D4. Vector theorems - Revision history

Eantem: /* ✏️ EXAMPLE D4.9: barycentric decomposition */

2026-02-24T16:59:02Z

✏️ EXAMPLE D4.9: barycentric decomposition

← Older revision		Revision as of 16:59, 24 February 2026
Line 795:		Line 795:

	:<math>\H{O}{}{RTO} = \H{O}{}{E} = \H{G}{}{RTG} + \H{O}{\oplus}{E} = (\otimes 4\ms\Ls^2\omega) + \OGvec\times 4\ms\vel{G}{E} =</math>		:<math>\H{O}{}{RTO} = \H{O}{}{E} = \H{G}{}{RTG} + \H{O}{\oplus}{E} = (\otimes 4\ms\Ls^2\omega) + \OGvec\times 4\ms\vel{G}{E} =</math>
	<math>\hspace{1cm} = (\otimes 4\ms\Ls^2\omega) + [(\rightarrow\xs) + (\downarrow\hs)]\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\downarrow\hs)\times 4\ms(\rightarrow\dot\xs) ~~=</math>~~		<math>\hspace{1cm} = (\otimes 4\ms\Ls^2\omega) + [(\rightarrow\xs) + (\downarrow\hs)]\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\downarrow\hs)\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\odot 4\ms\hs\dot\xs) = \otimes 4\ms(\Ls^2\omega - \hs\dot\xs)</math>
	~~<math>\hspace{1cm}~~= (\otimes4\ms\Ls^2\omega) + (\odot 4\ms\hs\dot\xs) = \otimes 4\ms(\Ls^2\omega - \hs\dot\xs)</math>

Eantem: /* ✏️ EXAMPLE D4.9: barycentric decomposition */

2026-02-24T16:57:26Z

✏️ EXAMPLE D4.9: barycentric decomposition

← Older revision		Revision as of 16:57, 24 February 2026
Line 794:		Line 794:
	:The angular momentum at '''O''' (fixed to the ground) can be calculated from <math>\H{G}{}{RTG}</math> through barycentric decomposition:		:The angular momentum at '''O''' (fixed to the ground) can be calculated from <math>\H{G}{}{RTG}</math> through barycentric decomposition:

	:<math>\H{O}{}{RTO} = \H{O}{}{T} = \H{G}{}{RTG} + \H{O}{\oplus}{T} = (\otimes 4\ms\Ls^2\omega) + \OGvec\times 4\ms\vel{G}{T} =</math>		:<math>\H{O}{}{RTO} = \H{O}{}{E} = \H{G}{}{RTG} + \H{O}{\oplus}{E} = (\otimes 4\ms\Ls^2\omega) + \OGvec\times 4\ms\vel{G}{E} =</math>
	<math>\hspace{1cm} = (\otimes 4\ms\Ls^2\omega) + [(\rightarrow\xs) + (\downarrow\hs)]\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\downarrow\hs)\times 4\ms(\rightarrow\dot\xs) =</math>		<math>\hspace{1cm} = (\otimes 4\ms\Ls^2\omega) + [(\rightarrow\xs) + (\downarrow\hs)]\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\downarrow\hs)\times 4\ms(\rightarrow\dot\xs) =</math>
	<math>\hspace{1cm}= (\otimes4\ms\Ls^2\omega) + (\odot 4\ms\hs\dot\xs) = \otimes 4\ms(\Ls^2\omega - \hs\dot\xs)</math>		<math>\hspace{1cm}= (\otimes4\ms\Ls^2\omega) + (\odot 4\ms\hs\dot\xs) = \otimes 4\ms(\Ls^2\omega - \hs\dot\xs)</math>

Eantem at 12:47, 24 February 2026

2026-02-24T12:47:05Z

← Older revision		Revision as of 12:47, 24 February 2026
Line 99:		Line 99:
	When addressing a problem, it is essential to identify the unknowns contained in the system under study and to analyze whether the vector theorems provide a sufficient number of equations to solve them all (one must know whether the problem is determinate or indeterminate!). The unknowns can be classified into two groups:		When addressing a problem, it is essential to identify the unknowns contained in the system under study and to analyze whether the vector theorems provide a sufficient number of equations to solve them all (one must know whether the problem is determinate or indeterminate!). The unknowns can be classified into two groups:

	-~~--------~~		<u>Unknowns associated with degrees of freedom <span style="text-decoration: underline;">[[D3._Interactions_between_rigid_bodies\|'''degrees of freedom''']]</span></u>
	~~---------~~

	* Time evolution of ~~the free~~ <span style="text-decoration: underline;">[[C2. ~~Movement of a mechanical system~~#C2.~~7 Degrees of freedom of a mechanical system~~\|'''~~degrees of freedom (DoF)~~''']]</span> ~~(not controlled by actuators) of the system~~. ~~The equations that govern these DoF are called~~ '''~~equations of motion~~'''~~. If the free DoF are described by time derivatives of coordinates (~~<~~math~~>~~\qs_i~~</~~math~~>, with i=~~1,2,3~~...), their ~~time evolution~~ is the ~~second time derivatives~~ of ~~these coordinates (accelerations)~~. ~~Their general aspect~~ is:		The DoF of a mechanical system can be <u> <span style="text-decoration: underline;">[[C2._Movement_of_a_mechanical_system#C2.7_Degrees_of_freedom\|'''free''']]</span></u> or <u> <span style="text-decoration: underline;">[[C2._Movement_of_a_mechanical_system#C2.7_Degrees_of_freedom\|'''actuated''']]</span></u> (or forced, associated with <u> <span style="text-decoration: underline;">[[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators\|'''actuators''']]</span></u>. In both cases, their initial value (the value when an experiment begins or the system is started) is known. In the case of free DoF, the evolution of those initial values is not known: it is an unknown. The problem is said to be one of '''direct dynamics'''.

	~~<center>~~<math>\~~dot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text~~{ ~~dynamic parameters, geometric parameters~~})</math></~~center~~>		This is also the case when dealing with actuated DoF if the action of the associated actuator is known (i.e. when the value of <math>\Fs_{ac}</math> - in the case of linear actuators – or of <math>\Gamma</math> – in the case of rotational actuators – is known). The problem is also one of direct dynamics.

	The ~~dependence~~ of the equations of motion on the ~~second~~ time derivatives ~~is always linear, while the dependence on the~~ coordinates ~~and velocities can be of any type. The dynamic parameters are the mass of the elements and those associated with their distribution in space~~, ~~and the parameters associated~~ with ~~the interactions on the system (spring and damper constants~~, ~~coefficients of friction~~, ~~gravitational field constants~~...); the ~~geometric parameters have to do with the shape~~ of ~~the elements of the system~~ (~~distances and angles~~).		The equations governing the evolution of the DoF are called '''equations of motion'''. If the DoF are described by time derivatives of coordinates (<math>\dot{\qs}_i </math>, with i=1,2,3...), their time evolution is the second time derivative of these coordinates (accelerations).

	* <~~span style="text-decoration: underline;"~~>~~[[D2. Interaction forces between particles#D2.6 Interaction through actuators\|'''Actuator''']]~~</~~span~~> '''actions''': these are the forces (in the case of linear actuators) and moments (in the case of rotatory actuators) required to ensure predetermined evolutions of the DoF they control. As discussed in <~~span style="text-decoration: underline;"~~>~~[[D2. Interaction forces between particles#D2.6 Interaction through actuators\|'''section D2.6''']]~~</~~span~~>, ~~in some cases~~ the ~~actuator actions can be considered as data, and then the associated unknowns are the evolutions of the DoF they control.~~		For a system with 2 free DoF described, for example, by (<math>\dot{\xs}</math>, <math>\dot{\theta}</math>), these equations would have the following general appearance:

	* <span style="text-decoration: underline;">[[D3. ~~Interaccions entre sòlids rígids~~\|'''~~Constraint forces and movements~~''']]</span>: ~~the~~ number of unknowns associated with the constraints depends on the description given of them (<span style="text-decoration: underline;">[[D3. ~~Interactions between rigid bodies~~#D3.~~4 Direct constraint interactions~~\|'''direct constraints''']]</span>, <span style="text-decoration: underline;">[[D3. ~~Interactions between rigid bodies~~#D3.~~5 Indirect constraint interactions~~: ~~Constraint Auxiliary Elements~~ (CAE)\|'''indirect constraints''']]</span>). When a dynamics problem is indeterminate, the indeterminacy always refers to the constraint torsor components, never to the DoF (free or forced).		<center><math>\ddot{\xs} = f_{\xs}(\xs,\theta,\dot{\xs},\dot{\theta},\ddot{\theta},\text{dynamic parameters, geometric parameters})</math></center>

			<center><math>\ddot{\theta} = f_{\theta}(\xs,\theta,\dot{\xs},\dot{\theta},\ddot{\xs},\text{dynamic parameters, geometric parameters})</math></center>

			The dependency on the second time derivatives (<math>\ddot{\xs}</math>, <math>\ddot{\theta}</math>) is always linear, while the dependence on the coordinates and velocities (<math>\xs</math>, <math>\theta</math>, <math>\dot{\xs}</math>, <math>\dot{\theta}</math>) can be of any type. The dynamic parameters are the mass of the elements and those associated with their distribution in space, and the parameters associated with the interactions on the system (spring and damper constants, coefficients of friction, gravitational field constants...); the geometric parameters have to do with the shape of the elements of the system (distances and angles).

			Sometimes, the actions of the actuators (<math>\Fs_{ac}</math>, <math>\Gamma</math>) are not known, while the time evolutions of the actuated DoF are given. Then, the unknowns associated with the actuated DoF are the value of the forces (in the case of linear actuators) and moments (in the case of rotational actuators) necessary to guarantee these given evolutions. The problem is one of '''inverse dynamics'''.

			For the case of the previous example, if <math>\ddot{\theta}</math> is given, the equations that describe the unknowns of the problem associated with the DoF are:

			<center><math>\ddot{\xs} = f_{\xs}(\xs,\theta,\dot{\xs},\dot{\theta},\ddot{\theta},\text{dynamic parameters, geometric parameters})</math></center>

			<center><math>\Gamma = f_{\theta}(\xs,\theta,\dot{\xs},\dot{\theta},\ddot{\xs},\ddot{\theta},\text{dynamic parameters, geometric parameters})</math></center>

			where <math>\Gamma</math> would be the motor torque of the rotatory actuator associated with the <math>\theta</math> motion.

			<u>Unknowns associated with degrees of freedom <span style="text-decoration: underline;">[[D3._Interactions_between_rigid_bodies\|'''constraints''']]</span></u>

			These unknowns are the values of the <u> <span style="text-decoration: underline;">[[D3._Interactions_between_rigid_bodies\|'''constraint torsor''']]</span></u> components. The number of unknowns associated with the constraints depends on the description given of them (<u> <span style="text-decoration: underline;">[[D3._Interactions_between_rigid_bodies#D3.4_Direct_constraint_interactions\|'''direct constraints''']]</span></u>, <u> <span style="text-decoration: underline;">[[D3._Interactions_between_rigid_bodies#D3.5_Indirect_constraint_interactions:_Constraint_Auxiliary_Elements_(CAE)\|'''indirect constraints''']]</span></u>). When a dynamics problem is indeterminate, the indeterminacy always refers to the constraint torsor components, never to the DoF (free or forced).

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

Eantem at 11:48, 24 February 2026

2026-02-24T11:48:58Z

← Older revision		Revision as of 11:48, 24 February 2026
Line 98:		Line 98:

	When addressing a problem, it is essential to identify the unknowns contained in the system under study and to analyze whether the vector theorems provide a sufficient number of equations to solve them all (one must know whether the problem is determinate or indeterminate!). The unknowns can be classified into two groups:		When addressing a problem, it is essential to identify the unknowns contained in the system under study and to analyze whether the vector theorems provide a sufficient number of equations to solve them all (one must know whether the problem is determinate or indeterminate!). The unknowns can be classified into two groups:

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

	* Time evolution of the free <span style="text-decoration: underline;">[[C2. Movement of a mechanical system#C2.7 Degrees of freedom of a mechanical system\|'''degrees of freedom (DoF)''']]</span> (not controlled by actuators) of the system. The equations that govern these DoF are called '''equations of motion'''. If the free DoF are described by time derivatives of coordinates (<math>\qs_i</math>, with i=1,2,3...), their time evolution is the second time derivatives of these coordinates (accelerations). Their general aspect is:		* Time evolution of the free <span style="text-decoration: underline;">[[C2. Movement of a mechanical system#C2.7 Degrees of freedom of a mechanical system\|'''degrees of freedom (DoF)''']]</span> (not controlled by actuators) of the system. The equations that govern these DoF are called '''equations of motion'''. If the free DoF are described by time derivatives of coordinates (<math>\qs_i</math>, with i=1,2,3...), their time evolution is the second time derivatives of these coordinates (accelerations). Their general aspect is:

Eantem at 15:32, 19 February 2026

2026-02-19T15:32:53Z

← Older revision		Revision as of 15:32, 19 February 2026
Line 97:		Line 97:
	In this course, only the <u>version of the theorems for the case of constant matter</u> systems is presented. Although this includes systems with fluids, the application examples in this course are essentially multibody systems made up of rigid bodies.		In this course, only the <u>version of the theorems for the case of constant matter</u> systems is presented. Although this includes systems with fluids, the application examples in this course are essentially multibody systems made up of rigid bodies.

	When addressing a problem, it is essential to identify the unknowns contained in the system under study and to analyze whether the vector theorems provide a sufficient number of equations to solve them all (one must know whether the problem is determinate or indeterminate!). The unknowns can be classified into ~~three~~ groups:		When addressing a problem, it is essential to identify the unknowns contained in the system under study and to analyze whether the vector theorems provide a sufficient number of equations to solve them all (one must know whether the problem is determinate or indeterminate!). The unknowns can be classified into two groups:

	* Time evolution of the free <span style="text-decoration: underline;">[[C2. Movement of a mechanical system#C2.7 Degrees of freedom of a mechanical system\|'''degrees of freedom (DoF)''']]</span> (not controlled by actuators) of the system. The equations that govern these DoF are called '''equations of motion'''. If the free DoF are described by time derivatives of coordinates (<math>\qs_i</math>, with i=1,2,3...), their time evolution is the second time derivatives of these coordinates (accelerations). Their general aspect is:		* Time evolution of the free <span style="text-decoration: underline;">[[C2. Movement of a mechanical system#C2.7 Degrees of freedom of a mechanical system\|'''degrees of freedom (DoF)''']]</span> (not controlled by actuators) of the system. The equations that govern these DoF are called '''equations of motion'''. If the free DoF are described by time derivatives of coordinates (<math>\qs_i</math>, with i=1,2,3...), their time evolution is the second time derivatives of these coordinates (accelerations). Their general aspect is:

Bros: /* D4.8 Barycentric decomposition of the angular momentum */

2025-05-14T07:14:54Z

D4.8 Barycentric decomposition of the angular momentum

← Older revision		Revision as of 07:14, 14 May 2025
Line 721:		Line 721:
	The general formulation presented for the AMT allows a free choice of the Q point of application. The criterion for choosing it is based on what we want to investigate (a constraint force, an equation of motion...). This criterion is discussed in <span style="text-decoration: underline;">[[D6. Examples of 2D dynamics\|'''unit D6''']]</span>.		The general formulation presented for the AMT allows a free choice of the Q point of application. The criterion for choosing it is based on what we want to investigate (a constraint force, an equation of motion...). This criterion is discussed in <span style="text-decoration: underline;">[[D6. Examples of 2D dynamics\|'''unit D6''']]</span>.

	In some cases, it may be interesting to choose a point '''Q''' that does not belong to any element of the system. It is then useful to calculate the angular momentum <math>\H{Q}{}{RTQ}</math> from the angular momentum at the system center of mass '''G''', <math>\H{Q}{}{RTG}</math>. This is known as the '''barycentric decomposition''' of the angular momentum: <math>\H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{Q}{\oplus}{RTQ}</math>, where the superscript <math>\oplus</math> indicates that the angular momentum is to be calculated as if the system had been reduced to a particle concentrated at '''G''' with mass equal to the total mass of the system (M): <math>\H{Q}{\oplus}{RTQ} = \QGvec\times\Ms\vel{G}{RTQ}</math>.		In some cases, it may be interesting to choose a point '''Q''' that does not belong to any element of the system. It is then useful to calculate the angular momentum <math>\H{Q}{}{RTQ}</math> from the angular momentum at the system center of mass '''G''', <math>\H{Q}{}{RTG}</math>. This is known as the '''barycentric decomposition''' of the angular momentum:

			<math>\H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{Q}{\oplus}{RTQ}</math>,
			where the superscript <math>\oplus</math> indicates that the angular momentum is to be calculated as if the system had been reduced to a particle concentrated at '''G''' with mass equal to the total mass of the system (M):

			<math>\H{Q}{\oplus}{RTQ} = \QGvec\times\Ms\vel{G}{RTQ}</math>.

Bros: /* D4.8 Barycentric decomposition of the angular momentum */

2025-05-14T07:14:15Z

D4.8 Barycentric decomposition of the angular momentum

← Older revision		Revision as of 07:14, 14 May 2025
Line 721:		Line 721:
	The general formulation presented for the AMT allows a free choice of the Q point of application. The criterion for choosing it is based on what we want to investigate (a constraint force, an equation of motion...). This criterion is discussed in <span style="text-decoration: underline;">[[D6. Examples of 2D dynamics\|'''unit D6''']]</span>.		The general formulation presented for the AMT allows a free choice of the Q point of application. The criterion for choosing it is based on what we want to investigate (a constraint force, an equation of motion...). This criterion is discussed in <span style="text-decoration: underline;">[[D6. Examples of 2D dynamics\|'''unit D6''']]</span>.

	In some cases, it may be interesting to choose a point '''Q''' that does not belong to any element of the system. It is then useful to calculate the angular momentum <math>\H{Q}{}{RTQ}</math> from the angular momentum at the system center of mass '''G''', <math>\H{Q}{}{RTG}</math>. This is known as the '''barycentric decomposition''' of the angular momentum: <math>\H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{G}{\oplus}{RTQ}</math>, where the superscript <math>\oplus</math> indicates that the angular momentum is to be calculated as if the system had been reduced to a particle concentrated at '''G''' with mass equal to the total mass of the system (M): <math>\H{G}{\oplus}{RTQ} = \QGvec\times\Ms\vel{G}{RTQ}</math>.		In some cases, it may be interesting to choose a point '''Q''' that does not belong to any element of the system. It is then useful to calculate the angular momentum <math>\H{Q}{}{RTQ}</math> from the angular momentum at the system center of mass '''G''', <math>\H{Q}{}{RTG}</math>. This is known as the '''barycentric decomposition''' of the angular momentum: <math>\H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{Q}{\oplus}{RTQ}</math>, where the superscript <math>\oplus</math> indicates that the angular momentum is to be calculated as if the system had been reduced to a particle concentrated at '''G''' with mass equal to the total mass of the system (M): <math>\H{Q}{\oplus}{RTQ} = \QGvec\times\Ms\vel{G}{RTQ}</math>.


Line 738:		Line 738:
	:The definition of <span style="text-decoration: underline;">[[D4. Vector theorems#D4.1 Linear Momentum Theorem (LMT) in Galilean reference frames\|'''centre of mass''']]</span> leads to rewriting the second term as:		:The definition of <span style="text-decoration: underline;">[[D4. Vector theorems#D4.1 Linear Momentum Theorem (LMT) in Galilean reference frames\|'''centre of mass''']]</span> leads to rewriting the second term as:

	:<math>\left[\int_\Ss\QPvec\ds\ms(\Ps)\right] = \Ms\QGvec\times\vel{P}{RTQ} = \QGvec\times\Ms\vel{P}{RTQ}</math>, which coincides with the angular momentum at '''Q''' of a particle of mass M located at '''G''': <math>\QGvec\times\Ms\vel{P}{RTQ} = \H{G}{\oplus}{RTQ}</math>.		:<math>\left[\int_\Ss\QPvec\ds\ms(\Ps)\right] = \Ms\QGvec\times\vel{P}{RTQ} = \QGvec\times\Ms\vel{P}{RTQ}</math>, which coincides with the angular momentum at '''Q''' of a particle of mass M located at '''G''': <math>\QGvec\times\Ms\vel{G}{RTQ} = \H{Q}{\oplus}{RTQ}</math>.

	:In the first term of the <math>\H{Q}{\Ss}{RTQ}</math> expression, <math>\QPvec</math> can be decomposed into two terms:		:In the first term of the <math>\H{Q}{\Ss}{RTQ}</math> expression, <math>\QPvec</math> can be decomposed into two terms:
Line 746:		Line 746:
	:From the definition of centre of mass: <math>\int_\Ss\QGvec\times\vel{P}{RTQ}\ds\ms(\Ps) = \QGvec\times\left[\int_\Ss\vel{P}{RTG}\ds\ms(\Ps)\right] = \QGvec\times\Ms\vel{G}{RTG} = \vec{0}</math>.		:From the definition of centre of mass: <math>\int_\Ss\QGvec\times\vel{P}{RTQ}\ds\ms(\Ps) = \QGvec\times\left[\int_\Ss\vel{P}{RTG}\ds\ms(\Ps)\right] = \QGvec\times\Ms\vel{G}{RTG} = \vec{0}</math>.

	:Therefore <math>\H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{G}{\oplus}{RTQ}</math>.		:Therefore <math>\H{Q}{}{RTQ} = \H{G}{}{RTG} + \H{Q}{\oplus}{RTQ}</math>.
	</small>		</small>
	</div>		</div>
Line 769:		Line 769:
	:The angular momentum at '''O''' (fixed to the ground) can be calculated from <math>\H{G}{}{RTG}</math> through barycentric decomposition:		:The angular momentum at '''O''' (fixed to the ground) can be calculated from <math>\H{G}{}{RTG}</math> through barycentric decomposition:

	:<math>\H{O}{}{RTO} = \H{O}{}{~~\epsilon~~} = \H{G}{}{RTG} + \H{G}{\oplus}{E} = (\otimes 4\ms\Ls^2\omega) + \OGvec\times\Ms\vel{G}{T} = (\otimes 4\ms\Ls^2\omega) + [(\rightarrow\xs) + (\downarrow\hs)]\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\downarrow\hs)\times\Ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\odot 4\ms\hs\dot\xs) = \otimes 4\ms(\Ls^2\omega - \hs\dot\xs)</math>		:<math>\H{O}{}{RTO} = \H{O}{}{T} = \H{G}{}{RTG} + \H{O}{\oplus}{T} = (\otimes 4\ms\Ls^2\omega) + \OGvec\times 4\ms\vel{G}{T} =</math>
			<math>\hspace{1cm} = (\otimes 4\ms\Ls^2\omega) + [(\rightarrow\xs) + (\downarrow\hs)]\times 4\ms(\rightarrow\dot\xs) = (\otimes4\ms\Ls^2\omega) + (\downarrow\hs)\times 4\ms(\rightarrow\dot\xs) =</math>
			<math>\hspace{1cm}= (\otimes4\ms\Ls^2\omega) + (\odot 4\ms\hs\dot\xs) = \otimes 4\ms(\Ls^2\omega - \hs\dot\xs)</math>

Bros: /* ✏️ EXAMPLE D4.7: longitudinal dynamics of a vehicle */

2025-05-06T12:20:07Z

✏️ EXAMPLE D4.7: longitudinal dynamics of a vehicle

← Older revision		Revision as of 12:20, 6 May 2025
Line 606:		Line 606:
	where <math>(\Ns_{\ds\vs}, \Ts_{\ds\vs})</math> and <math>(\Ns_{\ds\rs}, \Ts_{\ds\rs})</math> are the resultant normal and tangential forces exerted by the ground on the two front wheels and the two rear wheels, respectively.		where <math>(\Ns_{\ds\vs}, \Ts_{\ds\vs})</math> and <math>(\Ns_{\ds\rs}, \Ts_{\ds\rs})</math> are the resultant normal and tangential forces exerted by the ground on the two front wheels and the two rear wheels, respectively.

	The AMT at '''G''' yields: <math>\sum\vec{\Ms}_{ext}(\Gs) = \vec{H}_{RTG}(\Gs)\Rightarrow(\odot\Ls_{\ds\vs}\Ns_{\ds\vs}) + (\otimes\Ls_{\ds\rs}\Ns_{\ds\rs}) + [\odot\hs(\Ts_{\ds\vs}) + \Ts_{\ds\rs}] =0</math>		The AMT at '''G''' yields: <math>\sum\vec{\Ms}_{ext}(\Gs) = \vec{H}_{RTG}(\Gs)\Rightarrow(\odot\Ls_{\ds\vs}\Ns_{\ds\vs}) + (\otimes\Ls_{\ds\rs}\Ns_{\ds\rs}) + [\odot\hs(\Ts_{\ds\vs} + \Ts_{\ds\rs})] =0</math>
	(the angular momentum <math>\vec{H}_{RTG}(\Gs)</math> is permanently zero beacuse the wheels are massless, and the chassis motion relative to the RTG is zero).		(the angular momentum <math>\vec{H}_{RTG}(\Gs)</math> is permanently zero beacuse the wheels are massless, and the chassis motion relative to the RTG is zero).
	\|[[File:D4-Ex7-2-eng.png\|thumb\|center\|200px\|link=]]		\|[[File:D4-Ex7-2-eng.png\|thumb\|center\|200px\|link=]]

Bros: /* ✏️ EXAMPLE D4.6: Atwood machine */

2025-04-11T07:49:53Z

✏️ EXAMPLE D4.6: Atwood machine

← Older revision		Revision as of 07:49, 11 April 2025
Line 563:		Line 563:

	<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RTO}(\Os)=\overline{0} \quad \Rightarrow \quad (\odot\Ts_1 \cdot\rs) + (\otimes \Ts_2 \cdot 2 \rs)=0</math><br>		<math>\sum \overline{\mathbf{M}}_\mathrm{ext}(\Os) = \dot{\overline{\mathbf{H}}}_\mathrm{RTO}(\Os)=\overline{0} \quad \Rightarrow \quad (\odot\Ts_1 \cdot\rs) + (\otimes \Ts_2 \cdot 2 \rs)=0</math><br>
	[[~~Fitxer~~:D4-Ex6-4-neut.png\|thumb\|right\|220px\|link=]]		[[File:D4-Ex6-4-neut.png\|thumb\|right\|220px\|link=]]

			Applying the AMT to each of the blocks provides two more equations:<br>
	Applying the AMT to each of the blocks provides two more equations:<br>

	SYST: 10 kg block<br>		SYST: 10 kg block<br>
Line 585:		Line 584:

	<div>		<div>



	====✏️ EXAMPLE D4.7: longitudinal dynamics of a vehicle====		====✏️ EXAMPLE D4.7: longitudinal dynamics of a vehicle====

Bros: /* AMT at a contact point between two rigid bodies */

2025-04-11T07:47:33Z

AMT at a contact point between two rigid bodies

← Older revision		Revision as of 07:47, 11 April 2025
Line 477:		Line 477:
	</math><br>		</math><br>

			[[File:TMCaJ-3-eng.png.png\|thumb\|center\|520px\|link=]]

	~~[[Fitxer:TMCaJ-3-cat-esp.png\|thumb\|center\|520px\|link=]]~~
	<center>		<center>
	<math>\overline{\Js_\mathrm{geom}\Gs} \times \ms \overline{\as}_\mathrm{Gal}(\Js_\mathrm{geom})= (\otimes \ms \rs \ddot{\xs})\hspace{0.7cm} \left\{\begin{array}{l}		<math>\overline{\Js_\mathrm{geom}\Gs} \times \ms \overline{\as}_\mathrm{Gal}(\Js_\mathrm{geom})= (\otimes \ms \rs \ddot{\xs})\hspace{0.7cm} \left\{\begin{array}{l}