C3. Composition of movements - Revision history

Bros: /* ✏️ EXAMPLE C3-1.4: ferris wheel */

2025-05-08T12:14:11Z

✏️ EXAMPLE C3-1.4: ferris wheel

← Older revision		Revision as of 12:14, 8 May 2025
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	::The velocity of point <math>\Qs</math>, fixed to the ground, relative to the cabin, is:		::The velocity of point <math>\Qs</math>, fixed to the ground, relative to the cabin, is:

	::If <math>\textrm{AB} = \textrm{ground}</math> <math>(\Rs</math>) i <math>\textrm{REL} = \textrm{cabin}</math>		::If <math>\textrm{AB} = \textrm{ground}</math> <math>(\Rs</math>) and <math>\textrm{REL} = \textrm{cabin}</math>

	::<math>\velQrel=\velQab-\vel{Q}{tr}=-\velQar</math>		::<math>\velQrel=\velQab-\vel{Q}{tr}=-\velQar</math>

Bros: /* ✏️ EXAMPLE C3-1.4: ferris wheel */

2024-11-09T19:53:35Z

✏️ EXAMPLE C3-1.4: ferris wheel

← Older revision		Revision as of 19:53, 9 November 2024
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	::The velocity of point <math>\Qs</math>, fixed to the ground, relative to the cabin, is:		::The velocity of point <math>\Qs</math>, fixed to the ground, relative to the cabin, is:
	~~'''Revisar versio catala'''~~
	~~::If <math>\textrm{AB} = \textrm{terra}</math> <math>(\Rs</math>) i <math>\textrm{REL} = \textrm{cabina}</math>~~

	::<math>\~~velQrel~~=\~~velQab-~~\~~velQar~~=-\~~velQar~~</math>		::If <math>\textrm{AB} = \textrm{ground}</math> <math>(\Rs</math>) i <math>\textrm{REL} = \textrm{cabin}</math>

	::~~Si imaginem que~~ <math>\Qs</math> ~~were~~ fixed to the cabin, its motion relative to the ground would be identical to that of <math>\Ps</math> because the cabin does not rotate relative to the ground: circular, with radius r and value <math>\rs\Omega_0</math>. What is different is the location of the center of curvature: for the <math>\Ps</math> motion, it is point <math>\Os</math>; for the <math>\Qs</math> transportation motion, it is at a distance R from <math>\Qs</math> to the right. Hence:		::<math>\velQrel=\velQab-\vel{Q}{tr}=-\velQar</math>
	<center><math>\velQrel=-\~~velQar~~=-(\downarrow\rs\Omega_0)=\uparrow\rs\Omega_0</math></center>
			::If we imagine <math>\Qs</math> fixed to the cabin, its motion relative to the ground would be identical to that of <math>\Ps</math> because the cabin does not rotate relative to the ground: circular, with radius r and value <math>\rs\Omega_0</math>. What is different is the location of the center of curvature: for the <math>\Ps</math> motion, it is point <math>\Os</math>; for the <math>\Qs</math> transportation motion, it is at a distance R from <math>\Qs</math> to the right. Hence:
			<center><math>\velQrel=-\vel{Q}{tr}=-(\downarrow\rs\Omega_0)=\uparrow\rs\Omega_0</math></center>
	</small>		</small>

Bros at 18:26, 4 November 2024

2024-11-04T18:26:47Z

← Older revision		Revision as of 18:26, 4 November 2024
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	\newcommand{\velQab}{\vel{Q}{AB}}		\newcommand{\velQab}{\vel{Q}{AB}}
	\newcommand{\velQrel}{\vel{Q}{REL}}		\newcommand{\velQrel}{\vel{Q}{REL}}
	\newcommand{\velQar}{\vel{Q}{ar}}		\newcommand{\velQar}{\vel{Q}{tr}}
	\newcommand{\psio}{\dot{\psi_0}}		\newcommand{\psio}{\dot{\psi_0}}
	\newcommand{\accrel}[1]{\vecbf{a}_{\textrm{REL}} (\textbf{#1})}		\newcommand{\accrel}[1]{\vecbf{a}_{\textrm{REL}} (\textbf{#1})}
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	<center><math>\vel{Q}{AB} = \vel{Q}{REL}+\vel{Q}{ar} </math></center>		<center><math>\vel{Q}{AB} = \vel{Q}{REL}+\vel{Q}{tr} </math></center>

	The second term in the right hand side is the '''transportation velocity''', and it corresponds to the velocity of <math>\Qs</math> at that precise time instant relative to AB if it were fixed to REL (at the same position it has at that time instant):		The second term in the right hand side is the '''transportation velocity''', and it corresponds to the velocity of <math>\Qs</math> at that precise time instant relative to AB if it were fixed to REL (at the same position it has at that time instant):
	<center><math> \vel{Q}{ar} = \overline{\textbf{v}}_{\textrm{AB}}(\Qs \in \textrm{REL})</math></center>		<center><math> \vel{Q}{tr} = \overline{\textbf{v}}_{\textrm{AB}}(\Qs \in \textrm{REL})</math></center>

	<div>		<div>
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	</math>		</math>
	</center>		</center>
	:Though the last equation is correct (it has been proved!), it contains two terms with <math>\Orel</math>, a point not univocally defined (it may be any point fixed in REL). That drawback can be overcome if we introduce the transportation motion: <math>\vel{Q}{ar} = \vvec_\textrm{AB}(\Qs \in \textrm{REL})</math>. If we imagine that <math>\Qs</math> is a point fixed in REL, its velocity relative to AB is:		:Though the last equation is correct (it has been proved!), it contains two terms with <math>\Orel</math>, a point not univocally defined (it may be any point fixed in REL). That drawback can be overcome if we introduce the transportation motion: <math>\vel{Q}{tr} = \vvec_\textrm{AB}(\Qs \in \textrm{REL})</math>. If we imagine that <math>\Qs</math> is a point fixed in REL, its velocity relative to AB is:
	<center>		<center>
	<math>\vvec_\textrm{AB}(\Qs \in \textrm{REL}) = \vvec_\textrm{REL}(\Qs \in \textrm{REL}) + \vvec_\textrm{AB}(\Orel) + \omegarelab \times \OQrelvec = \vvec_\textrm{AB}(\Orel) + \omegarelab \times \OQrelvec</math>		<math>\vvec_\textrm{AB}(\Qs \in \textrm{REL}) = \vvec_\textrm{REL}(\Qs \in \textrm{REL}) + \vvec_\textrm{AB}(\Orel) + \omegarelab \times \OQrelvec = \vvec_\textrm{AB}(\Orel) + \omegarelab \times \OQrelvec</math>
	</center>		</center>

	:Finally, <math>\vel{Q}{AB}=\vel{Q}{REL}+\vel{Q}{ar}</math>		:Finally, <math>\vel{Q}{AB}=\vel{Q}{REL}+\vel{Q}{tr}</math>
	</small>		</small>
	</div>		</div>


	The equation of composition of velocities implies an instantaneous operation between vectors: <math>\velQab</math> (or <math>\velQrel</math>) eat time t is obtained from two vectors, <math>\velQrel</math> (or <math>\velQab</math>) and <math>\vel{Q}{ar}</math> at the same time t. That mathematical operation is simpler than the time derivative (which is not instantaneous, as it calls for the position vector at two close time instants to obtain the velocity). As will be seen in <span style="text-decoration: underline;">[[C3. Composition of movements#✏️ EXAMPLE C3-1.1: rotating ring\|'''example C3-1.1''']]</span>, time t may correspond to a particular configuration or to a generic configuration.		The equation of composition of velocities implies an instantaneous operation between vectors: <math>\velQab</math> (or <math>\velQrel</math>) eat time t is obtained from two vectors, <math>\velQrel</math> (or <math>\velQab</math>) and <math>\vel{Q}{tr}</math> at the same time t. That mathematical operation is simpler than the time derivative (which is not instantaneous, as it calls for the position vector at two close time instants to obtain the velocity). As will be seen in <span style="text-decoration: underline;">[[C3. Composition of movements#✏️ EXAMPLE C3-1.1: rotating ring\|'''example C3-1.1''']]</span>, time t may correspond to a particular configuration or to a generic configuration.

	In <span style="text-decoration: underline;">[[C3. Composition of movements#C3.3 Composició de moviments VS derivació temporal\|'''section C3.3''']]</span> we will comment farther on the differences between the composition of movements and the time derivative of vectors to calculate velocities and accelerations.		In <span style="text-decoration: underline;">[[C3. Composition of movements#C3.3 Composició de moviments VS derivació temporal\|'''section C3.3''']]</span> we will comment farther on the differences between the composition of movements and the time derivative of vectors to calculate velocities and accelerations.
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	::The time instant shown in the figure is generic as the <math>\theta</math> coordinate does not have a precise numerical value. The velocity of point <math>\Qs</math> relative to the ground can be obtained in a straightforward way through a composition:		::The time instant shown in the figure is generic as the <math>\theta</math> coordinate does not have a precise numerical value. The velocity of point <math>\Qs</math> relative to the ground can be obtained in a straightforward way through a composition:
	<center>		<center>
	<math>\vel{Q}{AB}=\vel{Q}{REL}+\vel{Q}{ar}		<math>\vel{Q}{AB}=\vel{Q}{REL}+\vel{Q}{tr}
	</math></center>		</math></center>
	::The motion <math>\Qs</math> relative to REL is circular with radius r, and the velocity is tangent to the guide. If <math>\Qs</math> were fixed to the guide, its motion relative to the ground (transportation motion) would be circular, with radius <math>\abs{\OQvec} \equiv \rho</math> and center <math>\Os</math>, and the velocity would be perpendicular to <math>\OQvec</math> with value <math>\rho\dot\psi</math> . As the <math>\OQvec</math> direction (hence the direction perpendicular to <math>\OQvec</math>)do not correspond to the directions suggested by the system (as that of the ring handle, or the <math>\vecbf{CQ}</math> direction), it is better to describe the transportation velocity as the addition of two vectors:		::The motion <math>\Qs</math> relative to REL is circular with radius r, and the velocity is tangent to the guide. If <math>\Qs</math> were fixed to the guide, its motion relative to the ground (transportation motion) would be circular, with radius <math>\abs{\OQvec} \equiv \rho</math> and center <math>\Os</math>, and the velocity would be perpendicular to <math>\OQvec</math> with value <math>\rho\dot\psi</math> . As the <math>\OQvec</math> direction (hence the direction perpendicular to <math>\OQvec</math>)do not correspond to the directions suggested by the system (as that of the ring handle, or the <math>\vecbf{CQ}</math> direction), it is better to describe the transportation velocity as the addition of two vectors:
	<center><math>		<center><math>
	\vel{Q}{ar} = \vvec_\textrm{AB}(\Qs \in \textrm{REL}) = \vec{\dot\psi} \times \OQvec = \vec{\dot\psi} \times \vecbf{OC} + \vec{\dot\psi} \times \vecbf{CQ}		\vel{Q}{tr} = \vvec_\textrm{AB}(\Qs \in \textrm{REL}) = \vec{\dot\psi} \times \OQvec = \vec{\dot\psi} \times \vecbf{OC} + \vec{\dot\psi} \times \vecbf{CQ}
	</math></center>		</math></center>

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	::At the time instant shown in the figure, the velocity of point <math>\Qs</math> relative to the ground is straightforward through composition:		::At the time instant shown in the figure, the velocity of point <math>\Qs</math> relative to the ground is straightforward through composition:
	<center><math>\vel{Q}{AB} = \velQrel + \vel{Q}{ar}</math></center>		<center><math>\vel{Q}{AB} = \velQrel + \vel{Q}{tr}</math></center>
	::The motion of <math>\Qs</math> with respect to REL is circular with radius R at all times. At the time instant shown in the figure, <math>\velQrel</math> is vertical pointing downwards and with value <math>\Rs\omega_0</math>.		::The motion of <math>\Qs</math> with respect to REL is circular with radius R at all times. At the time instant shown in the figure, <math>\velQrel</math> is vertical pointing downwards and with value <math>\Rs\omega_0</math>.
	::The transportation motion, at that particular time instant, is zero: if <math>\Qs</math> were fixed to the support, as it is located precisely on the rotation axis, its instantaneous velocity would be zero <math>(\vel{Q}{ar} = \vec{0})</math> . Hence:		::The transportation motion, at that particular time instant, is zero: if <math>\Qs</math> were fixed to the support, as it is located precisely on the rotation axis, its instantaneous velocity would be zero <math>(\vel{Q}{tr} = \vec{0})</math> . Hence:
	<center><math>\vel{Q}{AB} = \velQrel + \vel{Q}{ar}=\velQrel = \downarrow \Rs\omega_0</math></center>		<center><math>\vel{Q}{AB} = \velQrel + \vel{Q}{tr}=\velQrel = \downarrow \Rs\omega_0</math></center>
	::If the <math>\Qs</math> location were diametrically opposite to the one shown in the figure, the transportation motion would be circular on a horizontal plane, with radius 2R and center of curvature on the rotation axis pf the support.		::If the <math>\Qs</math> location were diametrically opposite to the one shown in the figure, the transportation motion would be circular on a horizontal plane, with radius 2R and center of curvature on the rotation axis pf the support.
	::The configuration considered in this example is not generic, as it corresponds just to the time instant when <math>\Qs</math> is located on the support rotation axis. For that reason, it does not contain the information along time needed to calculate <math>\acc{Q}{AB}</math> as time derivative of the velocity <math>\velQab</math> that has been obtained.		::The configuration considered in this example is not generic, as it corresponds just to the time instant when <math>\Qs</math> is located on the support rotation axis. For that reason, it does not contain the information along time needed to calculate <math>\acc{Q}{AB}</math> as time derivative of the velocity <math>\velQab</math> that has been obtained.
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	::For the time instant shown in the figure, the velocity of point <math>\Qs</math> relative to vehicle VP is straightforward through composition:		::For the time instant shown in the figure, the velocity of point <math>\Qs</math> relative to vehicle VP is straightforward through composition:
	::Si <math>\textrm{AB} = \Rs</math> i <math>\textrm{REL} = \textrm{VP}</math>		::Si <math>\textrm{AB} = \Rs</math> i <math>\textrm{REL} = \textrm{VP}</math>
	<center><math>\velQrel=\velQab-\vel{Q}{ar} = (\rightarrow\vs_0)-\vel{Q}{ar}</math></center>		<center><math>\velQrel=\velQab-\vel{Q}{tr} = (\rightarrow\vs_0)-\vel{Q}{tr}</math></center>
	::Vehicle VP moves with a simple rotation relative to the ground. From a kinematical point of view, it is totally equivalent to a rotating platform with any radius with center <math>\Os</math> fixed to the ground.		::Vehicle VP moves with a simple rotation relative to the ground. From a kinematical point of view, it is totally equivalent to a rotating platform with any radius with center <math>\Os</math> fixed to the ground.
	::If <math>\Qs</math> were fixed to the platform, its motion relative to the ground would be circular, with center <math>\Os</math> and speed twice that of <math>\Ps</math>, as it is at a distance 2r from <math>\Os</math>: <math>\vel{Q}{ar}=(\leftarrow 2\vs_0)</math>.		::If <math>\Qs</math> were fixed to the platform, its motion relative to the ground would be circular, with center <math>\Os</math> and speed twice that of <math>\Ps</math>, as it is at a distance 2r from <math>\Os</math>: <math>\vel{Q}{tr}=(\leftarrow 2\vs_0)</math>.
	::Finally:		::Finally:
	<center><math>\velQrel= (\rightarrow\vs_0)-(\leftarrow 2\vs_0)=\rightarrow 3\vs_0</math></center>		<center><math>\velQrel= (\rightarrow\vs_0)-(\leftarrow 2\vs_0)=\rightarrow 3\vs_0</math></center>
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	\dert{(\velang{REL}{AB}\times\OQrelvec)}{AB}		\dert{(\velang{REL}{AB}\times\OQrelvec)}{AB}
	& =\dert{(\velang{}{tr}\times\OQrelvec)}{AB}=		& =\dert{(\velang{}{tr}\times\OQrelvec)}{AB}=
	\dert{\velang{}{tr}}{AB}\times\OQrelvec+\velang{ }{ar}\times\dert{\OQrelvec}{AB}= \\		\dert{\velang{}{tr}}{AB}\times\OQrelvec+\velang{ }{tr}\times\dert{\OQrelvec}{AB}= \\
	& = \accang{}{tr}\times\OQrelvec+\velang{}{tr}\times\left(\dert{\OQrelvec}{REL}+\velang{}{ar}\times\OQrelvec\right)= \\		& = \accang{}{tr}\times\OQrelvec+\velang{}{tr}\times\left(\dert{\OQrelvec}{REL}+\velang{}{tr}\times\OQrelvec\right)= \\
	& = \accang{}{tr}\times\OQrelvec+\velang{}{tr}\times\vel{Q}{REL}+\velang{}{tr}\times\left(\velang{}{tr}\times\OQrelvec\right)		& = \accang{}{tr}\times\OQrelvec+\velang{}{tr}\times\vel{Q}{REL}+\velang{}{tr}\times\left(\velang{}{tr}\times\OQrelvec\right)
	\end{align}\\		\end{align}\\
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	:<math>\accrel{Q}=\accs{Q}{REL}=[\leftarrow 2\Ls(\ddot{\theta}\stheta+\dot{\theta}^2\ctheta)]</math>		:<math>\accrel{Q}=\accs{Q}{REL}=[\leftarrow 2\Ls(\ddot{\theta}\stheta+\dot{\theta}^2\ctheta)]</math>

	:<math>\acc{Q}{tr}=\acc{$\Qs_{\in REL}$}{AB}=\accs{Q}{ar}+\accn{Q}{ar}=(\otimes 2\Ls\ddot{\psi}\ctheta) + (\leftarrow 2\Ls\dot{\psi}^2 \ctheta)</math>		:<math>\acc{Q}{tr}=\acc{$\Qs_{\in REL}$}{AB}=\accs{Q}{tr}+\accn{Q}{tr}=(\otimes 2\Ls\ddot{\psi}\ctheta) + (\leftarrow 2\Ls\dot{\psi}^2 \ctheta)</math>

	:<math>\acc{Q}{Cor}=2\velang{REL}{AB}\times\vel{Q}{REL}=2(\Uparrow\dot{\psi})\times(\leftarrow 2\Ls\dot{\theta}\stheta)=(\odot 4\Ls\dot{\psi}\dot{\theta}\stheta)</math>		:<math>\acc{Q}{Cor}=2\velang{REL}{AB}\times\vel{Q}{REL}=2(\Uparrow\dot{\psi})\times(\leftarrow 2\Ls\dot{\theta}\stheta)=(\odot 4\Ls\dot{\psi}\dot{\theta}\stheta)</math>

Apons: /* 2. Find the velocity and the acceleration of point G relative to the ground */

2024-09-17T21:31:42Z

2. Find the velocity and the acceleration of point G relative to the ground

← Older revision		Revision as of 21:31, 17 September 2024
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	:<math>\acc{G}{Cor'}=2\velang{REL'}{AB'}\times\vel{G}{REL'}=2(\Uparrow\psio)\times\left[(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})\right]=(\otimes 2\Ls\psio\dot{\theta}\ctheta)</math>		:<math>\acc{G}{Cor'}=2\velang{REL'}{AB'}\times\vel{G}{REL'}=2(\Uparrow\psio)\times\left[(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})\right]=(\otimes 2\Ls\psio\dot{\theta}\ctheta)</math>

	:Hence, <math>\acc{G}{E}\equiv\acc{G}{AB'} = (\nearrow\Ls(\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\downarrow\ddot{x})~~+</math>~~		:Hence, <math>\acc{G}{E}\equiv\acc{G}{AB'} = (\nearrow\Ls(\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\downarrow\ddot{x})+(\leftarrow\Ls\psio^2\ctheta)+(\otimes 2\Ls\psio\dot{\theta}\ctheta)</math>
	~~:<math>~~+(\leftarrow\Ls\psio^2\ctheta)+(\otimes 2\Ls\psio\dot{\theta}\ctheta)</math>

	:The projection of those results in the B vector basis fixed to the ring is:		:The projection of those results in the B vector basis fixed to the ring is:

Apons: /* 2. Determina la velocitat i acceleració del punt G respecte del terra */

2024-09-17T21:30:10Z

2. Determina la velocitat i acceleració del punt G respecte del terra

← Older revision		Revision as of 21:30, 17 September 2024
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	</div>		</div>

	===== 2. ~~Determina la velocitat i acceleració del punt~~ G ~~respecte del <span style="text-decoration: underline;">terra</span>~~=====		===== 2. Find the velocity and the acceleration of point G relative to the ground =====
	<div>		<div>
	~~:A partir del resultat anterior~~, ~~es pot obtenir la cinemàtica de~~ <math>\Gs</math> <span style="text-decoration: underline;">~~respecte del terra~~</span> ~~amb una segona composició on~~ '''AB’<math>\equiv</math>~~terra~~''' i '''REL’<math>\equiv</math>~~guia~~'''. ~~El moviment d’arrossegament~~ (~~el que tindria <math>\Gs</math> si es fixés~~ a ~~la guia i es mirés des del terra~~) ~~és senzill~~: circular, ~~de radi~~ <math>\Ls\ctheta</math> ~~i velocitat~~ angular ~~associada~~ <math>\psio</math>.		From the previous results, the <math>\Gs</math> kinematics <span style="text-decoration: underline;">relative to the ground</span> may be obtained through a second composition with '''AB’<math>\equiv</math> ground''' and '''REL’<math>\equiv</math> guide'''. The transportation motion (its motion relative to the ground if it were a point fixed to the guide) is simple: circular, with radius <math>\Ls\ctheta</math> and associated angular velocity <math>\psio</math>.

	:<math>\vel{G}{AB'}=\vel{G}{REL'}+\vel{G}{ar'}</math>		:<math>\vel{G}{AB'}=\vel{G}{REL'}+\vel{G}{tr'}</math>

	:<math>\vel{G}{REL'}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})</math>		:<math>\vel{G}{REL'}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})</math>

	:<math>\vel{G}{ar'}=\vel{$\Gs_{\in\Rs\Es\Ls'}$}{AB'}=(\otimes\Ls\psio\ctheta)</math>		:<math>\vel{G}{tr'}=\vel{$\Gs_{\in\Rs\Es\Ls'}$}{AB'}=(\otimes\Ls\psio\ctheta)</math>

	:~~Per tant~~, <math>\vel{G}{T}\equiv\vel{G}{AB'}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})+(\otimes\Ls\psio\ctheta)</math>		:Hence, <math>\vel{G}{E}\equiv\vel{G}{AB'}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})+(\otimes\Ls\psio\ctheta)</math>

	[[~~Fitxer~~:C3-E-Ex3-5-~~cat-esp~~.png\|thumb\|right\|300px\|link=]]		[[File:C3-E-Ex3-5-eng.png\|thumb\|right\|300px\|link=]]

	:~~El moviment d’arrossegament, que és~~ circular ~~amb celeritat~~ constant~~, només té~~ component ~~normal d’acceleració~~.		:The acceleration of the transportation motion (which is circular with constant speed) has just a normal component.

	:<math>\acc{G}{AB'}=\acc{G}{REL'}+\acc{G}{ar'}+\acc{G}{Cor'}</math>		:<math>\acc{G}{AB'}=\acc{G}{REL'}+\acc{G}{tr'}+\acc{G}{Cor'}</math>

	:<math>\acc{G}{REL'}=(\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\downarrow\ddot{x})</math>		:<math>\acc{G}{REL'}=(\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\downarrow\ddot{x})</math>

	:<math>\acc{G}{ar'}=\acc{$\Gs_{\in\Rs\Es\Ls'}$}{AB'}=(\leftarrow\Ls\psio^2\ctheta)</math>		:<math>\acc{G}{tr'}=\acc{$\Gs_{\in\Rs\Es\Ls'}$}{AB'}=(\leftarrow\Ls\psio^2\ctheta)</math>

	:<math>\acc{G}{Cor'}=2\velang{REL'}{AB'}\times\vel{G}{REL'}=2(\Uparrow\psio)\times\left[(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})\right]=(\otimes 2\Ls\psio\dot{\theta}\ctheta)</math>		:<math>\acc{G}{Cor'}=2\velang{REL'}{AB'}\times\vel{G}{REL'}=2(\Uparrow\psio)\times\left[(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})\right]=(\otimes 2\Ls\psio\dot{\theta}\ctheta)</math>

	:~~Per tant~~, <math>\acc{G}{T}\equiv\acc{G}{AB'} = (\nearrow\Ls(\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\downarrow\ddot{x})+</math>		:Hence, <math>\acc{G}{E}\equiv\acc{G}{AB'} = (\nearrow\Ls(\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\downarrow\ddot{x})+</math>
	:<math>+(\leftarrow\Ls\psio^2\ctheta)+(\otimes 2\Ls\psio\dot{\theta}\ctheta)</math>		:<math>+(\leftarrow\Ls\psio^2\ctheta)+(\otimes 2\Ls\psio\dot{\theta}\ctheta)</math>

	:~~La projecció d’aquests resultats a la base~~ B ~~fixa a l’anella és~~:		:The projection of those results in the B vector basis fixed to the ring is:

	:<math>\braq{\vel{G}{T}}{B}=\vector{-\dot{x}\stheta+\Ls\dot{\theta}}{-\dot{x}\ctheta}{-\Ls\psio\stheta}</math>		:<math>\braq{\vel{G}{E}}{B}=\vector{-\dot{x}\stheta+\Ls\dot{\theta}}{-\dot{x}\ctheta}{-\Ls\psio\stheta}</math>

	:<math>\braq{\acc{G}{T}}{B}=\vector{-\ddot{x}\stheta+\Ls(\ddot{\theta}-\psio^2\stheta\ctheta)}{-\ddot{x}\ctheta+\Ls(\dot{\theta}^2+\psio^2\text{sin}^2\theta)}{-2\Ls\psio\dot{\theta}\ctheta}</math>		:<math>\braq{\acc{G}{E}}{B}=\vector{-\ddot{x}\stheta+\Ls(\ddot{\theta}-\psio^2\stheta\ctheta)}{-\ddot{x}\ctheta+\Ls(\dot{\theta}^2+\psio^2\text{sin}^2\theta)}{-2\Ls\psio\dot{\theta}\ctheta}</math>
	</div>		</div>

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	'''*~~NOTA~~:''' ~~En aquest~~ web (~~i per manca de símbols de fletxa més precisos~~), ~~tot i que les fletxes~~ <math>\nearrow</math>, <math>\swarrow</math>, <math>\nwarrow</math> i <math>\searrow</math> ~~semblen indicar que els~~ vectors ~~formen un~~ angle ~~de 45° amb la direcció~~ vertical, ~~no té per què ser així~~. ~~Cal interpretar les fletxes de manera qualitativa~~, ~~observant el dibuix que sempre acompanya aquest tipus de notació~~. ~~Per exemple~~, ~~l’apartat~~ 1 ~~de l’exercici~~ C3-E.1, ~~el vector~~ <math>\vel{P}{REL}</math> ~~forma un angle~~ <math>\theta</math> ~~genèric amb la direcció~~ vertical. ~~Si el valor de l’angle~~ <math>\theta</math> ~~és menor de~~ 90° (~~com a la figura~~), ~~el vector~~ <math>\vel{P}{REL}</math> té component ~~cap a dalt i una altra cap a la dreta~~.		'''*NOTE:''' In this web (for lack of more precise symbols), though the arrows <math>\nearrow</math>, <math>\swarrow</math>, <math>\nwarrow</math> and <math>\searrow</math> seem to indicate that the vectors form a 45° angle with the vertical direction, this does not have to be the case. The arrows must be interpreted qualitatively, observing the figure that is always included when using this type of notation. For instance, in section 1 of exercise C3-E.1, the <math>\vel{P}{REL}</math> vector forms a generic <math>\theta</math> angle with the vertical direction. If the value of <math>\theta</math> is less than 90° (as in the following figure), the <math>\vel{P}{REL}</math> vector has a downward and rightward component.

	<p align="right"><small>© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor \|All rights reserved]]</small></p>		<p align="right"><small>© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor \|All rights reserved]]</small></p>

Apons: /* 1. Find the velocity and the acceleration of point G relative to the guide */

2024-09-17T21:22:59Z

1. Find the velocity and the acceleration of point G relative to the guide

← Older revision		Revision as of 21:22, 17 September 2024
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	:Hence, <math>\vel{G}{guide}\equiv\vel{G}{AB}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})</math>		:Hence, <math>\vel{G}{guide}\equiv\vel{G}{AB}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})</math>

			:The <math>\Gs</math> motion relative to the support is circular, and its acceleration has both a tangential and a normal component. The acceleration of the transportation motion (which is rectilinear) has only a tangential component.


	:The <math>\Gs</math> motion relative to the support is circular, and its acceleration has both a tangential and a normal component. The acceleration of the transportation motion (which is rectilinear) has only a tangential component.

	:''Comment: If the trajectory of point relative to a reference frame has a circular span (that is, the curvature radius along that span is constant), we may use directly the normal acceleration and the tangential acceleration calculated in <span style="text-decoration: underline;">[[C2. Movement of a mechanical system#C2.3 Intrinsic components of the acceleration\|'''section C2.3''']]</span>.''		:''Comment: If the trajectory of point relative to a reference frame has a circular span (that is, the curvature radius along that span is constant), we may use directly the normal acceleration and the tangential acceleration calculated in <span style="text-decoration: underline;">[[C2. Movement of a mechanical system#C2.3 Intrinsic components of the acceleration\|'''section C2.3''']]</span>.''

	:''Comment: If the trajectory of a point in a reference frame has a rectilinear span, we may use directly the tangential acceleration calculated in <span style="text-decoration: underline;">[[C2. Movement of a mechanical system#C2.3 Intrinsic components of the acceleration\|'''section C2.3''']]</span>, and its normal acceleration is zero.''		:''Comment: If the trajectory of a point in a reference frame has a rectilinear span, we may use directly the tangential acceleration calculated in <span style="text-decoration: underline;">[[C2. Movement of a mechanical system#C2.3 Intrinsic components of the acceleration\|'''section C2.3''']]</span>, and its normal acceleration is zero.''


	[[File:C4-E-Ex3-3-neut.png\|thumb\|right\|300px\|link=]]		[[File:C4-E-Ex3-3-neut.png\|thumb\|right\|300px\|link=]]

	:<math>\acc{G}{AB}=\accrel{G}+\acc{G}{ar}+\acc{G}{Cor}</math>		:<math>\acc{G}{AB}=\accrel{G}+\acc{G}{tr}+\acc{G}{Cor}</math>

	:<math>\accrel{G}=\accs{G}{REL}+\accn{G}{REL}=(\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)</math>		:<math>\accrel{G}=\accs{G}{REL}+\accn{G}{REL}=(\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)</math>

	:<math>\acc{G}{ar}=\acc{$\Gs\in REL$}{AB}=(\downarrow\ddot{x})</math>		:<math>\acc{G}{tr}=\acc{$\Gs\in REL$}{AB}=(\downarrow\ddot{x})</math>

	:<math>\acc{G}{Cor}=2\velang{REL}{AB}\times\vel{G}{REL}=\vec{0}\:(\velang{REL}{AB}=\velang{~~suport~~}{~~guia~~}=\vec{0})</math>		:<math>\acc{G}{Cor}=2\velang{REL}{AB}\times\vel{G}{REL}=\vec{0}\:(\velang{REL}{AB}=\velang{support}{guide}=\vec{0})</math>

	:~~Per tant~~,		:Hence,

	:<math>\acc{G}{~~guia~~}\equiv\acc{G}{AB} = (\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\downarrow\ddot{x})</math>		:<math>\acc{G}{guide}\equiv\acc{G}{AB} = (\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\downarrow\ddot{x})</math>

	:~~Aquests resultats es poden projectar fàcilment a la base~~ B ~~fixa a l’anella~~:		:These results can be easily projected on the B vector basis fixed to the ring:

	:<math>\braq{\vel{G}{~~guia~~}}{B}=\vector{\Ls\dot{\theta}\ctheta}{-\dot{x}+\Ls\dot{\theta}\stheta}{0}</math>		:<math>\braq{\vel{G}{guide}}{B}=\vector{\Ls\dot{\theta}\ctheta}{-\dot{x}+\Ls\dot{\theta}\stheta}{0}</math>

	:<math>\braq{\acc{G}{~~guia~~}}{B}=\vector{\Ls(\ddot{\theta}\ctheta-\dot{\theta}^2\stheta)}{-\ddot{x}+\Ls(\ddot{\theta}\stheta+\dot{\theta}^2\ctheta)}{0}</math>		:<math>\braq{\acc{G}{guide}}{B}=\vector{\Ls(\ddot{\theta}\ctheta-\dot{\theta}^2\stheta)}{-\ddot{x}+\Ls(\ddot{\theta}\stheta+\dot{\theta}^2\ctheta)}{0}</math>

	</div>		</div>

Apons: /* 1. Find the velocity and the acceleration of point G relative to the guide */

2024-09-17T21:20:34Z

1. Find the velocity and the acceleration of point G relative to the guide

← Older revision		Revision as of 21:20, 17 September 2024
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	<div>		<div>

	[[File:C4-E-Ex3-2-eng.png\|thumb\|right\|250px\|link=]]		[[File:C3-E-Ex3-2-eng.png\|thumb\|right\|250px\|link=]]

	:The <math>\Gs</math> émotion is simple if assessed from the support: it is circular, with centre <math>\Os</math>, radius L and associated angular velocity <math>\dot{\theta}</math>.Taking '''AB <math>\equiv</math> guide''' and '''REL <math>\equiv</math> support''',the transportation motion (its motion relative to the support if it were a point fixed to the guide) is rectilinear (in the vertical direction and described by the time variation of coordinate x).		:The <math>\Gs</math> émotion is simple if assessed from the support: it is circular, with centre <math>\Os</math>, radius L and associated angular velocity <math>\dot{\theta}</math>.Taking '''AB <math>\equiv</math> guide''' and '''REL <math>\equiv</math> support''',the transportation motion (its motion relative to the support if it were a point fixed to the guide) is rectilinear (in the vertical direction and described by the time variation of coordinate x).

Apons: /* 1. Find the velocity and the acceleration of point G relative to the guide */

2024-09-17T21:17:19Z

1. Find the velocity and the acceleration of point G relative to the guide

← Older revision		Revision as of 21:17, 17 September 2024
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	[[File:C4-E-Ex3-3-~~eng~~.png\|thumb\|right\|300px\|link=]]		[[File:C4-E-Ex3-3-neut.png\|thumb\|right\|300px\|link=]]

	:<math>\acc{G}{AB}=\accrel{G}+\acc{G}{ar}+\acc{G}{Cor}</math>		:<math>\acc{G}{AB}=\accrel{G}+\acc{G}{ar}+\acc{G}{Cor}</math>

Apons: /* 1. Find the velocity and the acceleration of point G relative to the guide */

2024-09-17T21:14:39Z

1. Find the velocity and the acceleration of point G relative to the guide

← Older revision		Revision as of 21:14, 17 September 2024
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	<div>		<div>

	[[File:Cr-E-Ex3-2-eng.png\|thumb\|right\|250px\|link=]]		[[File:C4-E-Ex3-2-eng.png\|thumb\|right\|250px\|link=]]

	:The <math>\Gs</math> émotion is simple if assessed from the support: it is circular, with centre <math>\Os</math>, radius L and associated angular velocity <math>\dot{\theta}</math>.Taking '''AB <math>\equiv</math> guide''' and '''REL <math>\equiv</math> support''',the transportation motion (its motion relative to the support if it were a point fixed to the guide) is rectilinear (in the vertical direction and described by the time variation of coordinate x).		:The <math>\Gs</math> émotion is simple if assessed from the support: it is circular, with centre <math>\Os</math>, radius L and associated angular velocity <math>\dot{\theta}</math>.Taking '''AB <math>\equiv</math> guide''' and '''REL <math>\equiv</math> support''',the transportation motion (its motion relative to the support if it were a point fixed to the guide) is rectilinear (in the vertical direction and described by the time variation of coordinate x).

Apons: /* 🔎 Exercici C3-E.3: pèndol giratori amb punt de suspensió mòbil */

2024-09-17T21:14:23Z

🔎 Exercici C3-E.3: pèndol giratori amb punt de suspensió mòbil

← Older revision		Revision as of 21:14, 17 September 2024
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	</small>		</small>

	====🔎 ~~Exercici~~ C3-E.3: ~~pèndol giratori amb punt de suspensió mòbil~~====		====🔎 EXAMPLE C3-E.3: rotating pendulum with oscillating articulation point====
	---------		---------
	::{\|:		::{\|:
	<small>		<small>
	~~El pèndol format per una anella solidaria~~ a ~~una barra està articulat al suport, el qual té un enllaç prismàtic amb la guia~~. ~~La guia està articulada al sostre~~, ~~i la seva velocitat~~ angular ~~respecte d’aquest~~ (<math>\psio</math>) ~~es manté~~ constant. ~~La molla entre suport i guia evita que el primer no caigui a terra quan el sistema està aturat~~.		The ring-shaped pendulum is articulated to the support, which is linked to the guide through a prismatic joint. The guide is articulated to the ceiling, and its angular velocity relative to the ceiling (<math>\psio</math>) is constant. The spring between support and guide guarantees that the former does not fall to the ground when the system is at rest.
	[[~~Fitxer~~:C3-E-Ex3-1-~~cat~~.png\|thumb\|center\|350px\|link=]]		[[File:C4-E-Ex3-1-eng.png\|thumb\|center\|350px\|link=]]


	===== 1. ~~Determina la velocitat i acceleració del punt~~ G ~~respecte de la <span style="text-decoration: underline;">guia</span>~~ =====		===== 1. Find the velocity and the acceleration of point G relative to the guide =====
	<div>		<div>

	[[~~Fitxer~~:C3-E-Ex3-2-~~cat-esp~~.png\|thumb\|right\|250px\|link=]]		[[File:Cr-E-Ex3-2-eng.png\|thumb\|right\|250px\|link=]]

	:~~El moviment de~~ <math>\Gs</math> ~~és senzill si s’avalua respecte del suport~~: és circular, de centre <math>\Os</math>, ~~radi~~ L ~~i velocitat~~ angular ~~associada~~ <math>\dot{\theta}</math>. ~~Si es pren~~ '''AB <math>\equiv</math> ~~guia~~''' i '''REL <math>\equiv</math> ~~suport~~''', ~~el moviment d’arrossegament~~ (~~el que tindria <math>\Gs</math> si pertanyés al suport i s’observés des de la guia~~) ~~seria rectilini~~ (~~en la direcció~~ vertical ~~descrit pel canvi de la coordenada~~ x).		:The <math>\Gs</math> émotion is simple if assessed from the support: it is circular, with centre <math>\Os</math>, radius L and associated angular velocity <math>\dot{\theta}</math>.Taking '''AB <math>\equiv</math> guide''' and '''REL <math>\equiv</math> support''',the transportation motion (its motion relative to the support if it were a point fixed to the guide) is rectilinear (in the vertical direction and described by the time variation of coordinate x).

	:<math>\vel{G}{AB}=\vel{G}{REL}+\vel{G}{ar}</math>		:<math>\vel{G}{AB}=\vel{G}{REL}+\vel{G}{tr}</math>

	:<math>\vel{G}{REL}=(\nearrow\Ls\dot{\theta})^*</math>		:<math>\vel{G}{REL}=(\nearrow\Ls\dot{\theta})^*</math>

	:<math>\vel{G}{ar}=\vel{$\Gs_{\in\Rs\Es\Ls}$}{AB}=(\downarrow\dot{x})</math>		:<math>\vel{G}{tr}=\vel{$\Gs_{\in\Rs\Es\Ls}$}{AB}=(\downarrow\dot{x})</math>

	:~~Per tant~~, <math>\vel{G}{~~guia~~}\equiv\vel{G}{AB}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})</math>		:Hence, <math>\vel{G}{guide}\equiv\vel{G}{AB}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})</math>


	:~~El moviment respecte del suport és circular, i la seva acceleració té component tangencial i normal. El moviment d’arrossegament (el que tindria~~ <math>\Gs</math> ~~si pertanyés al suport i s’observés des de la guia~~), ~~que és rectilini~~, ~~només té component tangencial~~.		:The <math>\Gs</math> motion relative to the support is circular, and its acceleration has both a tangential and a normal component. The acceleration of the transportation motion (which is rectilinear) has only a tangential component.

			:''Comment: If the trajectory of point relative to a reference frame has a circular span (that is, the curvature radius along that span is constant), we may use directly the normal acceleration and the tangential acceleration calculated in <span style="text-decoration: underline;">[[C2. Movement of a mechanical system#C2.3 Intrinsic components of the acceleration\|'''section C2.3''']]</span>.''

			:''Comment: If the trajectory of a point in a reference frame has a rectilinear span, we may use directly the tangential acceleration calculated in <span style="text-decoration: underline;">[[C2. Movement of a mechanical system#C2.3 Intrinsic components of the acceleration\|'''section C2.3''']]</span>, and its normal acceleration is zero.''

	:''Observació: Si el moviment d’un punt en una referència és circular (és a dir, té una trajectòria de radi de curvatura constant en un arc de circumferència), són aplicables directament les definicions d'<span style="text-decoration: underline;">[[C2. Moviment d'un sistema mecànic#C2.3 Direccions intrínseques. Components intrínseques de l’acceleració\|'''acceleració normal''']]</span> i <span style="text-decoration: underline;">[[C2. Moviment d'un sistema mecànic#C2.3 Direccions intrínseques. Components intrínseques de l’acceleració\|'''acceleració tangencial''']]</span> ja que es coneixen el radi de curvatura, la celeritat (o la velocitat angular) i l’acceleració tangencial (o l’acceleració angular).''

			[[File:C4-E-Ex3-3-eng.png\|thumb\|right\|300px\|link=]]
	:''Observació: Si el moviment d’un punt en una referència és rectilini (és a dir, té una trajectòria de radi de curvatura infinit durant un cert interval temporal), són aplicables directament les definicions d'<span style="text-decoration: underline;">[[~~C2. Moviment d'un sistema mecànic#C2.3 Direccions intrínseques. Components intrínseques de l’acceleració\|'''acceleració normal''']]</span> i <span style="text-decoration~~: underline;">[[C2. Moviment d'un sistema mecànic#C2.3 Direccions intrínseques. Components intrínseques de l’acceleració\|'''acceleració tangencial''']]</span>, (l’acceleració normal resulta nul·la, ja que una recta matemàticament es pot entendre com una circunferència de radi infinit).''


	~~[[Fitxer:C3~~-E-Ex3-3-~~cat-esp~~.png\|thumb\|right\|300px\|link=]]

	:<math>\acc{G}{AB}=\accrel{G}+\acc{G}{ar}+\acc{G}{Cor}</math>		:<math>\acc{G}{AB}=\accrel{G}+\acc{G}{ar}+\acc{G}{Cor}</math>