D6. Examples of 2D dynamics - Revision history

Bros: /* ✏️ EXAMPLE D6.6: disk in a cylindrical cavity */

2024-11-30T20:14:05Z

✏️ EXAMPLE D6.6: disk in a cylindrical cavity

← Older revision		Revision as of 20:14, 30 November 2024
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	<math>\vel{J}{T}=\vel{G}{T}+\GJvec \times \velang{disk}{T}=(\nearrow\Rs\dot{\theta})+(\searrow \rs)\times[\otimes(\omega_0-\dot{\theta})]=(\swarrow [\rs\omega_0-(\Rs +\rs)\dot{\theta}]</math>		<math>\vel{J}{T}=\vel{G}{T}+\GJvec \times \velang{disk}{T}=(\nearrow\Rs\dot{\theta})+(\searrow \rs)\times[\otimes(\omega_0-\dot{\theta})]=(\swarrow [\rs\omega_0-(\Rs +\rs)\dot{\theta}]</math>

			<br>
			<br>
			<br>
			<br>
	<br>		<br>
	<br>		<br>

Bros: /* ✏️ EXAMPLE D6.6: disk in a cylindrical cavity */

2024-11-30T20:13:38Z

✏️ EXAMPLE D6.6: disk in a cylindrical cavity

← Older revision		Revision as of 20:13, 30 November 2024
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	<math>\vel{J}{T}=\vel{G}{T}+\GJvec \times \velang{disk}{T}=(\nearrow\Rs\dot{\theta})+(\searrow \rs)\times[\otimes(\omega_0-\dot{\theta})]=(\swarrow [\rs\omega_0-(\Rs +\rs)\dot{\theta}]</math>		<math>\vel{J}{T}=\vel{G}{T}+\GJvec \times \velang{disk}{T}=(\nearrow\Rs\dot{\theta})+(\searrow \rs)\times[\otimes(\omega_0-\dot{\theta})]=(\swarrow [\rs\omega_0-(\Rs +\rs)\dot{\theta}]</math>

	<~~math~~>~~\hspace{4cm}~~<~~/math~~>		<br>
			<br>
			<br>
			<br>

	<u>General diagram of interactions</u>		<u>General diagram of interactions</u>
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	unknowns: 2 associated with the DoF + 4 constraint unk. = 6 unk.		unknowns: 2 associated with the DoF + 4 constraint unk. = 6 unk.
	<~~math~~>~~\hspace{4cm}~~<~~/math~~>		<br>
			<br>
			<br>

	<u>Roadmap for the equation of motion</u>		<u>Roadmap for the equation of motion</u>

Bros: /* ✏️ EXAMPLE D6.6: disk in a cylindrical cavity */

2024-11-30T20:12:39Z

✏️ EXAMPLE D6.6: disk in a cylindrical cavity

← Older revision		Revision as of 20:12, 30 November 2024
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	<math>\vel{J}{T}=\vel{G}{T}+\GJvec \times \velang{disk}{T}=(\nearrow\Rs\dot{\theta})+(\searrow \rs)\times[\otimes(\omega_0-\dot{\theta})]=(\swarrow [\rs\omega_0-(\Rs +\rs)\dot{\theta}]</math>		<math>\vel{J}{T}=\vel{G}{T}+\GJvec \times \velang{disk}{T}=(\nearrow\Rs\dot{\theta})+(\searrow \rs)\times[\otimes(\omega_0-\dot{\theta})]=(\swarrow [\rs\omega_0-(\Rs +\rs)\dot{\theta}]</math>

	<math>\hspace{~~2cm~~}</math>		<math>\hspace{4cm}</math>

	<u>General diagram of interactions</u>		<u>General diagram of interactions</u>
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	unknowns: 2 associated with the DoF + 4 constraint unk. = 6 unk.		unknowns: 2 associated with the DoF + 4 constraint unk. = 6 unk.
			<math>\hspace{4cm}</math>

	<u>Roadmap for the equation of motion</u>		<u>Roadmap for the equation of motion</u>

Bros: /* ✏️ EXAMPLE D6.6: disk in a cylindrical cavity */

2024-11-30T20:10:53Z

✏️ EXAMPLE D6.6: disk in a cylindrical cavity

← Older revision		Revision as of 20:10, 30 November 2024
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	Rigid body kinematics (rigid body: disk):		Rigid body kinematics (rigid body: disk):
	<math>\vel{J}{T}=\vel{G}{T}+\GJvec \times \velang{disk}{T}=(\nearrow\Rs\dot{\theta})+(\searrow \rs)\times[\otimes(\omega_0-\dot{\theta})]=(\swarrow [\rs\omega_0-(\Rs +\rs)\dot{\theta}]</math>		<math>\vel{J}{T}=\vel{G}{T}+\GJvec \times \velang{disk}{T}=(\nearrow\Rs\dot{\theta})+(\searrow \rs)\times[\otimes(\omega_0-\dot{\theta})]=(\swarrow [\rs\omega_0-(\Rs +\rs)\dot{\theta}]</math>

			<math>\hspace{2cm}</math>

	<u>General diagram of interactions</u>		<u>General diagram of interactions</u>

Bros: /* ✏️ EXAMPLE D6.4: ring on a rough ground */

2024-11-30T20:07:19Z

✏️ EXAMPLE D6.4: ring on a rough ground

← Older revision		Revision as of 20:07, 30 November 2024
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	The kinematic description of the ring at three different time instants (initial, intermediate and final), and the dynamic description are:		The kinematic description of the ring at three different time instants (initial, intermediate and final), and the dynamic description are:
			<br>
			<br>
	[[File:ExD6-4-2-eng.png\|thumb\|center\|650px\|link=]]		[[File:ExD6-4-2-eng.png\|thumb\|center\|650px\|link=]]

Bros at 20:05, 30 November 2024

2024-11-30T20:05:47Z

Show changes

Bros: /* D6.1 2D kinematics and 2D dynamics */

2024-11-23T13:30:35Z

D6.1 2D kinematics and 2D dynamics

← Older revision		Revision as of 13:30, 23 November 2024
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	-------		-------
	==D6.1 2D kinematics and 2D dynamics==		==D6.1 2D kinematics and 2D dynamics==
	The condition for a system to exhibit 2D dynamics is not only that its kinematics be plane: it is also necessary that the angular momentum of each rigid body <math>\ss_\is</math> in the system relative to its center of inertia <math>\Gs_\is</math> be orthogonal to the plane of motion (<span style="text-decoration: underline;">[[D4. Vectorial theorems#D4.5 Teorema del Moment Cinètic (TMC): formulacions particulars\|'''~~secció~~ D4.5''']]</span>). This is equivalent to saying that the direction orthogonal to the plane of motion is a <span style="text-decoration: underline;">[[D5. Mass distribution#D5.2 Inertia tensor\|'''principal direction of inertia ''']]</span> for <math>\Gs_\is</math>.<br>		The condition for a system to exhibit 2D dynamics is not only that its kinematics be plane: it is also necessary that the angular momentum of each rigid body <math>\ss_\is</math> in the system relative to its center of inertia <math>\Gs_\is</math> be orthogonal to the plane of motion (<span style="text-decoration: underline;">[[D4. Vectorial theorems#D4.5 Teorema del Moment Cinètic (TMC): formulacions particulars\|'''section D4.5''']]</span>). This is equivalent to saying that the direction orthogonal to the plane of motion is a <span style="text-decoration: underline;">[[D5. Mass distribution#D5.2 Inertia tensor\|'''principal direction of inertia ''']]</span> for <math>\Gs_\is</math>.<br>

	====✏️ EXAMPLE D6.1: 2D kinematics and 3D dynamics ====		====✏️ EXAMPLE D6.1: 2D kinematics and 3D dynamics ====

Bros: /* ✏️ EXAMPLE D6.1: 2D kinematics and 3D dynamics */

2024-11-23T13:30:13Z

✏️ EXAMPLE D6.1: 2D kinematics and 3D dynamics

← Older revision		Revision as of 13:30, 23 November 2024
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	:The angular momentum of the bar at point <math>\Os</math> is:		:The angular momentum of the bar at point <math>\Os</math> is:
			[[File:ExD6-1-2-eng.png\|thumb\|right\|140px\|link=]]


	~~[[File:ExD6-1-2-eng.png\|thumb\|left\|140px\|link=]]~~
	<math>\braq{\overline{\mathbf{H}}_{\text {RTO=E}}(\Os)}{}= \matriz{\Is}{0}{0}{0}{0}{0}{0}{0}{\Is}\vector{\Omega_0 \sin \beta}{\Omega_0 \cos \beta}{0}= \vector{\Is\Omega_0 \sin \beta}{0}{0} .</math><br>		<math>\braq{\overline{\mathbf{H}}_{\text {RTO=E}}(\Os)}{}= \matriz{\Is}{0}{0}{0}{0}{0}{0}{0}{\Is}\vector{\Omega_0 \sin \beta}{\Omega_0 \cos \beta}{0}= \vector{\Is\Omega_0 \sin \beta}{0}{0} .</math><br>

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	----------		----------
	----------		----------

	==D6.2 Free-body diagram (FBD) and roadmap==		==D6.2 Free-body diagram (FBD) and roadmap==
	When a system consisting of only one rigid body has 2D dynamics, the number of equations to solve the problem is 3: the two components of the LMT in the plane of motion and the component of the AMT perpendicular to this plane. Depending on what we want to calculate (equation of motion, constraint force...), it may not be necessary to apply both theorems.<br>		When a system consisting of only one rigid body has 2D dynamics, the number of equations to solve the problem is 3: the two components of the LMT in the plane of motion and the component of the AMT perpendicular to this plane. Depending on what we want to calculate (equation of motion, constraint force...), it may not be necessary to apply both theorems.<br>

	Finding a quick way to achieve the desired result requires the development of a '''roadmap''' (which is a statement of the strategy to be followed). Exploring possible strategies and choosing an appropriate one requires a good understanding of the kinematics and dynamics of the rigid body. Therefore, it is always recommended to start by investigating its velocity distribution and the description of the interactions on it. The representation of these interactions constitutes the '''free-body diagram''' (FBD).		Finding a quick way to achieve the desired result requires the development of a '''roadmap''' (which is a statement of the strategy to be followed). Exploring possible strategies and choosing an appropriate one requires a good understanding of the kinematics and dynamics of the rigid body. Therefore, it is always recommended to start by investigating its velocity distribution and the description of the interactions on it. The representation of these interactions constitutes the '''free-body diagram''' (FBD).

Bros: Created page with "
TOC
$\newcommand{\uvec}{\overline{\textbf{u}}} \newcommand{\vvec}{\overline{\textbf{v}}} \newcommand{\evec}{\overline{\textbf{e}}} \newcommand{\Omegavec}{\overline{\mathbf{\Omega}}} \newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}} \newcommand{\Alfavec}{\overline{\mathbf{\alpha}}} \newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}} \newcommand{\ds}{\textrm{d}} \newcommand{\ps}{\textrm{p}} \newcommand{\hs}{\text..."$

2024-11-23T13:29:36Z

Created page with "<div class="noautonum">__TOC__</div> <math>\newcommand{\uvec}{\overline{\textbf{u}}} \newcommand{\vvec}{\overline{\textbf{v}}} \newcommand{\evec}{\overline{\textbf{e}}} \newcommand{\Omegavec}{\overline{\mathbf{\Omega}}} \newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}} \newcommand{\Alfavec}{\overline{\mathbf{\alpha}}} \newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}} \newcommand{\ds}{\textrm{d}} \newcommand{\ps}{\textrm{p}} \newcommand{\hs}{\text..."

New page

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\JQvec}{\vec{\Js\Qs}}
\newcommand{\GJvec}{\vec{\Gs\Js}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\GCvec}{\vec{\Gs\Cs}}
\newcommand{\PGvec}{\vec{\Ps\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
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\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
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{#1}\\
{#2}
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\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
</math>

This unit proposes a systematic method for solving problems in dynamics of systems consisting of any number of rigid bodies. If this number is high, defining the strategy to achieve the desired results is essential to avoid excessive development. <br>

The examples focus on problems in plane dynamics (2D). <span style="text-decoration: underline;">[[D7. Examples of 3D dynamics|'''Unit D7''']]</span> analyses examples of 3D dynamics.

-------
-------
==D6.1 2D kinematics and 2D dynamics==
The condition for a system to exhibit 2D dynamics is not only that its kinematics be plane: it is also necessary that the angular momentum of each rigid body <math>\ss_\is</math> in the system relative to its center of inertia <math>\Gs_\is</math> be orthogonal to the plane of motion (<span style="text-decoration: underline;">[[D4. Vectorial theorems#D4.5 Teorema del Moment Cinètic (TMC): formulacions particulars|'''secció D4.5''']]</span>). This is equivalent to saying that the direction orthogonal to the plane of motion is a <span style="text-decoration: underline;">[[D5. Mass distribution#D5.2 Inertia tensor|'''principal direction of inertia ''']]</span> for <math>\Gs_\is</math>.<br>

====✏️ EXAMPLE D6.1: 2D kinematics and 3D dynamics ====
------
:<small>
[[File:ExD6-1-1-eng.png|thumb|left|140px|link=]]

:The thin homogeneous bar is hinged at <math>\Os</math> to a support that rotates relative to the ground with constant vertical angular velocity <math>\Omega_0</math>. The bar has a single-point contact with the smooth ground, so its motion is planar: all its points (except point <math>\Os</math>) describe horizontal circular trajectories.<br>

:The fact that the kinematics of the bar is 2D may suggest that only the vertical component of the AMT is necessary to study its dynamics. But this is not the case.<br>

:The angular momentum of the bar at point <math>\Os</math> is:

[[File:ExD6-1-2-eng.png|thumb|left|140px|link=]]
<math>\braq{\overline{\mathbf{H}}_{\text {RTO=E}}(\Os)}{}= \matriz{\Is}{0}{0}{0}{0}{0}{0}{0}{\Is}\vector{\Omega_0 \sin \beta}{\Omega_0 \cos \beta}{0}= \vector{\Is\Omega_0 \sin \beta}{0}{0} .</math><br>

:Only the horizontal component of this vector is variable (it has a variable direction because of <math>\Omega_0</math>). Its geometrical time derivative yields:<br>

<math>\dot{\overline{\mathbf{H}}}_{\text {RTO}} (\Os)=(\Uparrow \Omega_0) \times (\Rightarrow \Is \Omega_0 \sin \beta \cos \beta)= \otimes\Is \Omega_0^2 \sin\beta \cos\beta.</math><br>

This time derivative can also be obtained through the vector basis:<br>

<math>\braq{\dot{\overline{\mathbf{H}}}_{\text {RTO}} (\Os)}{}= \vector{\Omega_0 \sin \beta}{\Omega_0 \cos \beta}{0} \times \vector{\Is \Omega_0 \sin \beta}{0}{0} =\vector{0}{0}{-\Is \Omega_0^2\sin\beta\cos\beta}.</math><br>

When applying the AMT at <math>\Os</math> to the bar, component 3, which is horizontal, cannot be ignored.
</small>

----------
----------
==D6.2 Free-body diagram (FBD) and roadmap==
When a system consisting of only one rigid body has 2D dynamics, the number of equations to solve the problem is 3: the two components of the LMT in the plane of motion and the component of the AMT perpendicular to this plane. Depending on what we want to calculate (equation of motion, constraint force...), it may not be necessary to apply both theorems.<br>

Finding a quick way to achieve the desired result requires the development of a '''roadmap''' (which is a statement of the strategy to be followed). Exploring possible strategies and choosing an appropriate one requires a good understanding of the kinematics and dynamics of the rigid body. Therefore, it is always recommended to start by investigating its velocity distribution and the description of the interactions on it. The representation of these interactions constitutes the '''free-body diagram''' (FBD).