reply to thread
Message
[QUOTE="etotheipi, post: 6432155"] 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##[SUP]th[/SUP] 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 [B]angular velocity[/B], ##\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 [B]rigid[/B], 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 [B]moment of inertia[/B] 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 [I]measured [/I]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 [B]general[/B] 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. [/QUOTE]
Insert quotes…
Post reply
Forums
Physics
Classical Physics
Mechanics
A doubt in the rotational analogue of F=ma
Back
Top