C3. Composition of movements
[math]\displaystyle{ \newcommand{\uvec}{\overline{\textbf{u}}} \newcommand{\vvec}{\overline{\textbf{v}}} \newcommand{\evec}{\overline{\textbf{e}}} \newcommand{\Omegavec}{\overline{\mathbf{\Omega}}} \newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}} \newcommand{\Alfavec}{\overline{\mathbf{\alpha}}} \newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}} \newcommand{\ds}{\textrm{d}} \newcommand{\ts}{\textrm{t}} \newcommand{\us}{\textrm{u}} \newcommand{\vs}{\textrm{v}} \newcommand{\Rs}{\textrm{R}} \newcommand{\Ts}{\textrm{T}} \newcommand{\Ls}{\textrm{L}} \newcommand{\Bs}{\textrm{B}} \newcommand{\Es}{\textrm{E}} \newcommand{\es}{\textrm{e}} \newcommand{\is}{\textrm{i}} \newcommand{\rs}{\textrm{r}} \newcommand{\Os}{\textbf{O}} \newcommand{\Cbf}{\textbf{C}} \newcommand{\Or}{\Os_\Rs} \newcommand{\Qs}{\textbf{Q}} \newcommand{\Cs}{\textbf{C}} \newcommand{\Ps}{\textbf{P}} \newcommand{\Ss}{\textbf{S}} \newcommand{\Gs}{\textbf{G}} \newcommand{\deg}{^\textsf{o}} \newcommand{\xs}{\textsf{x}} \newcommand{\ys}{\textsf{y}} \newcommand{\zs}{\textsf{z}} \newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}} \newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}} \newcommand{\vec}[1]{\overline{#1}} \newcommand{\vecbf}[1]{\overline{\textbf{#1}}} \newcommand{\OQvec}{\vec{\Os\Qs}} \newcommand{\OQrelvec}{\vec{\Os_{\textrm{REL}}\Qs}} \newcommand{\abs}[1]{\left|{#1}\right|} \newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}} \newcommand{\vector}[3]{ \begin{Bmatrix} {#1}\\ {#2}\\ {#3} \end{Bmatrix}} \newcommand{\vecdosd}[2]{ \begin{Bmatrix} {#1}\\ {#2} \end{Bmatrix}} \newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})} \newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})} \newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})} \newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})} \newcommand{\velo}[1]{\vvec_{\textrm{#1}}} \newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}} \newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}} \newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})} \newcommand{\Orel}{\Os_{\textrm{REL}}} \newcommand{\omegarelab}{\vec{\Omega}^{\textrm{REL}}_{\textrm{AB}}} \newcommand{\velQab}{\vel{Q}{AB}} \newcommand{\velQrel}{\vel{Q}{REL}} \newcommand{\velQar}{\vel{Q}{tr}} \newcommand{\psio}{\dot{\psi_0}} \newcommand{\accrel}[1]{\vecbf{a}_{\textrm{REL}} (\textbf{#1})} \newcommand{\stheta}{\text{sin}\theta} \newcommand{\ctheta}{\text{cos}\theta} }[/math]
Very often, the motion of a point [math]\displaystyle{ \Qs }[/math] looks complicated relative to a reference frame R (it is neither circular nor rectilinear) but it can be guessed when that motion is simple (rectilinear, circular or zero) relative to another reference frame R’, and that of R’ relative to R is also simple (for instance, it is a translational motion or a simple rotation). Combining the motion of [math]\displaystyle{ \Qs }[/math] relative to R’ and that of R’ relative to R to obtain the motion of [math]\displaystyle{ \Qs }[/math] relative to R is known as composition of movements (Figure C3.1).
In the example in Figure C3.2, the composition of movements can be applied to determine the motion of particle [math]\displaystyle{ \Qs }[/math] relative to the vehicle (R’), from the motion of [math]\displaystyle{ \Qs }[/math] relative to the ground (R), which is a simple one.
Traditionally, the reference frames R and R’ implied in the composition are called AB (absolute) and REL (relative). We will use these names from now on.
The relationship between [math]\displaystyle{ \vel{Q}{AB} }[/math] and [math]\displaystyle{ \vel{Q}{REL} }[/math] , and that between [math]\displaystyle{ \acc{Q}{AB} }[/math] and [math]\displaystyle{ \acc{Q}{REL} }[/math] presented in this unit are always valid, regardless the motions of [math]\displaystyle{ \Qs }[/math] relative to the frames REL or AB and the motion between AB and REL are simple or not. When they are simple, the composition of movements is an algebraic alternative (it does not imply time derivatives) to calculate velocities and accelerations (Figure C3.3). When they are not simple, it may be advisable to use other methods (as the time derivative).
C3.1 Composition of velocities
At every time instant, the equation relating the velocity of a point [math]\displaystyle{ \Qs }[/math] in two different reference frames AB and REL is:
The second term in the right hand side is the transportation velocity, and it corresponds to the velocity of [math]\displaystyle{ \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):
💭 Proof ➕
- The velocities [math]\displaystyle{ \vel{Q}{AB} }[/math] and [math]\displaystyle{ \vel{Q}{REL} }[/math] can be calculated as time derivatives of the corresponding position vectors:
[math]\displaystyle{ \newcommand{\punt}[2]{\textbf{#1}_\textrm{#2}} \vel{Q}{AB} = \dert{\vec{\punt{O}{AB}\punt{Q}{ }}}{AB} = \dert{\vec{\punt{O}{AB}\punt{O}{REL}}}{AB} + \dert{\OQrelvec}{AB} = \vel{$\punt{O}{REL}$}{AB} +\dert{\OQrelvec}{AB} }[/math]
- The last term in the equation has no physical interpretation: it is the time derivative of a REL position (as its origin, [math]\displaystyle{ \punt{O}{REL} }[/math] is a point belonging to REL), but the derivative is not performed in REL but in AB. Using the expression relating the time derivative of a same vector in two different reference frames:
[math]\displaystyle{ \dert{\OQrelvec}{AB} = \dert{\OQrelvec}{REL} + \omegarelab\times\OQrelvec = \vel{Q}{REL}+\omegarelab \times \OQrelvec }[/math]
[math]\displaystyle{ \vel{Q}{AB} = \vel{Q}{REL}+\vvec_\textrm{AB}(\Orel) + \omegarelab \times \OQrelvec }[/math]
- Though the last equation is correct (it has been proved!), it contains two terms with [math]\displaystyle{ \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]\displaystyle{ \vel{Q}{tr} = \vvec_\textrm{AB}(\Qs \in \textrm{REL}) }[/math]. If we imagine that [math]\displaystyle{ \Qs }[/math] is a point fixed in REL, its velocity relative to AB is:
[math]\displaystyle{ \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]
- Finally, [math]\displaystyle{ \vel{Q}{AB}=\vel{Q}{REL}+\vel{Q}{tr} }[/math]
The equation of composition of velocities implies an instantaneous operation between vectors: [math]\displaystyle{ \velQab }[/math] (or [math]\displaystyle{ \velQrel }[/math]) eat time t is obtained from two vectors, [math]\displaystyle{ \velQrel }[/math] (or [math]\displaystyle{ \velQab }[/math]) and [math]\displaystyle{ \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 example C3-1.1, time t may correspond to a particular configuration or to a generic configuration.
In section C3.3 we will comment farther on the differences between the composition of movements and the time derivative of vectors to calculate velocities and accelerations.
✏️ EXAMPLE C3-1.1: rotating ring
- The circular guide (REL) moves with a simple rotation relative to the ground (AB) , with value [math]\displaystyle{ \dot\psi }[/math] . The particle [math]\displaystyle{ \Qs }[/math] moves in the guide.
- The time instant shown in the figure is generic as the [math]\displaystyle{ \theta }[/math] coordinate does not have a precise numerical value. The velocity of point [math]\displaystyle{ \Qs }[/math] relative to the ground can be obtained in a straightforward way through a composition:
- The motion [math]\displaystyle{ \Qs }[/math] relative to REL is circular with radius r, and the velocity is tangent to the guide. If [math]\displaystyle{ \Qs }[/math] were fixed to the guide, its motion relative to the ground (transportation motion) would be circular, with radius [math]\displaystyle{ \abs{\OQvec} \equiv \rho }[/math] and center [math]\displaystyle{ \Os }[/math], and the velocity would be perpendicular to [math]\displaystyle{ \OQvec }[/math] with value [math]\displaystyle{ \rho\dot\psi }[/math] . As the [math]\displaystyle{ \OQvec }[/math] direction (hence the direction perpendicular to [math]\displaystyle{ \OQvec }[/math])do not correspond to the directions suggested by the system (as that of the ring handle, or the [math]\displaystyle{ \vecbf{CQ} }[/math] direction), it is better to describe the transportation velocity as the addition of two vectors:
- The result obtained for [math]\displaystyle{ \velQab }[/math] is generic, as [math]\displaystyle{ \theta }[/math] may have any value (thus, it is a result valid for all times). For that reason, the acceleration of [math]\displaystyle{ \Qs }[/math] relative to the ground ([math]\displaystyle{ \acc{Q}{AB} }[/math]) could be obtained as the time derivative of the [math]\displaystyle{ \velQab }[/math].
- In this system, the guide drags physically the particle [math]\displaystyle{ \Qs }[/math]. In general, though, when using the composition of movements, this is not always the case (see example C3-1.3 and example C3-1.4).
✏️ EXAMPLE C3-1.2: wheel on a rotating support
- The support (REL) moves with a simple rotation relative to the ground (AB) , about the vertical axis and with value [math]\displaystyle{ \omega_0 }[/math] . The wheel is linked to the support through a revolute joint, and rotates with angular velocity[math]\displaystyle{ \omega_0 }[/math] relative to the support.
- At the time instant shown in the figure, the velocity of point [math]\displaystyle{ \Qs }[/math] relative to the ground is straightforward through composition:
- The motion of [math]\displaystyle{ \Qs }[/math] with respect to REL is circular with radius R at all times. At the time instant shown in the figure, [math]\displaystyle{ \velQrel }[/math] is vertical pointing downwards and with value [math]\displaystyle{ \Rs\omega_0 }[/math].
- The transportation motion, at that particular time instant, is zero: if [math]\displaystyle{ \Qs }[/math] were fixed to the support, as it is located precisely on the rotation axis, its instantaneous velocity would be zero [math]\displaystyle{ (\vel{Q}{tr} = \vec{0}) }[/math] . Hence:
- If the [math]\displaystyle{ \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]\displaystyle{ \Qs }[/math] is located on the support rotation axis. For that reason, it does not contain the information along time needed to calculate [math]\displaystyle{ \acc{Q}{AB} }[/math] as time derivative of the velocity [math]\displaystyle{ \velQab }[/math] that has been obtained.
✏️ EXAMPLE C3-1.3: vehicles
- Points [math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math] of vehicles VP and VQ, respectively, describe circular trajectories relative to the ground (R), with the same radius r but different center of curvature. Their speed is [math]\displaystyle{ \vs_0 }[/math].
- For the time instant shown in the figure, the velocity of point [math]\displaystyle{ \Qs }[/math] relative to vehicle VP is straightforward through composition:
- Si [math]\displaystyle{ \textrm{AB} = \Rs }[/math] i [math]\displaystyle{ \textrm{REL} = \textrm{VP} }[/math]
- 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]\displaystyle{ \Os }[/math] fixed to the ground.
- If [math]\displaystyle{ \Qs }[/math] were fixed to the platform, its motion relative to the ground would be circular, with center [math]\displaystyle{ \Os }[/math] and speed twice that of [math]\displaystyle{ \Ps }[/math], as it is at a distance 2r from [math]\displaystyle{ \Os }[/math]: [math]\displaystyle{ \vel{Q}{tr}=(\leftarrow 2\vs_0) }[/math].
- Finally:
- In this case, the platform (vehicle VQ) does not drag physically point [math]\displaystyle{ \Qs }[/math].
✏️ EXAMPLE C3-1.4: ferris wheel
- The ring of the ferris wheel rotates with angular velocity [math]\displaystyle{ \Omega_0 }[/math] relative to the ground (R). The cabin is linked to the ring through a revolute joint. If we neglect the small oscillation allowed by the joint, the cabin does not rotate relative to the ground (example C2-6.4).
- The velocity of point [math]\displaystyle{ \Qs }[/math], fixed to the ground, relative to the cabin, is:
- If [math]\displaystyle{ \textrm{AB} = \textrm{ground} }[/math] [math]\displaystyle{ (\Rs }[/math]) and [math]\displaystyle{ \textrm{REL} = \textrm{cabin} }[/math]
- [math]\displaystyle{ \velQrel=\velQab-\vel{Q}{tr}=-\velQar }[/math]
- If we imagine [math]\displaystyle{ \Qs }[/math] fixed to the cabin, its motion relative to the ground would be identical to that of [math]\displaystyle{ \Ps }[/math] because the cabin does not rotate relative to the ground: circular, with radius r and value [math]\displaystyle{ \rs\Omega_0 }[/math]. What is different is the location of the center of curvature: for the [math]\displaystyle{ \Ps }[/math] motion, it is point [math]\displaystyle{ \Os }[/math]; for the [math]\displaystyle{ \Qs }[/math] transportation motion, it is at a distance R from [math]\displaystyle{ \Qs }[/math] to the right. Hence:
C3.2 Composition of accelerations
The equation relating the acceleration of a point [math]\displaystyle{ \Qs }[/math] in two different reference frames AB and REL is:
where [math]\displaystyle{ \acc{Q}{tr} }[/math] is the transportation acceleration, and [math]\displaystyle{ \acc{Q}{Cor} }[/math] is the Coriolis acceleration.
The transportation acceleration corresponds to the AB acceleration of [math]\displaystyle{ \Qs }[/math] at that time instant if it were fixed to REL (at the same location it has at that time instant):
The Coriolis acceleration is
and it does not have a simple physical interpretation. The velocity [math]\displaystyle{ \velang{REL}{AB} }[/math] may be called transportation angular velocity [math]\displaystyle{ (\velang{REL}{AB}\equiv\velang{ }{tr}) }[/math].
💭 Proof ➕
- The term [math]\displaystyle{ \dert{\vel{$\textbf{O}_{\textrm{REL}}$}{AB}}{AB} }[/math] corresponds to [math]\displaystyle{ \acc{$\textbf{O}_{\textrm{REL}}$}{AB} }[/math]. If we use the expression relating the time derivative of a same vector in two different reference frames, the other terms may be written as:
- A reorganization of the terms leads to:
- The three terms containing point [math]\displaystyle{ \Orel }[/math] correspond to the transportation acceleration of [math]\displaystyle{ \Qs }[/math]: if [math]\displaystyle{ \Qs }[/math] belonged to REL, [math]\displaystyle{ \overline{\textbf{v}}_{\textrm{REL}}(\Qs \in \textrm{REL})=\vec{0} }[/math], [math]\displaystyle{ \overline{\textbf{a}}_{\textrm{REL}}(\Qs \in \textrm{REL})=\vec{0} }[/math]. Substituting this into the preceding expression:
- Finally, [math]\displaystyle{ \acc{Q}{AB}=\acc{Q}{REL}+\acc{Q}{tr}+2\velang{}{tr}\times\vel{Q}{REL}\equiv\acc{Q}{REL}+\acc{Q}{tr}+\acc{Q}{Cor}. }[/math]
✏️ EXAMPLE C3-2.1: wheel on a rotating support
- For the system in example C3-1.2, the transportation motion of [math]\displaystyle{ \Qs }[/math] is zero. Hence:
- If we assume a constant [math]\displaystyle{ \omega_0 }[/math] the motion of [math]\displaystyle{ \Qs }[/math] with respect to REL (which is circular) has only a normal component of acceleration. Thus:
✏️ EXAMPLE C3-2.2: vehicles
- In example C3-1.3, the transportation angular velocity is vertical (perpendicular to the drawing) with value [math]\displaystyle{ \textrm{v}_0 /\textrm{r} }[/math]. If we assume that points [math]\displaystyle{ \Ps }[/math] i [math]\displaystyle{ \Qs }[/math] have constant speed ([math]\displaystyle{ \textrm{v}_0 }[/math] constant), acceleration of the relative motion and that of the transportation motion of [math]\displaystyle{ \Qs }[/math] (both circular) have only a normal component. Hence:
✏️ EXEMPLE C3-2.3: ferris wheel
- Let’s consider that the angular velocity [math]\displaystyle{ \Omega_0 }[/math] of the ring of the ferris wheel in example C3-1.4 is constant. Then, the transportation acceleration has only a normal component normal. Moreover, as the cabin does not rotate relative to the ground [math]\displaystyle{ \left(\Omega_{tr}=\Omega_{R}^{cabina}=0\right) }[/math], [math]\displaystyle{ \acc{Q}{Cor}=\vec{0} }[/math]. Thus:
C3.3 Composition versus time derivative
As already mentioned, the main advantage of the composition of movements is that it is based on an instantaneous operation between vectors (though the result may be valid for any time instant if the configuration is a generic one). Moreover, if the motion of the point relative to one of the two reference frames (AB or REL) is simple and the motion between the two reference frames is also simple, the composition of movements is a method more intuitive and direct than the time derivative.
Anyway, when it comes to doing kinematics, one has to assess which method is quicker and safer to obtain the result. That may lead to the use of composition of movements to calculate velocities and the time derivative to calculate accelerations (provided the expression of the velocity to be differentiated is generic, that is, valid at all times).
✏️ EXAMPLE C3-3.1: particle in a rotating rectilinear guide
Particle [math]\displaystyle{ \Ps }[/math] moves in a rectilinear guide that rotates relative to the ground with constant angular velocity [math]\displaystyle{ \Omega_0 }[/math]. If the guide is the REL frame and the ground is the AB frame, the composition of movements leads to the following velocity and acceleration:
- As that velocity corresponds to a general position of [math]\displaystyle{ \Ps }[/math] (as the r coordinate does not have a precise numerical value), that velocity is valid for any time instant. Thus, it can be differentiated to obtain the acceleration:
- The result is the same as that found through composition of accelerations.
✏️ EXAMPLE C3-3.2: Euler pendulum
- In the particular configuration of the Euler pendulum shown in the figure, if the block is the REL frame and the ground is the AB frame, the relative motion of the pendulum endpoint Q is circular with radius L (length of the pendulum) and center of curvature located at the revolute joint between pendulum and block. The transportation motion is rectilinear as the block has a translational motion relative to the ground [math]\displaystyle{ (\Omega_{AB}^{REL}=\Omega_{tr}=0) }[/math]. The composition of movements leads to the following velocity and acceleration:
- Calculating the acceleration as the time derivative of the velocity is risky. As the velocity corresponds to the particular vertical configuration of the pendulum, we could think erroneously that the velocity has a constant direction (always parallel to the ground) or that it changes its direction exactly in the same way as the pendulum (with a rate of change of orientation [math]\displaystyle{ \dot{\psi} }[/math]). That would yield the following wrong results:
- If we think properly, the geometric time derivative may yield the correct result. One has to bear in mind that a part of the velocity (the one that corresponds to the relative motion) changes its orientation at a rate [math]\displaystyle{ \dot{\psi} }[/math], whereas the other one (that associated with the transportation motion) is always horizontal.
- The analytical time derivative is more deceptive. If the velocity is projected on a vector basis with one axis parallel to the pendulum at that time instant, it is not clear whether [math]\displaystyle{ \velang{B}{R}=\vec{0} }[/math] o bé [math]\displaystyle{ \velang{B}{R}=\odot\;\dot{\psi} }[/math] (which corresponds exactly to the two errors leading to the previous figure).
C3.E General examples
🔎 EXAMPLE C3-E.1: rotating pendulum
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The plate is articulated at point [math]\displaystyle{ \Os }[/math] to a fork, which rotates with constant angular velocity [math]\displaystyle{ \psio }[/math] relative to the ground (E). Between fork and ground (ceiling), and between plate and fork there are revolute joints.
1. Find the velocity of point P relative to the ground
- The motion of [math]\displaystyle{ \Ps }[/math] relative to the ground is complicated, but that relative to the fork is simple: it is a circular motion with centre [math]\displaystyle{ \Os }[/math], radius L and associated angular velocity [math]\displaystyle{ \dot{\theta} }[/math]. The motion relative to the ground can be obtained through a composition of motions.
- Taking AB [math]\displaystyle{ \equiv }[/math] ground and REL [math]\displaystyle{ \equiv }[/math] fork, the transportation motion (the motion that [math]\displaystyle{ \Ps }[/math] would have if, in a generic configuration, it were a point of the fork and its motion were assessed from the ground) is also circular, with radius [math]\displaystyle{ \Ls\text{cos}\theta }[/math] iand associated angular velocity [math]\displaystyle{ \psio }[/math].
- [math]\displaystyle{ \vel{P}{AB}=\vel{P}{REL}+\vel{P}{tr} }[/math]
- [math]\displaystyle{ \vel{P}{REL}=(\nearrow\Ls\dot{\theta})^* }[/math]
- [math]\displaystyle{ \vel{P}{tr}=\vel{$\Ps_{\in REL}$}{AB}=(\otimes\Ls\psio\text{sin}\theta) }[/math]
- Hence, [math]\displaystyle{ \vel{P}{E}\equiv\vel{P}{AB}=(\nearrow\Ls\dot{\theta})+(\otimes\Ls\psio\text{sin}\theta) }[/math]
- This result can be easily projected on the B vector basis, fixed to the plate:
- [math]\displaystyle{ \braq{\vel{P}{E}}{B}=\vector{-\Ls\psio\text{sin}\theta}{\Ls\dot{\theta}}{0} }[/math]
2. Find the acceleration of point P relative to the ground
- Taking AB [math]\displaystyle{ \equiv }[/math] ground and REL [math]\displaystyle{ \equiv }[/math] fork, the [math]\displaystyle{ \Ps }[/math] motion relative to the support (relative motion) is circular with associated angular velocity [math]\displaystyle{ \dot{\theta} }[/math] (variable). The transportation motion (its motion relative to the ground if it were a point fixed to the fork) is also circular but with associated angular velocity [math]\displaystyle{ \psio }[/math] (constant). Hence, the speed is constant and the acceleration has only a normal 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 section C2.3.
- [math]\displaystyle{ \acc{P}{AB}=\accrel{P}+\acc{P}{tr}+\acc{P}{Cor} }[/math]
- [math]\displaystyle{ \accrel{P}=\accs{P}{REL}+\accn{P}{REL}=(\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2) }[/math]
- [math]\displaystyle{ \acc{P}{tr}=\acc{$\Ps_{\in \Rs\Es\Ls}$}{AB}=\accn{P}{tr}=(\leftarrow\Ls\psio^2\stheta) }[/math]
- [math]\displaystyle{ \acc{P}{Cor}=2\velang{REL}{AB}\times\vel{P}{REL}=2(\Uparrow\psio)\times(\nearrow\Ls\dot{\theta})=(\otimes 2\Ls\psio\dot{\theta}\ctheta) }[/math]
- Hence,
- [math]\displaystyle{ \acc{P}{E}\equiv\acc{P}{AB}=(\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\leftarrow\Ls\psio^2\stheta)+(\otimes 2\Ls\psio\dot{\theta}\ctheta) }[/math]
- Its projection on the B vector basis fixed to the plate is: [math]\displaystyle{ \braq{\acc{P}{E}}{B}=\vector{-2\Ls\psio\dot{\theta}\ctheta}{\Ls\ddot{\theta}-\Ls\psio^2\ctheta\stheta}{\Ls\dot{\theta}^2+\Ls\psio^2\text{sin}^2\theta} }[/math]
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The plate is articulated at point [math]\displaystyle{ \Os }[/math] to a fork, which rotates with constant angular velocity [math]\displaystyle{ \psio }[/math] relative to the ground (E). Between fork and ground (ceiling), and between plate and fork there are revolute joints.
🔎 EXAMPLE C3-E.2: rotating articulated plate
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The rectangular plate is joined to a rotation support through two bars with revolute joints at their endpoints. A third bar is joined to the plate through a spherical joint at [math]\displaystyle{ \Ps }[/math] iand to the support through a cylindrical joint. EThe support rotates with the variable angular velocity [math]\displaystyle{ \vec{\dot{\psi}} }[/math] relative to the ground (E).
1. Find the velocity of point Q relative to the ground
- The [math]\displaystyle{ \Qs }[/math] motion relative to the support is rectilinear with direction [math]\displaystyle{ \OQvec }[/math]. Taking AB [math]\displaystyle{ \equiv }[/math] ground and REL [math]\displaystyle{ \equiv }[/math] support, the transportation motion (its motion relative to the ground if it were a point fixed to the support) is circular with centre [math]\displaystyle{ \Os }[/math], radius [math]\displaystyle{ |\OQvec| }[/math] and associated angular velocity [math]\displaystyle{ \vec{\dot{\psi}} }[/math].
- [math]\displaystyle{ \vel{Q}{AB}=\vel{Q}{REL}+\vel{Q}{tr} }[/math]
- [math]\displaystyle{ \vel{Q}{REL}=(\leftarrow 2\Ls\dot{\theta}\stheta) }[/math]
- [math]\displaystyle{ \vel{Q}{tr}=\vel{$\Qs_{\in\Rs\Es\Ls}$}{AB}=(\otimes 2\Ls\dot{\psi}\ctheta) }[/math]
- Hence,
- [math]\displaystyle{ \vel{Q}{E}\equiv\vel{Q}{AB} = (\leftarrow 2\Ls\dot{\theta}\stheta)+(\otimes 2\Ls\dot{\psi}\ctheta) }[/math]
- This result can be easily projected on the B vector basis fixed to the support:
- [math]\displaystyle{ \braq{\vel{Q}{E}}{B}=\vector{2\Ls\dot{\theta}\stheta}{2\Ls\dot{\psi}\ctheta}{0} }[/math]
2. Find the acceleration of point Q relative to the ground
- The [math]\displaystyle{ \Qs }[/math] motion relative to the support is rectilinear, therefore its acceleration has only a tangential component. Taking AB [math]\displaystyle{ \equiv }[/math] ground and REL [math]\displaystyle{ \equiv }[/math] support,the [math]\displaystyle{ \Qs }[/math] transportation motion (its motion relative to the ground if it were a point fixed to the support) is circular with variable speed, so its acceleration has both a tangential and a normal component.
- Comment: If the trajectory of a point in a reference frame has a rectilinear span, we may use directly the tangential acceleration calculated in section C2.3, and its normal acceleration is zero.
- 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 section C2.3.
- [math]\displaystyle{ \acc{Q}{AB}=\accrel{Q}+\acc{Q}{tr}+\acc{Q}{Cor} }[/math]
- [math]\displaystyle{ \accrel{Q}=\accs{Q}{REL}=[\leftarrow 2\Ls(\ddot{\theta}\stheta+\dot{\theta}^2\ctheta)] }[/math]
- [math]\displaystyle{ \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]\displaystyle{ \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]
- Hence, [math]\displaystyle{ \acc{Q}{E}\equiv\acc{Q}{AB} = \left[\leftarrow 2\Ls(\ddot{\theta}\stheta+(\dot{\theta}^2+\dot{\psi}^2)\ctheta)\right]+\left[\otimes 2\Ls(\ddot{\psi}\ctheta-2\dot{\psi}\dot{\theta}\stheta)\right] }[/math]
- Its projection on the B vector basis fixed to the support is: [math]\displaystyle{ \braq{\acc{Q}{E}}{B}=2\Ls\vector{-\ddot{\theta}\stheta-(\dot{\theta}^2+\dot{\psi}^2)\ctheta}{\ddot{\psi}\ctheta-2\dot{\psi}\dot{\theta}\stheta}{0} }[/math]
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The rectangular plate is joined to a rotation support through two bars with revolute joints at their endpoints. A third bar is joined to the plate through a spherical joint at [math]\displaystyle{ \Ps }[/math] iand to the support through a cylindrical joint. EThe support rotates with the variable angular velocity [math]\displaystyle{ \vec{\dot{\psi}} }[/math] relative to the ground (E).
🔎 EXAMPLE C3-E.3: rotating pendulum with oscillating articulation point
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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]\displaystyle{ \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.
1. Find the velocity and the acceleration of point G relative to the guide
- The [math]\displaystyle{ \Gs }[/math] émotion is simple if assessed from the support: it is circular, with centre [math]\displaystyle{ \Os }[/math], radius L and associated angular velocity [math]\displaystyle{ \dot{\theta} }[/math].Taking AB [math]\displaystyle{ \equiv }[/math] guide and REL [math]\displaystyle{ \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]\displaystyle{ \vel{G}{AB}=\vel{G}{REL}+\vel{G}{tr} }[/math]
- [math]\displaystyle{ \vel{G}{REL}=(\nearrow\Ls\dot{\theta})^* }[/math]
- [math]\displaystyle{ \vel{G}{tr}=\vel{$\Gs_{\in\Rs\Es\Ls}$}{AB}=(\downarrow\dot{x}) }[/math]
- Hence, [math]\displaystyle{ \vel{G}{guide}\equiv\vel{G}{AB}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x}) }[/math]
- The [math]\displaystyle{ \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 section C2.3.
- Comment: If the trajectory of a point in a reference frame has a rectilinear span, we may use directly the tangential acceleration calculated in section C2.3, and its normal acceleration is zero.
- [math]\displaystyle{ \acc{G}{AB}=\accrel{G}+\acc{G}{tr}+\acc{G}{Cor} }[/math]
- [math]\displaystyle{ \accrel{G}=\accs{G}{REL}+\accn{G}{REL}=(\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2) }[/math]
- [math]\displaystyle{ \acc{G}{tr}=\acc{$\Gs\in REL$}{AB}=(\downarrow\ddot{x}) }[/math]
- [math]\displaystyle{ \acc{G}{Cor}=2\velang{REL}{AB}\times\vel{G}{REL}=\vec{0}\:(\velang{REL}{AB}=\velang{support}{guide}=\vec{0}) }[/math]
- Hence,
- [math]\displaystyle{ \acc{G}{guide}\equiv\acc{G}{AB} = (\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\downarrow\ddot{x}) }[/math]
- These results can be easily projected on the B vector basis fixed to the ring:
- [math]\displaystyle{ \braq{\vel{G}{guide}}{B}=\vector{\Ls\dot{\theta}\ctheta}{-\dot{x}+\Ls\dot{\theta}\stheta}{0} }[/math]
- [math]\displaystyle{ \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]
2. Find the velocity and the acceleration of point G relative to the ground
From the previous results, the [math]\displaystyle{ \Gs }[/math] kinematics relative to the ground may be obtained through a second composition with AB’[math]\displaystyle{ \equiv }[/math] ground and REL’[math]\displaystyle{ \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]\displaystyle{ \Ls\ctheta }[/math] and associated angular velocity [math]\displaystyle{ \psio }[/math].
- [math]\displaystyle{ \vel{G}{AB'}=\vel{G}{REL'}+\vel{G}{tr'} }[/math]
- [math]\displaystyle{ \vel{G}{REL'}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x}) }[/math]
- [math]\displaystyle{ \vel{G}{tr'}=\vel{$\Gs_{\in\Rs\Es\Ls'}$}{AB'}=(\otimes\Ls\psio\ctheta) }[/math]
- Hence, [math]\displaystyle{ \vel{G}{E}\equiv\vel{G}{AB'}=(\nearrow\Ls\dot{\theta})+(\downarrow\dot{x})+(\otimes\Ls\psio\ctheta) }[/math]
- The acceleration of the transportation motion (which is circular with constant speed) has just a normal component.
- [math]\displaystyle{ \acc{G}{AB'}=\acc{G}{REL'}+\acc{G}{tr'}+\acc{G}{Cor'} }[/math]
- [math]\displaystyle{ \acc{G}{REL'}=(\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\downarrow\ddot{x}) }[/math]
- [math]\displaystyle{ \acc{G}{tr'}=\acc{$\Gs_{\in\Rs\Es\Ls'}$}{AB'}=(\leftarrow\Ls\psio^2\ctheta) }[/math]
- [math]\displaystyle{ \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]\displaystyle{ \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]
- The projection of those results in the B vector basis fixed to the ring is:
- [math]\displaystyle{ \braq{\vel{G}{E}}{B}=\vector{-\dot{x}\stheta+\Ls\dot{\theta}}{-\dot{x}\ctheta}{-\Ls\psio\stheta} }[/math]
- [math]\displaystyle{ \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]
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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]\displaystyle{ \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.
*NOTE: In this web (for lack of more precise symbols), though the arrows [math]\displaystyle{ \nearrow }[/math], [math]\displaystyle{ \swarrow }[/math], [math]\displaystyle{ \nwarrow }[/math] and [math]\displaystyle{ \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]\displaystyle{ \vel{P}{REL} }[/math] vector forms a generic [math]\displaystyle{ \theta }[/math] angle with the vertical direction. If the value of [math]\displaystyle{ \theta }[/math] is less than 90° (as in the following figure), the [math]\displaystyle{ \vel{P}{REL} }[/math] vector has a downward and rightward component.
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