D8. Conservation of dynamic magnitudes - Revision history

Eantem: /* ✏️ EXAMPLE D8.8: bar in a smooth rotating circular guide */

2026-07-15T11:47:26Z

✏️ EXAMPLE D8.8: bar in a smooth rotating circular guide

← Older revision		Revision as of 11:47, 15 July 2026
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	:[[File:ExD8-8-2-neut.png\|thumb\|right\|270px\|link=]]		:[[File:ExD8-8-2-neut.png\|thumb\|right\|270px\|link=]]

	:<math>\overline{\Hs}_\mathrm{RTO}(\Os)=\Is\Is(\Os)\velang{~~bar~~}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\bigoplus(\Os)\right]\left(\overline{\dot{\psi}}+\overline{\dot{\theta}}\right)</math>		:<math>\overline{\Hs}_\mathrm{RTO}(\Os)=\Is\Is(\Os)\velang{barra}{RTO=T}=\left[\Is\Is(\Gs)+\Is\Is^\bigoplus(\Os)\right]\left(\overline{\dot{\psi}}+\overline{\dot{\theta}}\right)</math>


	:<math>\braq{\overline{\Hs}_\mathrm{RTO}(\Os)}{}=\left(\ms~~(\sqrt{3~~\rs})^2\diag{1}{0}{1}+\ms\left(\frac{\rs}{2}\right)^2\diag{1}{1}{0}\right)\vector{\dot{\theta}}{\dot{\psi}\sin\theta}{\dot{\psi}\cos\theta}</math>		:<math>\braq{\overline{\Hs}_\mathrm{RTO}(\Os)}{}=\left(\frac{1}{4}\ms\rs^2\diag{1}{0}{1}+\ms\left(\frac{\rs}{2}\right)^2\diag{1}{1}{0}\right)\vector{\dot{\theta}}{\dot{\psi}\sin\theta}{\dot{\psi}\cos\theta}</math>

	:<math>\braq{\overline{\Hs}_\mathrm{RTO}(\Os)}{}=\frac{1}{4}\ms\rs^2\vector{14\dot{\theta}}{\dot{\psi}\sin\theta}{13\dot{\psi}\cos\theta}</math>		:<math>\braq{\overline{\Hs}_\mathrm{RTO}(\Os)}{}=\frac{1}{4}\ms\rs^2\vector{2\dot{\theta}}{\dot{\psi}\sin\theta}{\dot{\psi}\cos\theta}</math>


	:<math>\left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_\mathrm{vert} =\left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_3\cos\theta+ \left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_2\sin\theta= \frac{1}{4}\ms\rs^2\dot{\psi}~~(13~~\cos^2\theta + \sin^2\theta)= \frac{1}{4}\ms\rs^2 \dot{\psi}~~(1+12\cos^2\theta)~~</math>		:<math>\left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_\mathrm{vert} =\left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_3\cos\theta+ \left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_2\sin\theta= \frac{1}{4}\ms\rs^2(\dot{\psi}\cos^2\theta + \dot{\psi}\sin^2\theta)= \frac{1}{4}\ms\rs^2 \dot{\psi}</math>

	:<math>\left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_\mathrm{vert} =\text{constant } \Rightarrow \quad \frac{\ds\left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_\mathrm{vert}}{\ds\ts}=0=\frac{1}{4}\ms\rs^2 \~~dot~~{\psi}~~(1+12\cos^2\theta)-6\ms\rs^2 \dot{\psi}\dot{\theta}\sin\theta\cos\theta~~ </math>		:<math>\left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_\mathrm{vert} =\text{constant } \Rightarrow \quad \frac{\ds\left.\overline{\Hs}_\mathrm{RTO}(\Os)\right]_\mathrm{vert}}{\ds\ts}=0=\frac{1}{4}\ms\rs^2 \ddot{\psi} </math>


	:<math>\boxed{\ddot{\psi}~~(3+7\cos^2\theta)-14\dot{\psi}\dot{\theta} \sin\theta \cos\theta~~ =0}</math>		:<math>\boxed{\ddot{\psi}=0}</math>

Bros at 18:38, 4 November 2024

2024-11-04T18:38:11Z

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	:The vertical angular momentum at <math>\Qs</math> and <math>\Ps</math> is not conserved since those two points are accelerated:		:The vertical angular momentum at <math>\Qs</math> and <math>\Ps</math> is not conserved since those two points are accelerated:

	:<math>\sum \overline{\bar{\Ms}}_{\text {ext }}(\mathbf{Q})-\overline{\mathbf{P G}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})=\dot{\overline{\mathrm{H}}}_{\text {RTQ }}(\mathbf{Q}), \quad \overline{\mathbf{QG}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})=\overline{\mathbf{QG}} \times \mathrm{m}\left[\overline{\mathrm{a}}_{\text {RЕL }}(\mathbf{Q})+\overline{\mathrm{a}}_{\mathrm{ar}}(\mathbf{Q})+\overline{\mathrm{a}}_{\text {cor }}(\mathbf{Q})\right]</math>		:<math>\sum \overline{\bar{\Ms}}_{\text {ext }}(\mathbf{Q})-\overline{\mathbf{P G}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})=\dot{\overline{\mathrm{H}}}_{\text {RTQ }}(\mathbf{Q}), \quad \overline{\mathbf{QG}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})=\overline{\mathbf{QG}} \times \mathrm{m}\left[\overline{\mathrm{a}}_{\text {RЕL }}(\mathbf{Q})+\overline{\mathrm{a}}_{\mathrm{tr}}(\mathbf{Q})+\overline{\mathrm{a}}_{\text {cor }}(\mathbf{Q})\right]</math>

	:<math> \left\{\overline{\mathbf{Q G}} \times \ms \bar{\mathrm{a}}_{\mathrm{Gal}}(\mathbf{Q})\right\}=\left\{\begin{array}{c}		:<math> \left\{\overline{\mathbf{Q G}} \times \ms \bar{\mathrm{a}}_{\mathrm{Gal}}(\mathbf{Q})\right\}=\left\{\begin{array}{c}
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	\mathrm{a}_{\mathrm{REL}}^{\mathrm{s}} \cos \left(30^{\circ}-\theta\right)+\mathrm{a}_{\mathrm{REL}}^{\mathrm{n}} \sin \left(30^{\circ}-\theta\right)		\mathrm{a}_{\mathrm{REL}}^{\mathrm{s}} \cos \left(30^{\circ}-\theta\right)+\mathrm{a}_{\mathrm{REL}}^{\mathrm{n}} \sin \left(30^{\circ}-\theta\right)
	\end{array}\right\}+\left\{\begin{array}{c}		\end{array}\right\}+\left\{\begin{array}{c}
	\mathrm{a}_{\mathrm{ar}}^{\mathrm{s}}-\mathrm{a}_{\mathrm{Cor}} \\		\mathrm{a}_{\mathrm{tr}}^{\mathrm{s}}-\mathrm{a}_{\mathrm{Cor}} \\
	-\mathrm{a}_{\mathrm{ar}}^{\mathrm{n}} \\		-\mathrm{a}_{\mathrm{tr}}^{\mathrm{n}} \\
	0		0
	\end{array}\right\}\right]		\end{array}\right\}\right]
	</math>		</math>

	:<math>\left.\left.\left.\overline{\mathbf{Q G}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})\right]_{\text {vert }}=\overline{\mathbf{QG}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})\right]_{3^{\prime}}=\sqrt{3} \mathrm{m}\left(\mathrm{a}_{\text {Cor }}-\mathrm{a}_{\mathrm{ar}}^{\mathrm{s}}\right) \sin \theta \neq 0 \quad \Rightarrow \quad \dot{\bar{\Hs}}_{\text {RTQ }}(\mathbf{Q})\right]_{\text {vert }} \neq 0		:<math>\left.\left.\left.\overline{\mathbf{Q G}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})\right]_{\text {vert }}=\overline{\mathbf{QG}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})\right]_{3^{\prime}}=\sqrt{3} \mathrm{m}\left(\mathrm{a}_{\text {Cor }}-\mathrm{a}_{\mathrm{tr}}^{\mathrm{s}}\right) \sin \theta \neq 0 \quad \Rightarrow \quad \dot{\bar{\Hs}}_{\text {RTQ }}(\mathbf{Q})\right]_{\text {vert }} \neq 0
	</math>		</math>


	</small>		</small>

Bros at 18:31, 4 November 2024

2024-11-04T18:31:07Z

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	</math>		</math>

	~~:<math>\mu=\text{datos.mean()}</math>~~

	~~:<math>\sigma=\text{datos.std()}</math>~~

	</small>		</small>

Bros at 18:24, 4 November 2024

2024-11-04T18:24:00Z

← Older revision		Revision as of 18:24, 4 November 2024
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	:For the disk SYSTEM, the resultant moment about its centre of inertia <math>\Gs</math> is zero in the direction of its axis. The motor can change the orientation of this axis relative to the ground (and relative to any reference frame with a translational motion relative to the ground), and therefore <u>it is not a direction fixed to the ground</u> (so not to the RTG either): the angular momentum is not conserved in this direction.<br>		:For the disk SYSTEM, the resultant moment about its centre of inertia <math>\Gs</math> is zero in the direction of its axis. The motor can change the orientation of this axis relative to the ground (and relative to any reference frame with a translational motion relative to the ground), and therefore <u>it is not a direction fixed to the ground</u> (so not to the RTG either): the angular momentum is not conserved in this direction.<br>

	:For the SYSTEM (~~disc~~ + support + fork), the ~~reulstant~~ moment relative to the ~~centre~~ of the disk <math>\Os</math> is zero in the vertical direction, which is fixed to the ground. Therefore:		:For the SYSTEM (disk + support + fork), the resulting moment relative to the center of the disk <math>\Os</math> is zero in the vertical direction, which is fixed to the ground. Therefore:

	:<math> \left.\sum \overline{\Ms}_\mathrm{ext}(\Os)\right]_\mathrm{vert} =0 \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTO=T}(\Os) \right]_\mathrm{vert} \text{ CONSTANT!} </math>		:<math> \left.\sum \overline{\Ms}_\mathrm{ext}(\Os)\right]_\mathrm{vert} =0 \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTO=T}(\Os) \right]_\mathrm{vert} \text{ CONSTANT!} </math>
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	</small>		</small>

	====✏️ ~~Exemple~~ D8.8: ~~barra dins de guia llisa giratòria~~ ====		====✏️ EXAMPLE D8.8: bar in a smooth rotating circular guide====
	------		------
	:<small>		:<small>
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	\|		\|
	:[[File:ExD8-8-1-eng.png\|thumb\|left\|180px\|link=]]		:[[File:ExD8-8-1-eng.png\|thumb\|left\|180px\|link=]]
	\|~~La barra~~ '''PQ''', ~~homogènia i de massa~~ m, ~~es mou mantenint els seus dos extrems dins d’una guia llisa~~ <math>(\mu=0)</math> , ~~de radi~~ r ~~i massa~~ negligible, ~~que pot girar lliurement al voltant de la direcció~~ vertical. ~~L’angle~~ POQ ~~és de~~ <math>120^o</math>. ~~Es tracta de determinar~~ <u>~~l’equació del moviment per a la coordenada</u>~~ <math>\psi</math> . ~~Es negligeix la rotació de la barra sobre el seu eix~~ (~~rotació pròpia~~ <math>\dot{\varphi}</math> ).		\|The homogeneous bar '''PQ''', with mass m, moves with its two endpoins inside a smooth guide <math>(\mu=0)</math>, with radius r and negligible mass, which can rotate freely around the vertical direction. The '''POQ''' angle is <math>120^o</math>. We want to <u>find the equation of motion for the coordinate <math>\psi</math></u>. The rotation of the bar on its axis (spin <math>\dot{\varphi}</math>) is neglected.
			\|}

	\|}		:<u>Is the linear momentum conserved?</u><br>
	:<u>~~Hi ha conservació de la quantitat de moviment~~?</u>.<br>		:The external forces on the bar are not zero: in addition to the weight, there are normal forces from the guide on the bar at <math>\Ps</math> and <math>\Qs</math> (directed towards <math>\Os</math>) and a force component perpendicular to the plane of the guide. Therefore, the resultant external force on the bar has components in the three directions of space, and the linear momentum is not conserved.
	:~~Les~~ forces ~~externes sobre la barra no són nul·les~~: ~~a més del pes~~, ~~hi ha~~ forces ~~de la guia sobre la barra a~~ <math>\Ps</math> ~~i a~~ <math>\Qs</math> ~~que només tenen component normal~~ (~~dirigida cap a~~ <math>\Os</math>) i component perpendicular ~~al pla de la guia~~. ~~En total, doncs~~, ~~la força externa~~ resultant ~~sobre la barra té~~ components ~~en les tres direccions de l’espai~~, ~~i la quantitat de moviment no es conserva~~.

	:~~Si s’analitzen les~~ forces ~~externes sobre el SISTEMA~~ (~~barra~~ + ~~guia~~), ~~la conclusió és la mateixa~~: ~~a més del pes de la barra~~, ~~hi ha la força d’enllaç associada al coixinet entre terra i guia~~, ~~que té tres~~ components ~~no nul·les en principi~~.		:If we analyze the external forces on the SYSTEM (bar + guide), the conclusion is the same: in addition to the weight of the bar, there is the constraint force associated with the bearing between the ground and the guide, which has three non-zero components in principle.

	:<u>~~Hi ha conservació del moment cinètic~~?</u>		:<u>Is the angular momentum conserved?</u>

	:~~Per al SISTEMA~~ (~~barra~~ + ~~guia~~), el moment ~~extern sobre qualsevol punt de l’eix de rotació de la guia~~ (en particular, ~~per al punt~~ <math>\Os</math>) té component ~~vertical nul·la~~ (~~doncs el~~ moment ~~d’enllaç del coixinet en aquesta direcció és nul~~, ~~i el pes no pot donar~~ moment ~~en direcció~~ vertical). ~~Per altra banda~~, ~~l’acceleració~~ angular ~~de la guia respecte del terra~~ <math>\left(\overline{\ddot{\psi}}\right)</math> ~~té aquesta direcció~~. ~~Per tant~~:		:For the SYSTEM (bar + guide), the external moment about any point on the axis of rotation of the guide (in particular, for point <math>\Os</math>) has a zero vertical component (since the constraint moment of the bearing in that direction is zero, and the weight cannot yield a moment in the vertical direction). On the other hand, the angular acceleration of the guide relative to the ground <math>\left(\overline{\ddot{\psi}}\right)</math> is vertical. Therefore:

	:<math>\boxed{\left.\text{~~Full de ruta~~: ~~SISTEMA~~ (~~barra~~+~~guia~~), ~~TMC a~~ }\Os\right]_\mathrm{vert}}</math>		:<math>\boxed{\left.\text{Roadmap: SYSTEM (bar+guide), AMT at }\Os\right]_\mathrm{vert}}</math>


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	:~~L’únic~~ element ~~amb massa és la barra~~, ~~i el punt~~ <math>\Os</math> ~~hi pertany cinemàticament~~ :		:The only element with nonzero mass is the bar, and point <math>\Os</math> is fixed to it:

	:[[File:ExD8-8-2-neut.png\|thumb\|right\|270px\|link=]]		:[[File:ExD8-8-2-neut.png\|thumb\|right\|270px\|link=]]

	:<math>\overline{\Hs}_\mathrm{RTO}(\Os)=\Is\Is(\Os)\velang{~~barra~~}{RTO=T}=\left[\Is\Is(\Gs)+\Is\Is^\bigoplus(\Os)\right]\left(\overline{\dot{\psi}}+\overline{\dot{\theta}}\right)</math>		:<math>\overline{\Hs}_\mathrm{RTO}(\Os)=\Is\Is(\Os)\velang{bar}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\bigoplus(\Os)\right]\left(\overline{\dot{\psi}}+\overline{\dot{\theta}}\right)</math>


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	:<u>~~Comentari rellevant~~</u>		:<u>Relevant comment</u>
	:[[File:ExD8-8-3-neut.png\|thumb\|right\|430px\|link=]]		:[[File:ExD8-8-3-neut.png\|thumb\|right\|430px\|link=]]

	:~~El moment cinètic~~ vertical ~~no es conserva ni a~~ <math>\Qs</math> ~~ni a~~ <math>\Ps</math> ~~perquè tots dos punts estan accelerats~~:		:The vertical angular momentum at <math>\Qs</math> and <math>\Ps</math> is not conserved since those two points are accelerated:

	:<math>\sum \overline{\bar{\Ms}}_{\text {ext }}(\mathbf{Q})-\overline{\mathbf{P G}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})=\dot{\overline{\mathrm{H}}}_{\text {RTQ }}(\mathbf{Q}), \quad \overline{\mathbf{QG}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})=\overline{\mathbf{QG}} \times \mathrm{m}\left[\overline{\mathrm{a}}_{\text {RЕL }}(\mathbf{Q})+\overline{\mathrm{a}}_{\mathrm{ar}}(\mathbf{Q})+\overline{\mathrm{a}}_{\text {cor }}(\mathbf{Q})\right]</math>		:<math>\sum \overline{\bar{\Ms}}_{\text {ext }}(\mathbf{Q})-\overline{\mathbf{P G}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})=\dot{\overline{\mathrm{H}}}_{\text {RTQ }}(\mathbf{Q}), \quad \overline{\mathbf{QG}} \times \ms \overline{\mathrm{a}}_{\text {Gal }}(\mathbf{Q})=\overline{\mathbf{QG}} \times \mathrm{m}\left[\overline{\mathrm{a}}_{\text {RЕL }}(\mathbf{Q})+\overline{\mathrm{a}}_{\mathrm{ar}}(\mathbf{Q})+\overline{\mathrm{a}}_{\text {cor }}(\mathbf{Q})\right]</math>

Bros at 18:18, 4 November 2024

2024-11-04T18:18:07Z

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	\|		\|
	:[[File:ExD8-7-1-eng.png\|thumb\|left\|180px\|link=]]		:[[File:ExD8-7-1-eng.png\|thumb\|left\|180px\|link=]]
	\|~~El sistema consta d’un disc homogeni, de massa~~ m ~~i radi~~ r, ~~que està articulat~~ a ~~un suport de massa negligible~~, ~~i d’una forquilla que pot girar lliurement al voltant de l’eix~~ vertical. ~~Entre suport i forquilla hi ha un~~ motor. El moment ~~d’inèrcia del suport respecte de l’eix~~ vertical ~~que passa pel centre del disc és~~ <math>\Is=(\lambda/2)\ms\rs^2</math>. ~~Inicialment~~ <math>(\ts_\mathrm{incial}</math> ~~el disc està paral·lel a terra~~, ~~i gira amb velocitat~~ angular ~~vertical~~ <math>\velang{disc}{T}=\psio</math> . ~~Es negligeixen les friccions associades a totes les articulacions~~. <u> ~~Es tracta d’investigar com es mou el suport quan el~~ motor ~~canvia l’orientació del disc respecte del terra~~ .</u>		\|The system consists of a homogeneous disk with mass m and radius r, articulated to a massless support, and a fork that can rotate freely about the vertical axis. There is a motor between the support and the fork. The inertia moment of the support with respect to the vertical axis through the center of the disk is <math>\Is=(\lambda/2)\ms\rs^2</math>. Initially <math>(\ts_\mathrm{incial}</math> the disk is parallel to the ground, and rotates with vertical angular velocity <math>\velang{disc}{T}=\psio</math>. The friction associated with all the joints is neglected. We want to <u> investigate how the support moves when the motor changes the disk orientation relative to the ground.</u>

	\|}		\|}
	:<u>~~Hi ha conservació del moment cinètic~~?</u>.<br>		:<u>Is the angular momentum conserved?</u><br>

	:~~Per al SISTEMA disc~~, el moment ~~total respecte del seu~~ centre ~~d’inèrcia~~ <math>\Gs</math> ~~és nul en la direcció del seu eix~~. El motor ~~pot canviar l’orientació d’aquest eix respecte del terra~~ (~~i respecte de qualsevol referència que es traslladi respecte del terra~~), ~~i per tant~~ <u>~~no es tracta d’una direcció fixa al terra~~</u> (~~per tant, tampoc a la~~ RTG): ~~no es conserva el moment cinètic en aquesta direcció~~.<br>		:For the disk SYSTEM, the resultant moment about its centre of inertia <math>\Gs</math> is zero in the direction of its axis. The motor can change the orientation of this axis relative to the ground (and relative to any reference frame with a translational motion relative to the ground), and therefore <u>it is not a direction fixed to the ground</u> (so not to the RTG either): the angular momentum is not conserved in this direction.<br>

	:~~Per al SISTEMA~~ (disc + ~~suport~~ + ~~forquilla~~), el moment ~~total respecte del~~ centre ~~del disc~~ <math>\Os</math> ~~és nul en la direcció~~ vertical, ~~que sí que és fixa a terra~~. ~~Per tant~~:		:For the SYSTEM (disc + support + fork), the reulstant moment relative to the centre of the disk <math>\Os</math> is zero in the vertical direction, which is fixed to the ground. Therefore:

	:<math> \left.\sum \overline{\Ms}_\mathrm{ext}(\Os)\right]_\mathrm{vert} =0 \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTO=T}(\Os) \right]_\mathrm{vert} \text{ CONSTANT!} </math>		:<math> \left.\sum \overline{\Ms}_\mathrm{ext}(\Os)\right]_\mathrm{vert} =0 \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTO=T}(\Os) \right]_\mathrm{vert} \text{ CONSTANT!} </math>

	:~~Mentre el~~ motor ~~canvia l’orientació del pla del disc~~ <math>(\dot{\theta} \neq 0)</math> , ~~apareix la rotació del suport al voltant de l’eix~~ vertical <math>\dot{\psi}</math> . ~~El moment cinètic a~~ <math>\Os</math> ~~en cada instant és~~:		:As the motor changes the orientation of the disk plane <math>(\dot{\theta} \neq 0)</math>, the support rotates around the vertical axis <math>\dot{\psi}</math>. The angular momentum at <math>\Os</math> is:
	:[[File:ExD8-7-2-neut.png\|thumb\|right\|180px\|link=]]		:[[File:ExD8-7-2-neut.png\|thumb\|right\|180px\|link=]]

	:<math>\overline{\Hs}_\mathrm{RTO=T}(\Os,\ts)=\Is\Is^\mathrm{~~forq~~}(\Os)\velang{~~forq~~}{T}(\ts)+\Is\Is^\mathrm{~~disc~~}(\Os)\velang{~~disc~~}{T}(\ts)=\Is\Is^\mathrm{~~forq~~}(\Os)\dot{\psi}+\Is\Is^\mathrm{~~forq~~}(\Os)\left( \overline{\dot{\psi}}+ \overline{\dot{\theta}}+ \overline{\dot{\varphi}}\right)</math>		:<math>\overline{\Hs}_\mathrm{RTO=E}(\Os,\ts)=\Is\Is^\mathrm{sup}(\Os)\velang{sup}{E}(\ts)+\Is\Is^\mathrm{disk}(\Os)\velang{disk}{E}(\ts)=\Is\Is^\mathrm{sup}(\Os)\dot{\psi}+\Is\Is^\mathrm{disk}(\Os)\left( \overline{\dot{\psi}}+ \overline{\dot{\theta}}+ \overline{\dot{\varphi}}\right)</math>


	:<math>\braq{\overline{\Hs}_\Ts^\mathrm{~~forq~~}(\Os,\ts)}{B}=\diag{\Is_{11}}{\Is_{22}}{(\lambda/2)\ms\rs^2}\vector{0}{0}{\dot{\psi}}=\vector{0}{0}{(\lambda/2)\ms\rs^2 \dot{\psi}},</math>		:<math>\braq{\overline{\Hs}_\Es^\mathrm{sup}(\Os,\ts)}{B}=\diag{\Is_{11}}{\Is_{22}}{(\lambda/2)\ms\rs^2}\vector{0}{0}{\dot{\psi}}=\vector{0}{0}{(\lambda/2)\ms\rs^2 \dot{\psi}},</math>


	:<math>\braq{\overline{\Hs}_\Ts^\mathrm{~~disc~~}(\Os,\ts)}{B'}=\frac{1}{4}\ms\rs^2\diag{1}{1}{2}\vector{-\dot{\theta}}{-\dot{\psi} \sin\theta}{\dot{\varphi}+\dot{\psi}\cos\theta}=\frac{1}{4}\ms\rs^2\vector{-\dot{\theta}}{-\dot{\psi} \sin\theta}{2(\dot{\varphi}+\dot{\psi}\cos\theta)}</math>		:<math>\braq{\overline{\Hs}_\Es^\mathrm{disk}(\Os,\ts)}{B'}=\frac{1}{4}\ms\rs^2\diag{1}{1}{2}\vector{-\dot{\theta}}{-\dot{\psi} \sin\theta}{\dot{\varphi}+\dot{\psi}\cos\theta}=\frac{1}{4}\ms\rs^2\vector{-\dot{\theta}}{-\dot{\psi} \sin\theta}{2(\dot{\varphi}+\dot{\psi}\cos\theta)}</math>


	:~~El moment cinètic en direcció~~ vertical ~~prové de la projecció de les~~ components 2 i 3:		:The angular momentum in the vertical direction comes from the projection of components 2 and 3:

	:<math>\left.\overline{\Hs}_\Ts(\Os,\ts)\right]_\mathrm{vert} =\overline{\Hs}_\Ts^\mathrm{sup}(\Os,\ts)+ \left.\overline{\Hs}_\Ts^ \mathrm{~~disc~~} (\Os,\ts) \right]_{3'}\cos\theta - \left.\overline{\Hs}_\Ts^\mathrm{~~disc~~}(\Os,\ts)\right]_{2'} \sin\theta ,</math>		:<math>\left.\overline{\Hs}_\Es(\Os,\ts)\right]_\mathrm{vert} =\overline{\Hs}_\Es^\mathrm{sup}(\Os,\ts)+ \left.\overline{\Hs}_\Es^ \mathrm{disk} (\Os,\ts) \right]_{3'}\cos\theta - \left.\overline{\Hs}_\Es^\mathrm{disk}(\Os,\ts)\right]_{2'} \sin\theta ,</math>

	:<math>\left.\overline{\Hs}_\Ts(\Os,\ts)\right]_\mathrm{vert} = \left(\Uparrow \frac{\lambda}{2}\ms\rs^2\dot{\psi}\right)+ \left(\Uparrow \frac{1}{4}\ms\rs^2\left[2\dot{\varphi}\cos\theta+\dot{\psi}(1+\cos^2\theta)\right]\right)</math>		:<math>\left.\overline{\Hs}_\Es(\Os,\ts)\right]_\mathrm{vert} = \left(\Uparrow \frac{\lambda}{2}\ms\rs^2\dot{\psi}\right)+ \left(\Uparrow \frac{1}{4}\ms\rs^2\left[2\dot{\varphi}\cos\theta+\dot{\psi}(1+\cos^2\theta)\right]\right)</math>


	:~~Imposant la conservació de moment cinètic~~ vertical:		:Taking into account the conservation of the vertical angular momentum:

	:<math>\left.\begin{array}{ll}		:<math>\left.\begin{array}{ll}
	\left.\overline{\mathrm{H}}_{\mathrm{T}}(\mathbf{O}, \mathrm{t})\right]_{\text {vert }}=\left(\Uparrow \frac{1}{4} \mathrm{mr}^2\left[2 \dot{\varphi} \cos \theta+\dot{\psi}\left(1+2 \lambda+\cos ^2 \theta\right)\right]\right) \\		\left.\overline{\mathrm{H}}_{\mathrm{E}}(\mathbf{O}, \mathrm{t})\right]_{\text {vert }}=\left(\Uparrow \frac{1}{4} \mathrm{mr}^2\left[2 \dot{\varphi} \cos \theta+\dot{\psi}\left(1+2 \lambda+\cos ^2 \theta\right)\right]\right) \\
	\left.\begin{array}{l}		\left.\begin{array}{l}
	\dot{\psi}\left(\mathrm{t}_{\text {~~inicial~~ }}\right)=0 \\		\dot{\psi}\left(\mathrm{t}_{\text {initial }}\right)=0 \\
	\dot{\varphi}\left(\mathrm{t}_{\text {~~inicial~~ }}\right)=\dot{\varphi}_0 \\		\dot{\varphi}\left(\mathrm{t}_{\text {initial }}\right)=\dot{\varphi}_0 \\
	\theta\left(\mathrm{t}_{\text {~~inicial~~ }}\right)=0		\theta\left(\mathrm{t}_{\text {initial }}\right)=0
	\end{array}\right\} \left.\Rightarrow \overline{\mathrm{H}}_{\mathrm{T}}\left(\mathbf{O}, \mathrm{t}_{\text {~~inicial~~ }}\right)\right]_{\text {vert }}=\left[\Uparrow\left(\frac{1}{2} \mathrm{mr}^2 \dot{\varphi}_0\right)\right] \\		\end{array}\right\} \left.\Rightarrow \overline{\mathrm{H}}_{\mathrm{T}}\left(\mathbf{O}, \mathrm{t}_{\text {initial }}\right)\right]_{\text {vert }}=\left[\Uparrow\left(\frac{1}{2} \mathrm{mr}^2 \dot{\varphi}_0\right)\right] \\
	\end{array}\right\}		\end{array}\right\}
	\Rightarrow 2 \dot{\varphi} \cos \theta+\dot{\psi}\left(1+2 \lambda+\cos ^2 \theta\right)=2 \dot{\varphi}_0</math>		\Rightarrow 2 \dot{\varphi} \cos \theta+\dot{\psi}\left(1+2 \lambda+\cos ^2 \theta\right)=2 \dot{\varphi}_0</math>


	:~~Per altra banda~~, ~~per al SISTEMA disc~~, el moment ~~extern en la direcció de l’eix del disc~~ (~~direcció 3’~~) ~~és nul~~.		:On the other hand, for the SYSTEM disk, the external moment in the direction of the disk axis (direction 3') is zero. Hence, <math>\left.\dot{\overline{\Hs}}_\mathrm{disk,T}(\Os)\right]_{3'}=0</math>

	~~:Per tant, <math>\left.\dot{\overline{\Hs}}_\mathrm{disc,T}(\Os)\right]_{3'}=0</math>~~		:<math>\braq{\dot{\overline{\Hs}}_\mathrm{disk,E}(\Os,\ts)}{B'}=\frac{1}{4}\ms\rs^2 \vector{-\ddot{\theta}}{-\ddot{\psi} \sin\theta -\dot{\psi}\dot{\theta}\cos\theta}{2(\ddot{\varphi}+\ddot{\psi}\cos\theta-\dot{\psi}\dot{\theta}\sin\theta)} + \vector{-\dot{\theta}}{-\dot{\psi}\sin\theta}{\dot{\psi}\cos\theta}\times\frac{1}{4}\ms\rs^2\vector{-\dot{\theta}}{-\dot{\psi}\sin\theta}{2(\dot{\varphi}+\dot{\psi}\cos\theta)}=</math>

	:<math>\braq{\dot{\overline{\Hs}}_\mathrm{~~disc~~,T}(\Os,\ts)}{B'}=\frac{1}{4}\ms\rs^2 \vector{-\ddot{\theta}}{-\ddot{\psi} \sin\theta -\dot{\psi}\dot{\theta}\cos\theta}{2(\ddot{\varphi}+\ddot{\psi}\cos\theta-\dot{\psi}\dot{\theta}\sin\theta)} + \vector{-\dot{\theta}}{-\dot{\psi}\sin\theta}{\dot{\psi}\cos\theta}\times\frac{1}{4}\ms\rs^2\vector{-\dot{\theta}}{-\dot{\psi}\sin\theta}{2(\dot{\varphi}+\dot{\psi}\cos\theta)}=</math>


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	:[[File:ExD8-7-3D-neut.png\|thumb\|right\|350px\|link=]]		:[[File:ExD8-7-3D-neut.png\|thumb\|right\|350px\|link=]]
	:~~La integració d’aquesta equació condueix a~~: <math>\dot{\varphi}+\dot{\psi}\cos\theta=\dot{\varphi}_0</math> , on <math>\dot{\varphi}_0</math> ~~és la~~ constant ~~d’integració~~, ~~que es determina imposant les condicions inicials~~.		:The integration of that equation yields: <math>\dot{\varphi}+\dot{\psi}\cos\theta=\dot{\varphi}_0</math>, where <math>\dot{\varphi}_0</math> is the integration constant, which may be found from the initial conditions.

	:~~Combinant aquest resultat amb l’anterior~~:		:Combination of this result with the previous one yields:

	:<math>\left.\begin{array}{l}		:<math>\left.\begin{array}{l}

Bros at 18:09, 4 November 2024

2024-11-04T18:09:41Z

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	\|}		\|}
	:<u>~~Hi ha conservació de la quantitat de moviment~~?</u>.<br>		:<u>Is the linear momentum conserved?</u><br>

	:~~El sòlid està sotmès a l’atracció gravitatòria terrestre com a única força externa~~. ~~Per tant~~, la component ~~vertical de la quantitat de moviment respecte del terra no es conserva~~, ~~però les horitzontals sí.~~.<br>		:The Earth's gravitational attraction as the only external force on the rigid body. Therefore, the vertical component of the linear momentum relative to the ground is not conserved, but the horizontal components are.<br>

	:~~Ja que la quantitat de moviment respecte del terra i la velocitat del centre d’inèrcia~~ <math>\vel{G=O}{T}</math> ~~són estrictament proporcionals~~, ~~les~~ components ~~horitzontals de~~ <math>\vel{G}{T}</math> ~~es mantenen constants.~~.		:Since the linear momentum relative to the ground and the velocity of the center of inertia <math>\vel{G=O}{E}</math> are strictly proportional, the horizontal components of <math>\vel{G}{E}</math> are also constant.

	:<u>~~Hi ha conservació del moment cinètic~~?</u>		:<u>Is the angular momentum conserved?</u>

	:El <span style="text-decoration: underline;"> [[D3. Interaccions entre sòlids rígids#D3.2 Atracció gravitatòria\|'''torsor ~~gravitatori~~''']]</span> al centre ~~de gravetat~~ <math>\Gs</math> (~~que coincideix amb el~~ centre ~~d’inèrcia~~ <math>\Os</math>) ~~es redueix~~ a ~~una força~~ resultant ~~i cap~~ moment. ~~Per tant~~:		:The <span style="text-decoration: underline;"> [[D3. Interaccions entre sòlids rígids#D3.2 Atracció gravitatòria\|'''gravitational torsor''']]</span> at the gravity centre <math>\Gs</math> (which is the same point as the inertia centre <math>\Os</math>) reduces to a resultant force and no moment. Therefore:

	:<math>\sum\overline{\mathrm{M}}_\mathrm{ext}(\Gs)=\overline{0}=\dot{\overline{\mathrm{H}}}_\mathrm{RTG} (\Gs) \quad \Rightarrow \quad \overline{\mathrm{H}}_\mathrm{RTG} (\Gs) \mathrm{CONSTANT!}</math>		:<math>\sum\overline{\mathrm{M}}_\mathrm{ext}(\Gs)=\overline{0}=\dot{\overline{\mathrm{H}}}_\mathrm{RTG} (\Gs) \quad \Rightarrow \quad \overline{\mathrm{H}}_\mathrm{RTG} (\Gs) \mathrm{CONSTANT!}</math>

	:~~Per al cas del sòlid que s’estudia~~: <math>\overline{\mathrm{H}}_\mathrm{RTG} (\Gs) = \Is\Is (\Gs) \velang{}{RTG} = \Is\Is (\Gs) \velang{}{T}.</math>		:For the rigid body under study: <math>\overline{\mathrm{H}}_\mathrm{RTG} (\Gs) = \Is\Is (\Gs) \velang{}{RTG} = \Is\Is (\Gs) \velang{}{T}.</math>

	:~~El moment cinètic i la velocitat~~ angular ~~no són proporcionals en~~ general (~~només ho són quan la direcció de la velocitat~~ angular ~~és una~~ <span style="text-decoration: underline;"> [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia\|'''~~direcció~~ principal ~~d’inèrcia~~''']]</span> ~~per al~~ centre ~~d’inèrcia~~ <math>\Gs</math> ), ~~i la conservació del primer no implica la constància de la segona~~.		:The angular momentum and the angular velocity are not proportional in general (they are only proportional when the direction of angular velocity is a <span style="text-decoration: underline;"> [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia\|'''a principal direction of inertia''']]</span> for the centre of inertia <math>\Gs</math> (section D5.3)), and the conservation of the former does not imply that of the latter.

	:<u> ~~Avaluació qualitativa del~~ tensor ~~d’inèrcia~~</u>		:<u>Qualitative assessment of the inertia tensor</u>

	:~~En tractar-se les dues plaques de sòlids plans simètrics~~:		:Since the two plates are symmetrical
			:<math>\left[\Is\Is(\Gs)\right]=\left[\Is\Is(\Gs)\right]_\text{lower plate} + \left[\Is\Is(\Gs)\right]_\text{upper plate} = \diag{\Is_{11}}{\Is_{11} + \Is_{33}}{\Is_{33}}+ \diag{2\Is}{\Is}{\Is} , \text{ amb} \Is_{11}<\Is_{33}.</math>

			:<u>Quantitative assessment of the inertia tensor</u>

	:<math>\left[\Is\Is(\Gs)\right]=\left[\Is\Is(\Gs)\right]_\text{placa inf} + \left[\Is\Is(\Gs)\right]_\text{placa sup.} = \diag{\Is_{11}}{\Is_{11} + \Is_{33}}{\Is_{33}}+ \diag{2\Is}{\Is}{\Is} , \text{ amb} \Is_{11}<\Is_{33}.</math>

	:<u>~~Avaluació quantitativa del tensor d’inèrcia~~</u>		:<math>\left[\Is\Is(\Gs)\right]=\frac{1}{3} \ms\Ls^2 \diag{1}{1+4}{4}+\frac{1}{3} \ms\Ls^2 \diag{2}{1}{1}=\frac{1}{3} \ms\Ls^2 \diag{3}{6}{5} \equiv \diag{\Is_\mathrm{low}}{\Is_\mathrm{high}}{\Is_\mathrm{medium}}</math>

			:The directions 1, 2 and 3 are the inertia principal directions for point <math>\Gs</math>.

	:<~~math~~>\left[\Is\Is(\Gs)\right]=\frac{1}{3} \ms\Ls^2 \diag{1}{1+4}{4}+\frac{1}{3} \ms\Ls^2 \diag{2}{1}{1}=\frac{1}{3} \ms\Ls^2 \diag{3}{6}{5} \equiv \diag{\Is_\mathrm{petit}}{\Is_\mathrm{gran}}{\Is_\mathrm{mitjà}}</~~math~~>		:<u>Angular momentum calculation</u>

	:~~Les direccions 1~~, ~~2 i 3 són les direccions principals d’inèrcia per a~~ <math>\Gs</math>.		:<math>\left\{\overline{\mathrm{H}}_\mathrm{RTG} (\Gs)\right\}=\diag{\Is_\mathrm{low}}{\Is_\mathrm{high}}{\Is_\mathrm{medium}} \vector{\Omega_1}{\Omega_2}{\Omega_3}=\vector{\Is_\mathrm{low}\Omega_1}{\Is_\mathrm{high}\Omega_2}{\Is_\mathrm{medium}\Omega_3}</math>, is not proportional to <math>\velang{}{T}</math> in principle.

	:~~<u>Cálcul del moment cinètic</u>~~		:If the initial angular velocity is exclusively in one of the three directions (that is, if its direction is a principal direction of inertia for <math>\Gs</math>), then there is proportionality between <math>\overline{\mathrm{H}}_\mathrm{RTG} (\Gs)</math> and <math>\velang{}{T}</math>,and the conservation of the former implies that of the latter.

	~~:<math>\left\{\overline{\mathrm{H}}_\mathrm{RTG}~~ (\Gs)\right\}=\diag{\Is_\mathrm{petit}}{\Is_\mathrm{gran}}{\Is_\mathrm{mitjà}} \vector{\Omega_1}{\Omega_2}{\Omega_3}=\vector{\Is_\mathrm{petit}\Omega_1}{\Is_\mathrm{gran}\Omega_2}{\Is_\mathrm{mitjà}\Omega_3}</math>,~~no és proporcional a <math>\velang{}{T}</math> en principi.~~

	~~:Si la velocitat angular inicial és exclusivament en una de les tres direccions (és~~ a ~~dir, si la seva direcció es~~ principal ~~d’inèrcia per a~~ <math>\Gs</math>), ~~llavors sí que hi ha proporcionalitat entre~~ <math>\overline{\mathrm{H}}_\mathrm{RTG} (\Gs)</math> i <math>\velang{}{T}</math>, ~~i la conservació del primer implica la de la segona~~.


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	</small>		</small>

	====✏️ ~~Exemple~~ D8.7: ~~giroscopi~~ ====		====✏️ EXAMPLE D8.7: gyroscope====
	------		------
	:<small>		:<small>

Bros at 17:36, 4 November 2024

2024-11-04T17:36:47Z

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	</small>		</small>

	====✏️ ~~Exemple~~ D8.5: ~~col·lisió d’una anella i un braç~~ ====		====✏️ EXAMPLE D8.5: collision of a ring and an articulated arm====
	------		------
	:<small>		:<small>
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	:A negative value for n means that the '''ICR''' is located at a distance e above <math>\Ps</math>.		:A negative value for n means that the '''ICR''' is located at a distance e above <math>\Ps</math>.

	:[[File:ExD8-5-3-eng~~-esp~~.png\|thumb\|center\|500px\|link=]]		:[[File:ExD8-5-3-eng.png\|thumb\|center\|500px\|link=]]

	:Before the collision <math>\ts_\mathrm{before}</math>, <math>\Os</math> is not a point fixed to the ring in general (it is not its '''ICR'''). Its angular momentum has to be calculated thorugh <span style="text-decoration: underline;"> [[D4. Teoremes vectorials#D4.8 Descomposició baricèntrica del moment cinètic\|'''barycentric decomposition ''']]</span>. The initial angular momentum of the arm is zero because it does not move:		:Before the collision <math>\ts_\mathrm{before}</math>, <math>\Os</math> is not a point fixed to the ring in general (it is not its '''ICR'''). Its angular momentum has to be calculated thorugh <span style="text-decoration: underline;"> [[D4. Teoremes vectorials#D4.8 Descomposició baricèntrica del moment cinètic\|'''barycentric decomposition ''']]</span>. The initial angular momentum of the arm is zero because it does not move:

Bros at 17:36, 4 November 2024

2024-11-04T17:36:03Z

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	\|		\|
	:[[File:ExD8-5-1-eng.png\|thumb\|left\|180px\|link=]]		:[[File:ExD8-5-1-eng.png\|thumb\|left\|180px\|link=]]
	\|~~L’anella~~, ~~de radi~~ L ~~i massa~~ m, ~~es mou sobre un terra~~ horizontal ~~llis amb velocitat~~ angular ~~respecte del terra~~ <math>\Omega_0=\ns\vs_0/\Ls</math> (on n ~~és un número enter~~), ~~i el seu~~ centre <math>\Ps</math> ~~s’apropa amb velocitat~~ <math>\~~vs_0~~</math> ~~cap a l’extrem~~ <math>\Qs</math> ~~del braç, de longitud~~ 2L ~~i massa~~ M, ~~que està articulat al punt~~ <math>\Os</math> ~~fix a terra i es troba inicialment en repòs~~. ~~Després de la col·lisió~~, ~~anella i braç queden enganxats~~. ~~Es negligeix la fricció associada a l’articulació a~~ <math>\Os</math>.<u> ~~Es tracta d’investigar el moviment del sistema~~ after ~~de la col·lisió~~.</u><br>		\|The ring, with radius L and mass m, moves on a smooth horizontal ground with angular velocity <math>\Omega_0=\ns\vs_0/\Ls</math> relative to the ground (where n is an integer), and its centre <math>\Ps</math> approaches the end <math>\Qs</math> of the arm with velocity <math>\vs_0</math>. The arm has a length 2L and a mass M, is articulated at point <math>\Os</math> fixed to the ground and is initially at rest. Because of the collision, the ring and arm stick together. The friction associated with the joint at <math>\Os</math> is neglected. We want to <u>investigate the motion of the system after the collision.</u><br>

	\|}		\|}
	:<u>~~Hi ha conservació de la quantitat de moviment~~?</u>.<br>		:<u>Is the linear momentum conserved?</u>.<br>

	:~~Per a cada~~ element ~~per separat~~ (~~anella i braç~~) ~~no hi ha conservació de la quantitat de moviment~~: ~~la col·lisió genera~~ forces ~~molt intenses entre elles en direcció horizontal que provoquen acceleració en els centres d’inèrcia respectius~~. ~~A més~~, ~~el braç rep una força intensa a l’articulació a~~ <math>\Os</math>.<br>		:The linear momentum of each element separately (ring and arm) is not conserved: the collision generates very intense horizontal forces between them that cause acceleration in their centers of inertia. In addition, the arm undergoes an intense force at the <math>\Os</math>-joint.<br>

	:~~Per al sistema~~ (~~anella~~ + ~~braç~~) ~~tampoc no n’hi ha~~: ~~l’articulació a~~ <math>\Os</math> ~~introdueix~~ forces ~~en el pla del moviment que provoquen acceleració en el centre d’inèrcia conjunt~~ <math>\Gs</math>.		:There is no conservation of the linear momentum for the system (ring + arm) either: the <math>\Os</math>-joint introduces forces into the plane of motion responsible for the acceleration of the system’s center of inertia <math>\Gs</math>.

	:<u>~~Conservació del moment cinètic~~?</u>		:<u>Is the angular momentum conserved?</u>

	:[[File:ExD8-5-2-eng.png\|thumb\|right\|200px\|link=]]		:[[File:ExD8-5-2-eng.png\|thumb\|right\|200px\|link=]]
	:~~Per al sistema~~ (~~anella~~ + ~~braç~~), ~~les~~ forces ~~associades~~ a ~~l’articulació provoquen~~ moment ~~extern vertical~~ (~~ortogonal al pla del moviment~~) ~~a qualsevol punt tret del punt~~ <math>\Os</math>:		:For the system (ring + arm), the forces associated with the joint yield a nonzero vertical external moment (orthogonal to the plane of motion) at any point except at <math>\Os</math>:

	:<math>\left.\sum\overline{\Ms_\mathrm{ext}}(\Os)\right]_\mathrm{vertical}=0=\left.\dot{\overline{\Hs}}_\mathrm{RTG}(\Os)\right]_\mathrm{vertical} \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTG}(\Os)\right]_\mathrm{vertical} \text{CONSTANT!}</math><br>		:<math>\left.\sum\overline{\Ms_\mathrm{ext}}(\Os)\right]_\mathrm{vertical}=0=\left.\dot{\overline{\Hs}}_\mathrm{RTG}(\Os)\right]_\mathrm{vertical} \quad \Rightarrow \quad \left.\overline{\Hs}_\mathrm{RTG}(\Os)\right]_\mathrm{vertical} \text{CONSTANT!}</math><br>

	:~~En ser~~ <math>\Os</math> ~~permanentment fix a terra~~, ~~ha de ser el~~ '''~~CIR~~''' ~~permanent respecte del terra del sòlid únic format~~ after ~~de la col·lisió~~. ~~Abans de la col·lisió~~, el '''~~CIR~~''' ~~de l’anella es troba a~~ <math>\Ls/2</math> ~~per sota del centre~~ <math>\Ps</math>:		:Since <math>\Os</math> is permanently fixed to the ground, it must be the permanent '''ICR''' relative to the ground of the rigid formed by the ring and the arm stuck together after the collision. Before the collision, the ring '''ICR''' is located at <math>\Ls/2</math> below the center <math>\Ps</math>:

	:<math>\vel{S}{T}=\vel{P}{T}+\velang{}{0}\times \PSvec = (\rightarrow \vs_0) + \left(\otimes \frac{\ns\vs_0}{\Ls}\right)\times (\downarrow \Ls)=\left[\leftarrow (\ns-1)\vs_0\right].</math>		:<math>\vel{S}{T}=\vel{P}{T}+\velang{}{0}\times \PSvec = (\rightarrow \vs_0) + \left(\otimes \frac{\ns\vs_0}{\Ls}\right)\times (\downarrow \Ls)=\left[\leftarrow (\ns-1)\vs_0\right].</math>

	:~~La distància~~ e ~~entre~~ <math>\Ps</math> ~~i el~~ '''~~CIR~~''' ~~de l’anella respecte del terra~~ before ~~de la col·lisió es pot trobar a partir de~~:		:The distance e between <math>\Ps</math> and the ring '''ICR''' relative to the ground before collision can be found from:

	:<math>\left.\begin{array}{l}		:<math>\left.\begin{array}{l}
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	\end{array}\right\} \Rightarrow \es=\frac{\Ls}{\ns} .</math>		\end{array}\right\} \Rightarrow \es=\frac{\Ls}{\ns} .</math>

	:~~Un valor negatiu de~~ n ~~voldria dir que el~~ '''~~CIR~~''' ~~es troba~~ a ~~una distància~~ e ~~pel damunt de~~ <math>\Ps</math>.		:A negative value for n means that the '''ICR''' is located at a distance e above <math>\Ps</math>.

	:[[File:ExD8-5-3-eng-esp.png\|thumb\|center\|500px\|link=]]		:[[File:ExD8-5-3-eng-esp.png\|thumb\|center\|500px\|link=]]

	:~~Abans de la col·lisió~~ <math>\ts_\mathrm{before}</math>, <math>\Os</math> ~~no pertany en~~ general ~~cinemàticament a l’anella~~ (~~no és el seu~~ '''~~CIR~~'''). ~~El càlcul del seu moment cinètic s’ha de fer per~~ <span style="text-decoration: underline;"> [[D4. Teoremes vectorials#D4.8 Descomposició baricèntrica del moment cinètic\|'''~~descomposició baricèntrica~~ ''']]</span>. ~~El moment cinètic inicial del braç és nul perquè no es mou~~:		:Before the collision <math>\ts_\mathrm{before}</math>, <math>\Os</math> is not a point fixed to the ring in general (it is not its '''ICR'''). Its angular momentum has to be calculated thorugh <span style="text-decoration: underline;"> [[D4. Teoremes vectorials#D4.8 Descomposició baricèntrica del moment cinètic\|'''barycentric decomposition ''']]</span>. The initial angular momentum of the arm is zero because it does not move:

	:<math>\overline{\Hs}_\mathrm{RTO=T}(\Os,\ts_\mathrm{before})=\left.\overline{\Hs}_\mathrm{T}(\Os,\ts_\mathrm{before})\right]_\mathrm{anella}= \left.\overline{\Hs}_\mathrm{RTP}(\Ps,\ts_\mathrm{before})\right]_\mathrm{anella} + \OPvec \times 2\ms\vel{P}{T}= </math><br>		:<math>\overline{\Hs}_\mathrm{RTO=T}(\Os,\ts_\mathrm{before})=\left.\overline{\Hs}_\mathrm{T}(\Os,\ts_\mathrm{before})\right]_\mathrm{anella}= \left.\overline{\Hs}_\mathrm{RTP}(\Ps,\ts_\mathrm{before})\right]_\mathrm{anella} + \OPvec \times 2\ms\vel{P}{T}= </math><br>
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	:<math>\hspace{3.4cm}=\Is_\mathrm{P} \left(\otimes \frac{\ns\vs_0}{\Ls} \right) + \left(\downarrow 2\Ls \right)\times \ms\left(\rightarrow \vs_0 \right) = \left(\otimes \ms\Ls^2 \frac{\ns\vs_0}{\Ls}\right) + \left(\odot 2\ms\Ls\vs_0 \right) = \left[ \otimes (\ns-2)\ms\Ls\vs_0 \right]</math><br>		:<math>\hspace{3.4cm}=\Is_\mathrm{P} \left(\otimes \frac{\ns\vs_0}{\Ls} \right) + \left(\downarrow 2\Ls \right)\times \ms\left(\rightarrow \vs_0 \right) = \left(\otimes \ms\Ls^2 \frac{\ns\vs_0}{\Ls}\right) + \left(\odot 2\ms\Ls\vs_0 \right) = \left[ \otimes (\ns-2)\ms\Ls\vs_0 \right]</math><br>

	:~~Després de la col·lisió~~:<br>		:After the collision:<br>

	:<math>\overline{\Hs}_\mathrm{RTO}(\Os,\ts_\mathrm{after})= \Is_\mathrm{O}\overline{\Omega}_\Ts=\left( \Is_\mathrm{O}^\mathrm{anella}+ \Is_\mathrm{O}^\mathrm{braç} \right) \overline{\Omega}_\Ts= \left( \Is_\mathrm{G}^\mathrm{anella}+ \Is_\mathrm{O}^{\mathrm{anella}\otimes}+\Is_\mathrm{O}^\mathrm{braç} \right) \overline{\Omega}_\Ts = \left[ \ms\Ls^2 + \ms(2\Ls)^2+ \frac{4}{3} \ms\Ls^2\right]\overline{\Omega}_\Ts</math>		:<math>\overline{\Hs}_\mathrm{RTO}(\Os,\ts_\mathrm{after})= \Is_\mathrm{O}\overline{\Omega}_\Ts=\left( \Is_\mathrm{O}^\mathrm{anella}+ \Is_\mathrm{O}^\mathrm{braç} \right) \overline{\Omega}_\Ts= \left( \Is_\mathrm{G}^\mathrm{anella}+ \Is_\mathrm{O}^{\mathrm{anella}\otimes}+\Is_\mathrm{O}^\mathrm{braç} \right) \overline{\Omega}_\Ts = \left[ \ms\Ls^2 + \ms(2\Ls)^2+ \frac{4}{3} \ms\Ls^2\right]\overline{\Omega}_\Ts</math>
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	\end{array}\right\} \Rightarrow \bar{\Omega}_{\mathrm{T}}=\left[\otimes(\mathrm{n}-2) \frac{19}{3} \frac{\mathrm{v}_0}{\mathrm{~L}}\right]</math>		\end{array}\right\} \Rightarrow \bar{\Omega}_{\mathrm{T}}=\left[\otimes(\mathrm{n}-2) \frac{19}{3} \frac{\mathrm{v}_0}{\mathrm{~L}}\right]</math>

	:~~Per a~~ n>2 , ~~el sistema gira en sentit horari~~. ~~Per a valors de~~ n<2, ~~el sistema gira en sentit antihorari~~. ~~Per a~~ n=2 , ~~el sistema queda en repòs~~.		:For <math>n>2</math>, the system rotation is clockwise. For <math>n<2</math>, it is counterclockwise. For <math>n=2</math>, the system is at rest.
	:'''ANIMACIONS'''		:'''ANIMACIONS'''
	</small>		</small>

	====✏️ ~~Exemple~~ D8.6: ~~sòlid lliure a l’espai~~ ====		====✏️ EXAMPLE D8.6: free rigid body in space====
	------		------
	:<small>		:<small>
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	\|		\|
	:[[File:ExD8-6-1-neut.png\|thumb\|left\|180px\|link=]]		:[[File:ExD8-6-1-neut.png\|thumb\|left\|180px\|link=]]
	\|~~El sòlid està format per dues plaques homogènies~~, ~~de la mateixa massa i alçària però amplàries diferents~~, ~~enganxades pel punt~~ <math>\Os</math>.~~<u> Es tracta d’investigar si alguna magnitud dinàmica es conserva quan es llança enlaire~~.~~</u>Es negligeixen les interaccions aerodinàmiques~~.<br>		\|The rigid body consists of two homogeneous plates, with the same mass and height but different widths, glued at point <math>\Os</math>. We want to investigate whether any dynamic magnitude is conserved when it is launched in the air. Aerodynamic interactions are neglected.<br>

	\|}		\|}

Bros at 17:26, 4 November 2024

2024-11-04T17:26:47Z

← Older revision		Revision as of 17:26, 4 November 2024
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	:After collision <math>\ts_\mathrm{after}</math>, <math>\left.\overline{\mathrm{QM}}\right]_\mathrm{horizontal}</math> of the system has to be zero, therefore the velocity of the system’s <span style="text-decoration: underline;"> [[D5. Geometria de masses#D5.1 Centre d'inèrcia\|'''center of inertia ''']]</span> is also zero: <math>\overline{\vs}_\Ts(\Gs,\ts_\mathrm{after})</math>. Hence, after collision, the rigid body formed by the two bars will have the '''ICR''' relative to the ground permanently located at <math>\Gs</math>:		:After collision <math>\ts_\mathrm{after}</math>, <math>\left.\overline{\mathrm{QM}}\right]_\mathrm{horizontal}</math> of the system has to be zero, therefore the velocity of the system’s <span style="text-decoration: underline;"> [[D5. Geometria de masses#D5.1 Centre d'inèrcia\|'''center of inertia ''']]</span> is also zero: <math>\overline{\vs}_\Ts(\Gs,\ts_\mathrm{after})</math>. Hence, after collision, the rigid body formed by the two bars will have the '''ICR''' relative to the ground permanently located at <math>\Gs</math>:
	[[File:ExD8-4-2-~~cat~~.png\|thumb\|right\|350px\|link=]]		[[File:ExD8-4-2-eng.png\|thumb\|right\|350px\|link=]]

	:Position of the inertia center <math>\Gs</math>: on the line <math>\Ps</math> and <math>\Qs</math>, at a distance 4L below <math>\Qs</math>.<br>		:Position of the inertia center <math>\Gs</math>: on the line <math>\Ps</math> and <math>\Qs</math>, at a distance 4L below <math>\Qs</math>.<br>
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	:[[File:ExD8-5-1-~~cat~~.png\|thumb\|left\|180px\|link=]]		:[[File:ExD8-5-1-eng.png\|thumb\|left\|180px\|link=]]
	\|L’anella, de radi L i massa m, es mou sobre un terra horizontal llis amb velocitat angular respecte del terra <math>\Omega_0=\ns\vs_0/\Ls</math> (on n és un número enter), i el seu centre <math>\Ps</math> s’apropa amb velocitat <math>\vs_0</math> cap a l’extrem <math>\Qs</math> del braç, de longitud 2L i massa M, que està articulat al punt <math>\Os</math> fix a terra i es troba inicialment en repòs. Després de la col·lisió, anella i braç queden enganxats. Es negligeix la fricció associada a l’articulació a <math>\Os</math>.<u> Es tracta d’investigar el moviment del sistema after de la col·lisió.</u><br>		\|L’anella, de radi L i massa m, es mou sobre un terra horizontal llis amb velocitat angular respecte del terra <math>\Omega_0=\ns\vs_0/\Ls</math> (on n és un número enter), i el seu centre <math>\Ps</math> s’apropa amb velocitat <math>\vs_0</math> cap a l’extrem <math>\Qs</math> del braç, de longitud 2L i massa M, que està articulat al punt <math>\Os</math> fix a terra i es troba inicialment en repòs. Després de la col·lisió, anella i braç queden enganxats. Es negligeix la fricció associada a l’articulació a <math>\Os</math>.<u> Es tracta d’investigar el moviment del sistema after de la col·lisió.</u><br>

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	:<u>Conservació del moment cinètic?</u>		:<u>Conservació del moment cinètic?</u>

	:[[File:ExD8-5-2-~~cat~~.png\|thumb\|right\|200px\|link=]]		:[[File:ExD8-5-2-eng.png\|thumb\|right\|200px\|link=]]
	:Per al sistema (anella + braç), les forces associades a l’articulació provoquen moment extern vertical (ortogonal al pla del moviment) a qualsevol punt tret del punt <math>\Os</math>:		:Per al sistema (anella + braç), les forces associades a l’articulació provoquen moment extern vertical (ortogonal al pla del moviment) a qualsevol punt tret del punt <math>\Os</math>:

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	:Un valor negatiu de n voldria dir que el '''CIR''' es troba a una distància e pel damunt de <math>\Ps</math>.		:Un valor negatiu de n voldria dir que el '''CIR''' es troba a una distància e pel damunt de <math>\Ps</math>.

	:[[File:ExD8-5-3-~~cat~~-esp.png\|thumb\|center\|500px\|link=]]		:[[File:ExD8-5-3-eng-esp.png\|thumb\|center\|500px\|link=]]

	:Abans de la col·lisió <math>\ts_\mathrm{before}</math>, <math>\Os</math> no pertany en general cinemàticament a l’anella (no és el seu '''CIR'''). El càlcul del seu moment cinètic s’ha de fer per <span style="text-decoration: underline;"> [[D4. Teoremes vectorials#D4.8 Descomposició baricèntrica del moment cinètic\|'''descomposició baricèntrica ''']]</span>. El moment cinètic inicial del braç és nul perquè no es mou:		:Abans de la col·lisió <math>\ts_\mathrm{before}</math>, <math>\Os</math> no pertany en general cinemàticament a l’anella (no és el seu '''CIR'''). El càlcul del seu moment cinètic s’ha de fer per <span style="text-decoration: underline;"> [[D4. Teoremes vectorials#D4.8 Descomposició baricèntrica del moment cinètic\|'''descomposició baricèntrica ''']]</span>. El moment cinètic inicial del braç és nul perquè no es mou:
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	:[[File:ExD8-7-1-~~cat~~.png\|thumb\|left\|180px\|link=]]		:[[File:ExD8-7-1-eng.png\|thumb\|left\|180px\|link=]]
	\|El sistema consta d’un disc homogeni, de massa m i radi r, que està articulat a un suport de massa negligible, i d’una forquilla que pot girar lliurement al voltant de l’eix vertical. Entre suport i forquilla hi ha un motor. El moment d’inèrcia del suport respecte de l’eix vertical que passa pel centre del disc és <math>\Is=(\lambda/2)\ms\rs^2</math>. Inicialment <math>(\ts_\mathrm{incial}</math> el disc està paral·lel a terra, i gira amb velocitat angular vertical <math>\velang{disc}{T}=\psio</math> . Es negligeixen les friccions associades a totes les articulacions. <u> Es tracta d’investigar com es mou el suport quan el motor canvia l’orientació del disc respecte del terra .</u>		\|El sistema consta d’un disc homogeni, de massa m i radi r, que està articulat a un suport de massa negligible, i d’una forquilla que pot girar lliurement al voltant de l’eix vertical. Entre suport i forquilla hi ha un motor. El moment d’inèrcia del suport respecte de l’eix vertical que passa pel centre del disc és <math>\Is=(\lambda/2)\ms\rs^2</math>. Inicialment <math>(\ts_\mathrm{incial}</math> el disc està paral·lel a terra, i gira amb velocitat angular vertical <math>\velang{disc}{T}=\psio</math> . Es negligeixen les friccions associades a totes les articulacions. <u> Es tracta d’investigar com es mou el suport quan el motor canvia l’orientació del disc respecte del terra .</u>

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	:[[File:ExD8-8-1-~~cat~~.png\|thumb\|left\|180px\|link=]]		:[[File:ExD8-8-1-eng.png\|thumb\|left\|180px\|link=]]
	\|La barra '''PQ''', homogènia i de massa m, es mou mantenint els seus dos extrems dins d’una guia llisa <math>(\mu=0)</math> , de radi r i massa negligible, que pot girar lliurement al voltant de la direcció vertical. L’angle POQ és de <math>120^o</math>. Es tracta de determinar <u>l’equació del moviment per a la coordenada</u> <math>\psi</math> . Es negligeix la rotació de la barra sobre el seu eix (rotació pròpia <math>\dot{\varphi}</math> ).		\|La barra '''PQ''', homogènia i de massa m, es mou mantenint els seus dos extrems dins d’una guia llisa <math>(\mu=0)</math> , de radi r i massa negligible, que pot girar lliurement al voltant de la direcció vertical. L’angle POQ és de <math>120^o</math>. Es tracta de determinar <u>l’equació del moviment per a la coordenada</u> <math>\psi</math> . Es negligeix la rotació de la barra sobre el seu eix (rotació pròpia <math>\dot{\varphi}</math> ).

Bros at 17:22, 4 November 2024

2024-11-04T17:22:50Z

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