C4. Rigid body kinematics

From Mechanics

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A rigid body is a set of points whose mutual distances are constant. As a consequence, the motion of different points in a rigid body is related (though not necessarily the same) (Figure C4.1).

C4-1-cat,eng.png
Figure C4.1 Velocities of points of a same rigid body for two different movements

The constant distance between any pair of points [math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math] of a same rigid body S is the reason why, in a general motion of S, the component in the [math]\displaystyle{ \QPvec }[/math], direction has to be the same, though that in the direction perpendicular to [math]\displaystyle{ \QPvec }[/math] may be different (Figure C4.2): [math]\displaystyle{ \braqII{\vel{P}{R}}=\braqII{\vel{Q}{R}} }[/math].

If this were not the case, points would be approaching [math]\displaystyle{ \left(\braqII{\vel{P}{R}}\lt \braqII{\vel{Q}{R}}\right) }[/math] or separating [math]\displaystyle{ \left(\braqII{\vel{P}{R}}\gt \braqII{\vel{Q}{R}}\right) }[/math]. This property is known as equiprojectivity.

C4-2-eng.png
Figure C4.2 Equiprojectivity in a general motion of two points of a same rigid body

Thus unit presents the relationships among velocities and accelerations of different points of a same rigid body (equations of velocity and acceleration distribution), and explores the geometry of the velocity distribution. That of the acceleration distribution is not as simple and useful, and is not included here.


C4.1 Velocity distribution

The equation relating the velocity of two points [math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math] of a same rigid body S (Figure C4.2) is:

[math]\displaystyle{ \vel{P}{R}=\vel{Q}{R}+\velang{S}{R}\times\QPvec }[/math]

This equation implies instantaneous operations between vectors, thus it is a method to obtain [math]\displaystyle{ \vel{P}{R} }[/math] simpler than the time derivative. If the operations are done for a generic configuration, the result is valid at all times. As R may be any reference frame, the subscript will be suppressed from now on once that frame has been clearly identified.

C4-3-neut.png
Figure C4.3 Necessary information for the calculation of the velocity distribution in a rigid body. The subscript S in both points emphasizes that they belong to S. If there is no possible confusion, it may be omitted.
💭 Proof ➕

The velocity [math]\displaystyle{ \vel{P}{R} }[/math] may be obtained as the time derivative of a position vector:
[math]\displaystyle{ \vel{P}{R}=\dert{\overline{\textbf{O}_\textrm{R} \textbf{P}}}{R}=\dert{\overline{\textbf{O}_\textrm{R} \textbf{Q}}}{R}+\dert{\QPvec}{R}=\vel{Q}{R}+\dert{\QPvec}{R} }[/math]
The [math]\displaystyle{ \QPvec }[/math] value is constant because, as both points belong to the same rigid body, their mutual distance is constant. As far as direction is concerned, as [math]\displaystyle{ \QPvec }[/math] is a vector fixed to the rigid body, its rate of change of orientation relative to R [math]\displaystyle{ \left(\velang{$\QPvec$}{R}\right) }[/math] is the same as that of the rigid body: [math]\displaystyle{ \velang{$\QPvec$}{R}=\velang{S}{R} }[/math]. Hence:
[math]\displaystyle{ \dert{\QPvec}{R}=\velang{S}{R}\times\QPvec\Rightarrow\vel{P}{R}=\vel{Q}{R}+\velang{S}{R}\times\QPvec }[/math]


The velocity equation shows that we just need to know the velocity of one of its points [math]\displaystyle{ \left(\vel{Q}{}\right) }[/math] and the angular velocity [math]\displaystyle{ \left(\velang{s}{}\right) }[/math]. to calculate the velocity of any point in the rigid body. In the most general case (rigid body moving in space without restrictions), that information consists of six scalar independent variables (6 DoF) that have to be provided as data of the problem. When there are constraints acting on the rigid body (and has less than 6 DoF), those two velocities may be inferred from the associated kinematic restrictions .


The equiprojectivity shown in Figure C4.2 can be proved from the equation of velocity Distribution. Both [math]\displaystyle{ \vel{P}{R} }[/math] and [math]\displaystyle{ \vel{Q}{R} }[/math] ecan be decomposed into two components, one parallel to [math]\displaystyle{ \QPvec }[/math] and another one perpendicular to [math]\displaystyle{ \QPvec }[/math]:

[math]\displaystyle{ \vel{P}{R}=\vel{Q}{R}+\velang{S}{R}\times\QPvec\Rightarrow\braqII{\vel{P}{R}}+\braqL{\vel{P}{R}}=\braqII{\vel{Q}{R}}+\braqL{\vel{Q}{R}}+\velang{S}{R}\times\QPvec }[/math]

The term [math]\displaystyle{ \velang{S}{R}\times\QPvec }[/math] is always perpendicular to [math]\displaystyle{ \QPvec }[/math] because it is a cross product involving [math]\displaystyle{ \QPvec }[/math]. Thus:

  • [math]\displaystyle{ \braqII{\vel{P}{R}}=\braqII{\vel{Q}{R}}\Leftrightarrow }[/math] equal [math]\displaystyle{ ||\QPvec }[/math] components,
  • [math]\displaystyle{ \braqL{\vel{P}{R}}=\braqL{\vel{Q}{R}}+\velang{S}{R}\times\QPvec\Leftrightarrow }[/math] different [math]\displaystyle{ \perp\QPvec }[/math] components in principle.


✏️ EXAMPLE C4-1.1: wheel on a rotating support

C4-Ex1-1-1-eng.png
Between ground and support, and between support and wheel, there is a revolute joint. Because of those links, the support and the wheel can only move according to a simple rotation relative to the ground (R) and to the support, respectively. The angular velocity of the wheel relative to the ground is the superposition of those two rotations, which correspond to a first and a second Euler rotations.
The movement of the wheel center [math]\displaystyle{ \Qs }[/math] relative to the ground is circular with radius r about a vertical axis. Its velocity relative to the ground is straightforward, with value [math]\displaystyle{ r\dot\psi_0 }[/math].
The velocity of [math]\displaystyle{ \Ps }[/math] relative to the ground for the configuration shown in the figure can be obtained through the velocity equation for the wheel:
[math]\displaystyle{ \vel{P}{}=\vel{Q}{}+\vec{\Omega}\times\QPvec=(\otimes\;\;r\dot\psi_0)+(\Uparrow\dot\psi_0+\odot\;\;\dot\theta_0)\times(\rightarrow r)=(\otimes\;\;r\dot\psi_0)+(\otimes\;\;\dot\psi_0)+(\uparrow r\dot\theta_0)=(\otimes\;\;2r\dot\psi_0)+(\uparrow r\dot\theta_0) }[/math]
C4-Ex1-1-2-eng.png
Though vectors [math]\displaystyle{ \vel{Q}{} }[/math] and [math]\displaystyle{ \velang{}{} }[/math] appearing in the equation are valid at all times, the result obtained for [math]\displaystyle{ \vel{P}{} }[/math] is not because vector [math]\displaystyle{ \QPvec }[/math] is not always perpendicular to [math]\displaystyle{ \velang{}{} }[/math].
For instance, later on [math]\displaystyle{ \QPvec }[/math] is vertical, and then:
[math]\displaystyle{ \vel{P}{}=\vel{Q}{}+\velang{}{}\times\QPvec=(\otimes\;\;r\dot\psi_0)+(\Uparrow\dot\psi_0+\odot\;\;\dot\theta_0)\times(\uparrow\QPvec)=(\otimes\;\;r\dot\psi_0)+(\leftarrow r\dot\theta_0) }[/math]
C4-Ex1-1-3-eng.png
Analytical calculation ➕
If we choose a vector basis fixed to the support: [math]\displaystyle{ \left\{\vel{P}{}\right\}=\left\{\vel{Q}{}\right\}+\{\velang{}{}\}\times\{\QPvec\}=\vector{-r\dot\psi_0}{0}{0}+\vector{\dot\theta_0}{0}{\dot\psi_0}\times\{\QPvec\} }[/math]
C4-Ex2-1-2-eng.png
[math]\displaystyle{ \left\{\vel{P}{}\right\}=\vector{-r\dot\psi_0}{0}{0}+\vector{\dot\theta_0}{0}{\dot\psi_0}\times\vector{0}{r}{0}=\vector{-2r\dot\psi_0}{0}{r\dot\theta_0} }[/math]          [math]\displaystyle{ \left\{\vel{P}{}\right\}=\vector{-r\dot\psi_0}{0}{0}+\vector{\dot\theta_0}{0}{\dot\psi_0}\times\vector{0}{0}{r}=\vector{-r\dot\psi_0}{-r\dot\theta_0}{0}. }[/math]

✏️ EXAMPLE C4-1.2: wheel perpendicular to the ground and not sliding

C4-Ex2-2-1-eng.png
The constraints on the wheel provide information on its angular velocity and on the velocity of one of its points:
  • Perpendicular to the ground: the second Euler angle (inclination relative to the ground) is constant, so [math]\displaystyle{ \velang{}{}=\vec{\dot\psi}+\vec{\dot\varphi} }[/math].
  • Nonsliding contact with the ground: the velocity of the wheel point touching the ground has to be instantaneously zero (section C2.8), [math]\displaystyle{ \vel{J}{}=\vec{0} }[/math].
The velocity of [math]\displaystyle{ \Cs }[/math] can be calculated from that information:

[math]\displaystyle{ \vel{C}{}=\vel{J}{}+\velang{}{}\times\JCvec=(\vec{\dot\psi}+\vec{\dot\varphi})\times\JCvec=\vec{\dot\varphi}\times\JCvec }[/math], as [math]\displaystyle{ \vec{\dot\psi} }[/math] and [math]\displaystyle{ \JCvec }[/math] are always orthogonal. As [math]\displaystyle{ \vec{\dot\varphi} }[/math] is always perpendicular to the ground and horizontal, and [math]\displaystyle{ \JCvec }[/math] is always vertical, the cross product has the direction of the horizontal diameter of the wheel.

C4-Ex1-2-2-eng.png
Similarly:
[math]\displaystyle{ \begin{align} \vel{P}{} & =\vel{J}{}+\velang{}{}\times\JQvec=(\Uparrow\vec{\dot\psi}+\otimes\;\;\vec{\dot\varphi})\times(\nwarrow r) =(\Uparrow\vec{\dot\psi}+\otimes\;\;\vec{\dot\varphi})\times\left(\uparrow\frac{r}{\sqrt{2}}+\leftarrow\frac{r}{\sqrt{2}}\right)=\\ & =(\Uparrow\vec{\dot\psi})\times\left(\leftarrow\frac{r}{\sqrt{2}}\right)+(\otimes\;\;\vec{\dot\varphi})\times\left(\uparrow\frac{r}{\sqrt{2}}+\leftarrow\frac{r}{\sqrt{2}}\right)=(\odot\;\; r\dot\psi)+\left(\rightarrow\frac{r\dot\varphi}{\sqrt{2}}\right)+\left(\uparrow\frac{r\dot\varphi}{\sqrt{2}}\right) \end{align} }[/math]
Analytical calculation ➕
If we choose a vector basis fixed to the vertical plane containing the wheel (that is, with [math]\displaystyle{ \velang{B}{T}=\vec{\dot\psi} }[/math]):
[math]\displaystyle{ \left\{\vel{C}{}\right\}=\left\{\velang{}{}\right\}\times\left\{\JCvec\right\}=\vector{-\dot\varphi}{0}{\dot\psi}\times\vector{0}{0}{r}=\vector{0}{r\dot\varphi}{0} }[/math]
[math]\displaystyle{ \left\{\vel{Q}{}\right\}=\left\{\velang{}{}\right\}\times\left\{\JQvec\right\}=\vector{-\dot\varphi}{0}{\dot\psi}\times\vector{0}{-r/\sqrt{2}}{r/\sqrt{2}}=\vector{r\dot\psi/\sqrt{2}}{r\dot\varphi/\sqrt{2}}{r\dot\varphi/\sqrt{2}} }[/math]
C4-Ex2-2-2-eng.png


C4.2 Acceleration distribution

The equation relating the acceleration of two points [math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math] of a same rigid body S (Figure C4.3) is:

[math]\displaystyle{ \acc{P}{R}=\acc{Q}{R}+\velang{S}{R}\times\left(\velang{S}{R}\times\QPvec\right)+\accang{S}{R}\times\QPvec }[/math]

It is also an equation implying instantaneous operations (as the velocity equation), but it requires more information to calculate [math]\displaystyle{ \acc{P}{R} }[/math]: the acceleration of a point [math]\displaystyle{ \left(\acc{Q}{R}\right) }[/math], the angular velocity [math]\displaystyle{ \left(\velang{S}{R}\right) }[/math] and the angular acceleration of the rigid body [math]\displaystyle{ \left(\accang{S}{R}\right) }[/math]. Though constraints yield direct information on linear and angular velocities (as seen in the previous example), this is not so when it comes to accelerations. In general, the angular acceleration can be obtained as the time derivative of [math]\displaystyle{ \velang{S}{R} }[/math], but discovering a point whose acceleration is straightforward is not that evident. That and the fact that the number of required operations to calculate accelerations (two additions and three cross products) is much higher than that needed for the calculation of velocities (one addition and one cross product), is the reason why obtaining [math]\displaystyle{ \vel{P}{R} }[/math] with rígid body kinematics and then calculate [math]\displaystyle{ \acc{P}{R} }[/math] as time derivative of [math]\displaystyle{ \vel{P}{R} }[/math] is a good alternative (whenever the result obtained for[math]\displaystyle{ \vel{P}{R} }[/math] is generic).

C4-4-neut.png
Figure C4.4: Information required for the calculation of the acceleration distribution in a rigid body
💭 Proof ➕

The equation of acceleration distribution can be obtained as the time derivative of the velocity distribution:
[math]\displaystyle{ \begin{align} \acc{P}{R} & =\dert{\vel{P}{R}}{R}=\dert{\vel{Q}{R}}{R}+\dert{\velang{S}{R}\times\QPvec}{R}=\acc{Q}{R}+\velang{S}{R}\times\dert{\QPvec}{R}+\dert{\velang{S}{R}}{R}\times\QPvec=\\ & =\acc{Q}{R}+\velang{S}{R}\times\left(\velang{S}{R}\times\QPvec\right)+\accang{S}{R}\times\QPvec \end{align} }[/math]


✏️ EXAMPLE C4-2.1: wheel on a rotating support

C4-Ex2-1-1-eng.png
The angular acceleration of the wheel relative to the ground can be obtained through the geometric time derivative of the angular velocity:
[math]\displaystyle{ \accang{wheel}{T}=\dert{\velang{wheel}{T}}{T}=\dert{\left(\vec{\dot\psi_0}+\vec{\dot\theta_0}\right)}{T}=\dert{\Uparrow\dot\psi_0}{T}+\dert{\odot\;\dot\theta_0}{T} }[/math]
If we assume that the values [math]\displaystyle{ \dot\psi_0 }[/math] and [math]\displaystyle{ \dot\theta_0 }[/math] are constant, the first term in the time derivative is zero because its direction is constant (it is always vertical). The second one, though, is variable: its direction is always perpendicular to the vertical plane containing the wheel, and so rotates at a rate [math]\displaystyle{ \dot\psi_0 }[/math] relative to the ground:
[math]\displaystyle{ \accang{wheel}{T}=\left(\vec{\dot\psi_0}\times\vec{\dot\theta_0}\right)=\left(\Uparrow\dot\psi_0\right)\times\left(\odot\;\dot\theta_0\right)=\left(\Rightarrow\dot\theta_0\dot\psi_0\right) }[/math]
The acceleration of [math]\displaystyle{ \Cs }[/math] relative to the ground is straightforward because it is a circular motion with constant speed: it has only a normal component with value [math]\displaystyle{ r\dot\psi^{2}_0 }[/math] pointing to the center of curvature: [math]\displaystyle{ \acc{C}{}=\left(\leftarrow r\dot\psi^2_0\right) }[/math].
Hence,
[math]\displaystyle{ \begin{align}\acc{P}{}& =\acc{C}{}+\Omegavec\times(\velang{}{}\times\CPvec)+\Alfavec\times\CPvec=(\leftarrow r\dot\psi^{2}_0)+(\Uparrow\dot\psi_0+\odot\;\dot\theta_0)\times\left[(\Uparrow\dot\psi_0+\odot\;\dot\theta_0)\times(\rightarrow r)\right] +(\Rightarrow\dot\psi_0\dot\theta_0)\times(\rightarrow r)=\\ &=(\leftarrow r\dot\psi^{2}_0)+(\Uparrow\dot\psi_0+\odot\;\dot\theta_0)\times[\otimes\; r\dot\psi_0+\uparrow r\dot\theta_0]=(\leftarrow r\dot\psi^{2}_0)+(\Uparrow\dot\psi_0)\times(\otimes\; r\dot\psi_0)+(\odot\;\dot\theta_0)\times(\uparrow r\dot\theta_0)=(\leftarrow r\dot\psi^{2}_0)+(\leftarrow r\dot\psi^{2}_0)+(\leftarrow r\dot\theta^{2}_0)=\\ &=[\leftarrow r(2\dot\psi^{2}_0+\dot\theta^{2}_0)]\end{align} }[/math]
Analytical calculation ➕
C4-Ex2-1-2-eng.png
If we choose a vector basis fixed to the support:
[math]\displaystyle{ \{\accang{wheel}{T}(\Ps)\}=\frac{\ds}{\ds\ts}=\{\velang{wheel}{T}\}+\{\velang{B}{T}\}\times\{\velang{wheel}{T}\}=\vector{0}{0}{\dot\psi_0}\times\vector{\dot\theta_0}{0}{\dot\psi_0}=\vector{0}{\dot\theta_0\dot\psi_0}{0} }[/math]
[math]\displaystyle{ \{\acc{P}{}\}=\{\acc{C}{}\}+\{\Omegavec\times(\Omegavec\times\CPvec)\}+\{\Alfavec\times\CPvec\}=\vector{0}{-r\dot\psi_0}{0}+\vector{\dot\theta_0}{0}{\dot\psi_0}\times\left(\vector{\dot\theta_0}{0}{\dot\psi_0}\times\vector{0}{r}{0}\right)+\vector{0}{\dot\theta_0\dot\psi_0}{0}\times\vector{0}{r}{0} }[/math]
[math]\displaystyle{ \{\acc{P}{}\}=\vector{0}{-r\dot\psi_0}{0}+\vector{\dot\theta_0}{0}{\dot\psi_0}\times\left(\vector{\dot\theta_0}{0}{\dot\psi_0}\times\vector{0}{r}{0}\right)=\vector{0}{-r\dot\psi_0}{0}+\vector{\dot\theta_0}{0}{\dot\psi_0}\times\vector{-r\dot\psi_0}{0}{r\dot\theta_0}=\vector{0}{-2r\dot\psi^2_0-r\dot\theta^2_0}{0} }[/math]
The acceleration of [math]\displaystyle{ \Ps }[/math] cannot be obtained through the analytical time derivative of its velocity as the latter is only valid for one time instant (the one represented in the figure). Indeed, if we do perform that time derivative, the result is wrong:
[math]\displaystyle{ \vel{P}{}=\vector{-2r\dot\psi_0}{0}{r\dot\theta_0}\Rightarrow\left\{\dert{\vel{P}{}}{T}\right\}=\frac{\ds}{\ds\ts}\{\vel{P}{}\} +\{\velang{B}{T}\}\times\{\vel{P}{}\}=\vector{0}{0}{\dot\psi_0}\times\vector{-2r\dot\psi_0}{0}{r\dot\theta_0}=\vector{0}{-2r\dot\psi^2_0}{0}\neq\{\acc{P}{}\} }[/math]

✏️ EXAMPLE C4-2.2: wheel perpendicular to the ground and not sliding

C4-Ex2-2-1-eng.png
Let’s assume that the wheel in example C4-1.2 has an angular velocity relative to the ground with constant value [math]\displaystyle{ \velang{wheel}{T}=\vec{\dot\psi_0}+\vec{\dot\varphi_0}. }[/math]
The angular acceleration of the wheel relative to the ground may be obtained through the geometric time derivative of its angular velocity. If we draw the vectors on the plane perpendicular to [math]\displaystyle{ \vec{\dot\varphi_0}: }[/math]
[math]\displaystyle{ \accang{wheel}{T}=\dert{\velang{wheel}{T}}{T}=\dert{(\vec{\dot\psi_0}+\vec{\dot\varphi_0})}{T}=\dert{\left(\Uparrow\dot\psi_0\right)}{T}+\dert{\left(\otimes\;\dot\varphi_0\right)}{T} }[/math]
The first term of the time derivative is zero because [math]\displaystyle{ \vec{\dot\psi_0} }[/math] has a constant direction (vertical), whereas the second one is not as the [math]\displaystyle{ \vec{\dot\varphi_0} }[/math] direction is variable (because of [math]\displaystyle{ \vec{\dot\psi_0} }[/math]):
[math]\displaystyle{ \accang{wheel}{T}=\left(\Uparrow\dot\psi_0\right)\times\left(\otimes\;\dot\varphi_0\right)=(\Leftarrow\dot\psi_0\dot\varphi_0). }[/math]
There is no point with a simple motion (rectilinear or circular) whose acceleration is straightforward.
A usual error is considering that, as the velocity of point [math]\displaystyle{ \Js }[/math] of the wheel in contact with the ground is zero, its acceleration will also be zero: [math]\displaystyle{ \vel{J}{}=\vec{0}\Rightarrow\acc{J}{}=\vec{0} }[/math]. That is wrong. The velocity is instantaneously zero: just after (or just before) touching the ground, it is not, and it is a different point of the wheel periphery the one in contact with the ground. That means that the velocity of [math]\displaystyle{ \Js }[/math] goes from being zero to being nonzero (or form being nonzero to being zero). If the velocity changes, the acceleration is not zero.
The velocity of [math]\displaystyle{ \Cs }[/math] calculated in example C4-1.2 is valid at all times (is not instantaneous): the [math]\displaystyle{ \psi }[/math] and [math]\displaystyle{ \varphi }[/math] values have no consequences on [math]\displaystyle{ \vel{C}{} }[/math]. Hence, the acceleration [math]\displaystyle{ \acc{C}{} }[/math] can be obtained through a time derivative. The [math]\displaystyle{ \vel{C}{} }[/math] value is constant, but its direction is not: it is the direction of the horizontal diameter (contained in the wheel plane), thus it rotates relative to the ground because of [math]\displaystyle{ \vec{\dot\psi_0} }[/math] (but not because of [math]\displaystyle{ \vec{\dot\varphi_0} }[/math] : if this rotation did affect [math]\displaystyle{ \vel{C}{} }[/math], that velocity would not be horizontal). Hence:
[math]\displaystyle{ \acc{C}{}=\dert{\vel{C}{}}{T}=\dert{(\rightarrow r\dot\varphi_0)}{T}=(\Uparrow\dot\psi_0)\times(\rightarrow r\dot\varphi_0)=(\otimes\; r\dot\psi_0\dot\varphi_0) }[/math]
Regarding point [math]\displaystyle{ \Js }[/math], as its motion slows down as it approaches the ground –where its velocity becomes zero- and then moves away while increasing its separating speed, its acceleration has a vertical component pointing upwards. Thus can be checked though the acceleration equation from the acceleration of [math]\displaystyle{ \Cs }[/math]:
[math]\displaystyle{ \begin{align} \acc{J}{} & =\acc{C}{}+\Omegavec\times(\Omegavec\times\CJvec)+\Alfavec\times\CJvec=(\otimes\; r\dot\psi_0\dot\theta_0)+(\Uparrow\dot\psi_0+\odot\;\dot\theta_0)\times[(\Uparrow\dot\psi_0+\odot\;\dot\theta_0)\times(\downarrow r)]+[(\Leftarrow\dot\psi_0\dot\theta_0)]\times(\downarrow r)=\\ & =(\otimes\; r\dot\psi_0\dot\theta_0)+(\Uparrow\dot\psi_0+\odot\;\dot\theta_0)\times[(\rightarrow r\dot\theta_0)]+(\odot\; r\dot\psi_0\dot\theta_0)=(\otimes\; r\dot\psi_0\dot\theta_0)+(\uparrow r\dot\theta^2_0) \end{align} }[/math]
Analytical calculation ➕
If we choose a vector basis fixed to the vertical vertical plane containing the wheel [math]\displaystyle{ (\velang{B}{T}=\vec{\dot\psi_0}): }[/math]
  • Calculation of [math]\displaystyle{ \accang{wheel}{T} }[/math] through analytical time derivative:
[math]\displaystyle{ \{\accang{wheel}{T}\}=\left\{\dert{\velang{wheel}{T}}{T}\right\}=\{\velang{B}{T}\}\times\{\velang{wheel}{T}\}=\vector{0}{0}{\dot\psi_0}\times\vector{-\dot\varphi_0}{0}{\dot\psi_0}=\vector{0}{-\dot\psi_0\dot\varphi_0}{0} }[/math]
  • Calculation of [math]\displaystyle{ \acc{C}{} }[/math] through analytical time derivative::
[math]\displaystyle{ \{\acc{C}{}\}=\left\{\dert{\vel{C}{}}{T}\right\}=\{\velang{B}{T}\}\times\{\vel{C}{}\}=\vector{0}{0}{\dot\psi_0}\times\vector{-\dot\varphi_0}{0}{\dot\psi_0}=\vector{0}{-\dot\psi_0\dot\varphi_0}{0} }[/math]
C4-Ex2-2-2-eng.png
  • Calculation of [math]\displaystyle{ \acc{J}{} }[/math] through rigid body kinematics:
[math]\displaystyle{ \{\acc{J}{}\}=\{\acc{C}{}\}+\{\Omegavec\}\times(\{\Omegavec\}\times\{\CJvec\})+\{\Alfavec\}\times\{\CJvec\} }[/math]
[math]\displaystyle{ \{\acc{J}{}\}=\vector{-r\dot\psi_0\dot\varphi_0}{0}{0}+\vector{-\dot\varphi_0}{0}{\dot\psi_0}\times\left(\vector{-\dot\varphi_0}{0}{\dot\psi_0}\times\vector{0}{0}{-r}\right)+\vector{0}{-\dot\psi_0\dot\varphi_0}{0}\times\vector{0}{0}{-r}=\vector{-r\dot\psi_0\dot\varphi_0}{0}{0}+\vector{-\dot\varphi_0}{0}{\dot\psi_0}\times\vector{0}{-r\dot\varphi_0}{0}+\vector{r\dot\psi_0\dot\varphi_0}{0}{0} }[/math]
[math]\displaystyle{ \{\acc{J}{}\}=\vector{r\dot\psi_0\dot\varphi_0}{0}{r\dot\varphi^2_0} }[/math]


C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)

The equation of velocity distribution of a rigid body (that may be applied both in particular and generic configurations) seems to suggest that all points have different velocity when the rigid body rotates:

[math]\displaystyle{ \vel{P}{}=\vel{Q}{}+\omeg{S}{}\times \vecbf{QP} \: ; \omeg{S}{} \neq \vec{0} }[/math]

However, there are common features among the velocities, as pointed out in the introduction (Figure C4.2), because of the constant mutual distances between points. For instance, it implies that the velocity components in the direction of the straight line going through any pair of points have to be equal (equiprojectivity property). Nevertheless, that component is different in principle when different pairs of points are considered (Figure C4.5), and its value may change constantly.

C4-5-neut.png
Figure C4.5 Equiprojectivity in the velocity Distribution in a rigid body

This section explores other instantaneous similarities among the velocities of different points of a rigid body from the equation of velocity distribution.

Let’s start by asking whether it is possible that points in a same rigid body have the same velocity (not just one component) when the rigid body rotates: [math]\displaystyle{ \vel{P}{}=\vel{Q}{} }[/math] even if [math]\displaystyle{ \omeg{S}{} \neq \vecbf{0} }[/math] ? This equality requires that [math]\displaystyle{ \omeg{S}{} \times \vecbf{QP} = \vec{0} }[/math] , and this is the case whenever [math]\displaystyle{ \omeg{S}{} }[/math] and [math]\displaystyle{ \vecbf{QP} }[/math] are parallel. Hence, all points located on a straight line parallel to [math]\displaystyle{ \omeg{S}{} }[/math] have the same velocity instantaneously. It is not just pair of points that have the same velocity, it is an infinite number of points (Figure C4.6). We must not forget, though, that each straight line parallel to [math]\displaystyle{ \omeg{S}{} }[/math] corresponds to a different velocity.

C4-6-neut.png
Figure C4.6 Equal velocity for points located on straight lines parallel to [math]\displaystyle{ \omeg{S}{} }[/math]

The [math]\displaystyle{ \omeg{S}{} }[/math] direction, then, seems to be of paramount importance when it comes to discovering the geometry of the velocity distribution. For that reason, it is interesting to project the velocity equation on that direction. In general, the velocity of any point may be decomposed on a component parallel to [math]\displaystyle{ \omeg{S}{} }[/math] and another one perpendicular to [math]\displaystyle{ \omeg{S}{} }[/math]: ,[math]\displaystyle{ \vel{P}{} = \left.\vel{P}{}\right]_{\parallel\omeg{S}{}}+\left.\vel{P}{}\right]_{\perp\omeg{S}{}} }[/math] i [math]\displaystyle{ \vel{Q}{} = \left.\vel{Q}{}\right]_{\parallel\omeg{S}{}}+\left.\vel{Q}{}\right]_{\perp\omeg{S}{}} }[/math] . If we introduce that decomposition in the velocity equation:

[math]\displaystyle{ \left.\vel{P}{}\right]_{\parallel\omeg{S}{}}+\left.\vel{P}{}\right]_{\perp\omeg{S}{}}=\left.\vel{Q}{}\right]_{\parallel\omeg{S}{}}+\left.\vel{Q}{}\right]_{\perp\omeg{S}{}}+\omeg{S}{}\times \vecbf{QP}\implies \begin{equation} \left\{\begin{array}{@{}l@{}} \left.\vel{P}{}\right]_{\parallel\omeg{S}{}}=\left.\vel{Q}{}\right]_{\parallel\omeg{S}{}}\\ \left.\vel{P}{}\right]_{\perp\omeg{S}{}}=\left.\vel{Q}{} \right]_{\perp\omeg{S}{}}+\omeg{S}{}\times \vecbf{QP} \end{array} \right.\,. \end{equation} }[/math]

That equation proves that the projection of the velocity of all points in the rigid body on the [math]\displaystyle{ \omeg{S}{} }[/math] direction is the same instantaneously (Figure C4.7), though that value may change with time: [math]\displaystyle{ \left.\vel{P}{}\right]_{\parallel\omeg{S}{}}=\left.\vel{Q}{}\right]_{\parallel\omeg{S}{}}=\textrm{v}_\Omega }[/math]. It is a result totally different from that shown in Figure C4.5.

C4-7-neut.png
Figure C4.7 The component of the velocity of any point on the [math]\displaystyle{ \omeg{S}{} }[/math] direction is the same

The component of the velocity perpendicular to [math]\displaystyle{ \omeg{S}{} }[/math] is different in principle because of the [math]\displaystyle{ \omeg{S}{}\times \vecbf{QP} }[/math] . For any point, the velocity modulus is:

[math]\displaystyle{ \abs{\vel{P}{}}= \sqrt{ \left( \left. \vel{P}{} \right]_{\parallel\omeg{S}{}} \right)^2 + \left( \left. \vel{P}{} \right]_{\perp\omeg{S}{}} \right)^2}= \sqrt{\textrm{v}^2_\Omega+\left(\left.\vel{P}{}\right]_{\perp\omeg{S}{}}\right)^2}\ge \textrm{v}_\Omega }[/math]

That equation provides an interpretation for [math]\displaystyle{ \textrm{v}_\Omega }[/math]: velocity: it is the minimum speed of points in the rigid body. Do not forget, though, that this value may change with time, as we are performing an instantaneous analysis.

If we discover a point [math]\displaystyle{ \Is }[/math] whose speed is exactly that value [math]\displaystyle{ (|\vel{I}{}|=\textrm{v}_\Omega) }[/math] and apply the property presented in Figure C4.7, we will see immediately that all points located on the straight line parallel to [math]\displaystyle{ \omeg{S}{} }[/math] through [math]\displaystyle{ \Is }[/math] have that minimum speed. That straight line is the Instantaneous Screw Axis (ISA), and the [math]\displaystyle{ \vs_\Omega }[/math], velocity, whose direction is that of the ISA, is the sliding velocity along the ISA. From now on, [math]\displaystyle{ \vs_\Omega }[/math] will be written as [math]\displaystyle{ \vs_\textrm{ISA} }[/math].

Both the ISA and the [math]\displaystyle{ \vs_\textrm{ISA} }[/math] are concepts associated with velocities, so they depend on the reference frame R from which the motion is observed. Strictly speaking, we should add a subscript R denoting that reference frame. Si the frame is clear enough when describing the problems, it may be omitted. The velocity of any point can be readily calculated from the ISA (Figure C4.8).

C4-8-eng.png
Figure C4.8 Calculation of the velocity of a point from the ISA

The [math]\displaystyle{ \vs_\textrm{ISA} }[/math] term is called sliding velocity along the ISA. The [math]\displaystyle{ \omeg{S}{}\times\vecbf{IP} }[/math] s’anomena velocitat de rotació al turnnt de l’ISARL, term is called ”rotation” velocity about the ISA, and its value is the distance from [math]\displaystyle{ \Ps }[/math] to the ISA times the angular velocity [math]\displaystyle{ \omeg{S}{} }[/math]:

[math]\displaystyle{ \vel{P}{}=\vel{I}{}+\omeg{S}{}\times\vecbf{IP}=\vvec_\textrm{ISA}+\overline{\textbf{v}}_{\textrm{rotació}}(\Ps) }[/math]


✏️ EXAMPLE C4-3.1: screw

C4-Ex3-1-1-eng.png
The ISA of a screw relative to the female thread (R) is straightforward as [math]\displaystyle{ \omeg{screw}{R} }[/math] is evident: it has always the direction of the screw axis (actually, in this case it is a permanent rotation axis).
The velocity of the points on the ISA [math]\displaystyle{ \vs_\textrm{ISA} }[/math] is not zero, and in this case it is proportional to [math]\displaystyle{ \omeg{screw}{R} }[/math] through the thread pitch '[math]\displaystyle{ \textrm{e} }[/math]'. If '[math]\displaystyle{ \textrm{e} }[/math]' is given in (mm/turn), it is necessary to transform the units to the international system:
[math]\displaystyle{ \textrm{e}\left(\frac{mm}{turn}\right)\cdot \left(\frac{1m}{10^3mm}\right)\cdot \left(\frac{1 turn}{2\pi rad}\right)=\frac{\textrm{e}}{10^3\cdot 2\pi}(\frac{m}{rad}) }[/math]

[math]\displaystyle{ \vs\left(\frac{m}{s}\right)=\frac{\textrm{e}}{10^3\cdot 2\pi}\left(\frac{m}{rad}\right)\cdot\omeg{screw}{R}\left(\frac{rad}{s}\right)=\frac{\textrm{e}}{10^3\cdot 2\pi}\cdot\omeg{screw}{R}\left(\frac{m}{s}\right) }[/math]

The velocity of point [math]\displaystyle{ \Ps }[/math] is: [math]\displaystyle{ \vel{P}{R}=(\uparrow\vs_\textrm{ISA})+(\otimes\;\; r\omeg{screw}{R}) }[/math]


Video C4.1 Exemples de la geometria de la distribució de velocitats en un sòlid rígid

✏️ EXAMPLE C4-3.2: ball on a circular guide

C4-Ex3-2-1-eng.png

The ball moves without sliding on a circular guide fixed to the ground (E). Its rotation is not simple: it is a 3D motion.
As the ball moves on a restricted zone of the space because of the guide, whose symmetry axis [math]\displaystyle{ \vecbf{OO'} }[/math] is vertical, one may think that its rotation is about that axis, and that its angular velocity is vertical. But that is a mistake.
If this were the case, the ball would have a planar motion, and all its points would describe circular trajectories with center of curvature on the [math]\displaystyle{ \vecbf{OO'} }[/math] axis. This is inconsistent with the hypothesis of nonsliding motion between ball and ground: points [math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math] would slide on the ground.
The only point with a circular trajectory is the center [math]\displaystyle{ \Cs }[/math]: because of its contact with the guide, its distance to the ground and to axis [math]\displaystyle{ \vecbf{OO'} }[/math] is constant. We may associate a vertical angular velocity [math]\displaystyle{ \vec{\dot\psi} }[/math] : This, that point belongs to two different rigid bodies: that vertical plane and the ball. As a point of the ball, its velocity can be calculated from [math]\displaystyle{ \omeg{plane}{T} }[/math]:
[math]\displaystyle{ \vvec(\Cbf_\textrm{plane})=\vel{O}{}+\omeg{plane}{}\times\vecbf{OC}=\vec{\dot\psi}\times\vecbf{OC} }[/math]
But remember: [math]\displaystyle{ \omeg{ball}{T}\neq \vec{\dot\psi} }[/math]. Anyway, both velocities are related: if we block the plane rotation [math]\displaystyle{ \vec{\dot\psi}=\vec{0} }[/math] and assume nonsliding [math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math], the ball is also blocked [math]\displaystyle{ (\omeg{ball}{T}=\vec{0}) }[/math].
The ISA is useful to discover the [math]\displaystyle{ \omeg{ball}{T} }[/math]. Because of the nonsliding contact at [math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math], their instantaneous velocity relative to the ground us zero: [math]\displaystyle{ \vel{P}{}=\vel{Q}{}=\vec{0} }[/math]. As there is no velocity lower than zero, [math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math] have the minimum velocity, [math]\displaystyle{ \vs_\textrm{ISA}=\vec{0} }[/math] , and the ISA is the radial straight line through [math]\displaystyle{ \Ps }[/math], [math]\displaystyle{ \Qs }[/math] i O’. The velocity of [math]\displaystyle{ \Cbf }[/math] can be calculated as a point of the ball: it all comes from the rotation about the ball’s ISA. The result has to be the same as that obtained when applying rigid body kinematics of the rotating vertical plane.
C4-Ex3-2-2-eng.png
Hence:
[math]\displaystyle{ h\Omega=R\dot\psi\implies\Omega=\frac{R}{h}\dot\psi }[/math]


All points of the ball located on the straight line [math]\displaystyle{ \vecbf{CO} }[/math] have the same velocity as [math]\displaystyle{ \Cs }[/math] (the distance to the ISA is also h). The velocity of the point of the ball located on the highest position has the same direction as [math]\displaystyle{ \vel{C}{} }[/math] , and the value is [math]\displaystyle{ (h+r)\Omega=\frac{R}{h}(h+r)\dot\psi }[/math] .


Video C4.2 Moviment d'una ball sobre una pista circular

✏️ EXAMPLE C4-3.3: wheel on a rotating platform

C4-Ex3-3-1-eng.png

The wheel does not slide in the rotating platform, whose angular velocity relative to the ground is [math]\displaystyle{ \omega }[/math]. The wheel axis [math]\displaystyle{ \Os\Cbf }[/math] rotates with angular velocity [math]\displaystyle{ \omega }[/math] under the action of a motor. The angular velocity of the wheel is neither that of its axis, nor that of the platform, but it is related to both.
The distance between point [math]\displaystyle{ \Os }[/math] and all points of the wheel is constant. Thus, it belongs to the wheel. Moreover, as its velocity is zero (permanently), the wheel ISA goes through [math]\displaystyle{ \Os }[/math], and [math]\displaystyle{ \vvec_\textrm{ISA}=\vec{0} }[/math]. Now we need to find out another point with zero velocity to determine the ISA.
Points [math]\displaystyle{ \Js_\textrm{plate} }[/math] and [math]\displaystyle{ \Js_\textrm{wheel} }[/math] of the platform and the wheel that are instantaneously in contact have the same velocity if there is no sliding (section C2.8). On the other hand, [math]\displaystyle{ \Cbf }[/math] belongs to the [math]\displaystyle{ \Os\Cbf }[/math] axis, and its velocity can be calculated from the axis rotation: [math]\displaystyle{ \vvec_\textrm{T}(\Cbf_\textrm{ISAx})=\otimes\;\Rs\omega }[/math] . Finally, as [math]\displaystyle{ \Cbf }[/math] and [math]\displaystyle{ \Js }[/math] have the same speed but opposite direction, the midpoint between them ([math]\displaystyle{ \Qs }[/math]) has zero velocity:
[math]\displaystyle{ \begin{equation} \left.\begin{array}{lr} \vvec_\textrm{T}(\Cbf_\textrm{ISAx})=\otimes\; \Rs \omega \\ \vvec_\textrm{T}(\Cbf_\textrm{roda})=\otimes\; \textrm{s}\Omega=\otimes\;(\Rs sin\beta_0)\Omega \\ sin\beta_0=(\rs/2)\left/\sqrt{\Rs^2+(\rs/2)^2}\right. \end{array} \right\} \end{equation}\implies \Omega=\frac{\omega}{sin\beta_0}=\omega\sqrt{\left(\frac{\Rs}{textrm{s}/2}\right)^2+1} }[/math]

C4-Ex3-3-2-eng.png


The preceding examples deal with rigid bodies with just 1 degree of freedom. The ISA is not univocally defined when a rigid body has more than 1 rotational DoF, and in that case is not very interesting.


C4.4 Fixed axode and moving axode

The preceding examples (example C4-3.1, example C4-3.2) show the interest of the ISA to determine the angular velocity [math]\displaystyle{ \omeg{S}{R} }[/math] of a rigid body. Any angular velocity can be described as a superposition of Euler rotations, but that is not always straightforward. When the object has a spherical symmetry (as the ball in example C4-3.2), choosing the third axis (fixed to the object) is not evident: because of too much symmetry, there are no singular directions fixed to the object that are easy to be remembered. The concepts fixed axode and moving axode are very useful when it comes to describing [math]\displaystyle{ \omeg{S}{R} }[/math] superposition of Euler rotations. The axodes are surfaces generated by the succession of straight lines containing the points that instantaneously are on the ISA. When those points are fixed to the reference frame R, it is the fixed axode. When they belong to the rigid body S, it is the moving axode. They are fixed to R and to S, respectively.


✏️ EXAMPLE C4-4.1: ball on a circular guide

C4-Ex4-1-1-eng.png

Let’s take the system in example C4-3.2, and let’s draw the straight lines in the reference frame E that define the ISA along time. All of them go through [math]\displaystyle{ \Os' }[/math], and cover a horizontal plane. Me may say, then, that the fixed axode is a planar surface.
However, that description does not include an essential information: that planar Surface has been generated by straight lines intersecting at a same point. In order to add that information, the fixed axode is described as a conical surface, with vertex [math]\displaystyle{ \Os' }[/math], vertical axis and half-aperture [math]\displaystyle{ 90\deg }[/math].
To visualize the moving axode, it is helpful to find out a way of marking the ISA in the ball. For instance, we may think of drilling a straight hole between the two contact points with the guide ([math]\displaystyle{ \Ps }[/math] and [math]\displaystyle{ \Qs }[/math]). along time. We will end up with a ball full of holes. The surface they define is the moving axode.
If one remembers that the two axodes share a straight line at all times (or, what is the same, that the moving axodes rotates without sliding on the fixed one), it is evident that the moving axode is also a conical surface with vertex [math]\displaystyle{ \Os' }[/math], axis [math]\displaystyle{ \Cbf\Os' }[/math] and half-aperture [math]\displaystyle{ \beta_0=atan(\textrm{h}/\Rs) }[/math].
We may replace the guide and the ball by the two axodes: the kinematics is exactly the same, but we have got rid of the spherical symmetry of the ball (which made difficult choosing the 3rd Euler axis). The moving axode can be seen as a spinning top FALTA LINK that has already fallen to the ground (hence, its inclination cannot change any more: [math]\displaystyle{ \dot\theta=0 }[/math]) and does not slide on it (hence, the precession and the spin are proportional [math]\displaystyle{ \dot\varphi\propto\dot\psi }[/math]). So described, the horizontal angular velocity becomes “understandable”: it is the addition of the first and the third Euler rotations.
C4-Ex4-1-2-eng.png
[math]\displaystyle{ \omeg{ball}{T}\equiv \omeg{}{}=\vec{\dot\psi}+\vec{\dot\varphi},\textrm{with} \begin{equation}\left\{\begin{array}\dot\varphi=\frac{\Omega}{cos\beta_0} \\ \dot\psi = \dot\varphi sin\beta_0=\Omega tan\beta_0\end{array}\right.\end{equation} }[/math]


Video C4.3 Axoides d'una bola sobre una pista circular


Video C4.4 Control d'una bola sobre una pista circular

✏️ EXAMPLE C4-4.2: wheel on a rotating platform

Knowing that the ISA of the wheel in exemple C4-3.3 relative to the ground (E) is the [math]\displaystyle{ \vecbf{OQ} }[/math], line, we can visualize the axodes very easily:
  • fixed axode: conical surface with vertex [math]\displaystyle{ \Os }[/math], vertical axis and half-aperture [math]\displaystyle{ \left(\frac{\pi}{2}-\beta_0\right) }[/math] ,
  • moving axode: conical surface with vertex [math]\displaystyle{ \Os }[/math], horizontal axis and half-aperture a [math]\displaystyle{ \beta_0 }[/math] .
The distance between point [math]\displaystyle{ \Os }[/math] and all points in the wheel is constant. Thus, it belongs to the wheel. Moreover, as its velocity is zero (permanently), the ISA goes through [math]\displaystyle{ \Os }[/math], and [math]\displaystyle{ \vvec_\textrm{EI}=\vec{0} }[/math]. We need a second point with zero velocity to determine the ISA.
C4-Ex4-2-eng.png
Again, this problem is equivalent to that of a spinning top rotating without sliding on a conical surface. The angular velocity ca be described as the superposition of a precession and a spin (whose addition is a vector parallel to the ISA):
[math]\displaystyle{ \omeg{wheel}{T}\equiv \Omegavec=\vec{\dot\psi}+\vec{\dot\varphi},\textrm{with} \begin{equation}\left\{\begin{array} \dot{\varphi}=\Omega cos\beta_0=\frac{\Omega\Rs}{\sqrt{ \Rs^2+(\rs/2)^2}} \\ \dot\psi = \Omega sin\beta_0=\frac{\Omega\rs}{\sqrt{\Rs^2+(\rs/2)^2}}\end{array}\right.\end{equation} }[/math]
If the kinematic analysis is done from the platform reference frame, the ISA goes through [math]\displaystyle{ \Os }[/math] and through [math]\displaystyle{ \Cs }[/math], and he axodes are:
  • fixed axode: conical surface with vertex [math]\displaystyle{ \Os }[/math], vertical axis and half-aperture [math]\displaystyle{ \frac{\pi}{2}-\gamma_0 }[/math] ,
  • moving axode: conical surface with vertex [math]\displaystyle{ \Os }[/math], horizontal axis and half-aperture a [math]\displaystyle{ \gamma_0 }[/math], with [math]\displaystyle{ \gamma_0=atan(\rs/Rs) }[/math] .


C4.E General examples

🔎 EXAMPLE C4-E.1: rotating pendulum

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.
C4-E-Ex1-1-eng.png
1. Find the velocity and the acceleration of point P relative to the ground.
The [math]\displaystyle{ \Ps }[/math] motion relative to the ground may be obtained from that of point [math]\displaystyle{ \Os }[/math] by applying the equations of rigid body kinematics (RBK) to the plate:
[math]\displaystyle{ \vel{P}{E}=\vel{O}{E}+\velang{plate}{E}\times\OPvec }[/math]
[math]\displaystyle{ \acc{P}{E}=\acc{O}{E}+\velang{plate}{E}\times(\velang{plate}{E}\times\OPvec)+\accang{plate}{E}\times\OPvec }[/math]
As point [math]\displaystyle{ \Os }[/math] is located on the fork rotating axis, it is permanently at rest relative to the ground, therefore [math]\displaystyle{ \vel{O}{E}=\vec{0} }[/math] and [math]\displaystyle{ \acc{O}{E}=\vec{0} }[/math]. The angular velocity of the plate is the superposition of [math]\displaystyle{ \vec{\psio} }[/math] i [math]\displaystyle{ \vec{\dth} }[/math]:
[math]\displaystyle{ \velang{plate}{E}= \velang{plate}{fork}+\velang{fork}{E}=\vec{\dth}+\vec{\psio}=(\odot\dth)+(\Uparrow\psio) }[/math]
[math]\displaystyle{ \accang{plate}{E}=\dert{\velang{plate}{E}}{E}=\dert{(\vec{\psio}+\vec{\dth})}{E}=\dert{(\Uparrow\psio)}{E}+\dert{(\odot\dth)}{E} }[/math]
The angular acceleration of the plate is associated exclusively to the change of value and direction of [math]\displaystyle{ \vec{\dth} }[/math] (as [math]\displaystyle{ \vec{\psio} }[/math] is constant both in value and direction):
[math]\displaystyle{ \accang{plate}{E}=\dert{(\odot\dth)}{E}=[\text{change of value}]+[\text{change of direction}]_\Ts=[\odot\ddot{\theta}]+[(\Uparrow\psio)+(\odot\dth)]=(\odot\ddot{\theta})+(\Rightarrow\psio\dth) }[/math]
Calculation of the [math]\displaystyle{ \Ps }[/math] velocity relative to the ground
C4-E-Ex1-2-neut.png
[math]\displaystyle{ \vel{P}{E}=\vel{O}{E}+\velang{plate}{E}\times\OPvec=\vec{0}+\left[(\Uparrow\psio)+(\odot\dth)\right]\times(\searrow\Ls)^*= }[/math]
[math]\displaystyle{ =(\Uparrow\psio)\times(\rightarrow\Ls\stheta)+(\odot\dth)\times(\searrow\Ls)^*=(\otimes\Ls\psio\stheta)+(\nearrow\Ls\dth)^* }[/math]
Alternatively, we may perform the same operation through the vector basis fixed to the plate:
[math]\displaystyle{ \braq{\vel{P}{E}}{B}=\vector{0}{0}{0}+\vector{\dth}{\psio\stheta}{\psio\ctheta}\times\vector{0}{0}{-\Ls}=\vector{-\Ls\psio\stheta}{\Ls\dth}{0} }[/math]
Calculation of the [math]\displaystyle{ \Ps }[/math] acceleration relative to the ground
[math]\displaystyle{ \acc{P}{E}=\velang{plate}{E}\times(\velang{plate}{E}\times\OPvec)+\accang{plate}{E}\times\OPvec=\velang{plate}{E}\times\vel{P}{E}+\accang{plate}{E}\times\OPvec= }[/math]
[math]\displaystyle{ =\left[(\Uparrow\psio)+(\odot\dth)\right]\times\left[(\otimes\Ls\psio\stheta)+(\nearrow\Ls\dth)\right]+\left[(\odot\ddot{\theta})+(\Rightarrow\psio\dth)\right]\times(\searrow\Ls) }[/math]
The number of required operations is rather high, therefore it is advisable to perform them through the vector basis:
[math]\displaystyle{ \braq{\acc{P}{E}}{B}=\vector{0}{0}{0}+\vector{\dot{\theta}}{\psio\stheta}{\psio\ctheta}\times\left(\vector{\dot{\theta}}{\psio\stheta}{\psio\ctheta}\times\vector{0}{0}{-\Ls}\right)+\vector{\ddot{\theta}}{-\psio\dth\ctheta}{\psio\dth\stheta}\times\vector{0}{0}{-\Ls}=\vector{-2\Ls\psio\dth\ctheta}{\Ls\ddot{\theta}-\Ls\psio^2\stheta\ctheta}{\Ls\dth^2+\Ls\psio^2\text{sin}^2\theta} }[/math]

🔎 EXAMPLE C4-E.2: rotating articulated plate

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], and to the support through a cylindrical joint. The support rotates with the variable angular velocity [math]\displaystyle{ \vec{\dpsi} }[/math] relative to the ground (E).
C4-E-Ex2-1-eng.png
1. Find the velocity and the acceleration of point Q relative to the ground.
The [math]\displaystyle{ \Qs }[/math] motion relative to the ground may be obtained through RBD applied to the plate, from that of point [math]\displaystyle{ \Ps }[/math], whose motion relative to the ground is rectilinear. The velocity and acceleration of point [math]\displaystyle{ \Ps }[/math]may be obtained as the first and second time derivatives, respectively, of the position vector [math]\displaystyle{ \OPvec }[/math]. Both [math]\displaystyle{ \OPvec }[/math] and [math]\displaystyle{ \vel{P}{E} }[/math] are vectors with variable value and constant direction:
C4-E-Ex2-2-eng.png
[math]\displaystyle{ \OPvec = (\uparrow 2\Ls\stheta) }[/math]
[math]\displaystyle{ \vel{P}{E}=\dert{\OPvec}{E}=[\text{change of direction}]_\Es=(\uparrow 2\Ls\dth\ctheta) }[/math]
[math]\displaystyle{ \acc{P}{E}=\dert{\vel{P}{E}}{E}=[\text{change of direction}]_\Es=[\uparrow 2\Ls(\ddth\ctheta-\dth^2\stheta)] }[/math]
The plate angular velocity is the superposition of [math]\displaystyle{ \vec{\dpsi} }[/math] and [math]\displaystyle{ \vec{\dot{\theta}} }[/math], and its angular acceleration is associated with the change of value of [math]\displaystyle{ \vec{\dpsi} }[/math], and the change of value and direction of [math]\displaystyle{ \vec{\dot{\theta}} }[/math] (example C2-E.2):
[math]\displaystyle{ \velang{plate}{E}=\velang{plate}{support}+\velang{support}{E}=(\otimes\dth)+(\Uparrow\dpsi) }[/math]
[math]\displaystyle{ \accang{plate}{E}=\dert{\velang{plate}{E}}{E}=(\Uparrow\ddpsi)+(\otimes\ddth)+(\Leftarrow\dpsi\dth) }[/math]
Calculation of the [math]\displaystyle{ \Qs }[/math] velocity relative to the ground
[math]\displaystyle{ \vel{Q}{E}=\vel{P}{E}+\velang{plate}{E}\times\PQvec=(\uparrow 2\Ls\dth\ctheta)+[(\otimes\dth)+(\Uparrow\dpsi)]\times(\searrow 2\Ls)^*=(\uparrow 2\Ls\dth\ctheta)+(\Uparrow\dpsi)\times(\rightarrow 2\Ls\ctheta)+(\otimes\dth)\times(\searrow 2\Ls)^*= }[/math]
[math]\displaystyle{ =(\uparrow 2\Ls\dth\ctheta)+(\otimes 2\Ls\dpsi\ctheta)+(\swarrow 2\Ls\dth)^*=(\otimes 2\Ls\dpsi\ctheta)+(\leftarrow 2\Ls\dth\stheta) }[/math]
C4-E-Ex2-3-neut.png
Alternatively, the same operation may be done through the vector basis fixed to the support:
[math]\displaystyle{ \braq{\vel{Q}{E}}{B}=\vector{0}{0}{2\Ls\dth\ctheta}+\vector{0}{\dth}{\dpsi}\times\vector{2\Ls\ctheta}{0}{-2\Ls\stheta}=\vector{-2\Ls\dth\stheta}{2\Ls\dpsi\ctheta}{0} }[/math]
Calculation of the [math]\displaystyle{ \Qs }[/math] acceleration relative to the ground

[math]\displaystyle{ \acc{Q}{E}=\acc{P}{E}+\velang{plate}{E}\times(\velang{plate}{E}\times\PQvec)+\accang{plate}{E}\times\PQvec= }[/math]
[math]\displaystyle{ =[\uparrow 2\Ls(\ddth\ctheta-\dth^2\stheta)]+[(\Uparrow\dpsi)+(\otimes\dth)]\times\left([(\Uparrow\dpsi)+(\otimes\dth)]\times(\searrow 2\Ls)\right)+ }[/math]
[math]\displaystyle{ +[(\Uparrow\ddpsi)+(\otimes\ddth)+(\Leftarrow\dpsi\dth)]\times(\searrow 2\Ls) }[/math]
The number of required operations is rather high, therefore it is advisable to perform them through the vector basis:
[math]\displaystyle{ \braq{\acc{Q}{E}}{B}=\vector{-2\Ls(\ddth\ctheta-\dth^2\stheta)}{0}{0}+\vector{0}{\dth}{\dpsi}\times\left(\vector{0}{\dth}{\dpsi}\times\vector{2\Ls\ctheta}{0}{-2\Ls\stheta}\right)+\vector{-\dpsi\dth}{\ddth}{\ddpsi}\times\vector{2\Ls\ctheta}{0}{-2\Ls\stheta}=2\Ls\vector{-\ddth\stheta-(\dpsi^2+\dth^2)\ctheta)}{\ddpsi\ctheta-2\dpsi\dth\stheta}{0} }[/math]

🔎EXAMPLE C4-E.3: rotating pendulum with oscillating articulation point

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{ (\vec{\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.
C4-E-Ex3-1-eng.png
1. Find the velocity and the acceleration of point [math]\displaystyle{ \Gs }[/math] relative to the ground.
C4-E-Ex3-2-eng.png

The[math]\displaystyle{ \Gs }[/math] motion relative to the ground can be obtained through RBK applied to the ring, as the motion of point [math]\displaystyle{ \Os }[/math](which belongs to the ring) relative to the ground is straightforward: it is a vertical rectilinear one.

[math]\displaystyle{ \vel{O}{E}=(\downarrow\dot{x});\:\:\:\:\:\:\acc{O}{E}=(\downarrow\ddot{x}) }[/math]
The plate angular velocity is the superposition of [math]\displaystyle{ \vec{\psio} }[/math] and [math]\displaystyle{ \vec{\dth} }[/math], and the angular acceleration is associated with the change of value and direction of [math]\displaystyle{ \vec{\dth} }[/math] (example C2-E.3):
[math]\displaystyle{ \velang{ring}{E}=\velang{ring}{support}+\velang{support}{guide}+\velang{guide}{E}=(\odot\dth)+\vec{0}+(\Uparrow\psio) }[/math]
[math]\displaystyle{ \accang{ring}{E}=\dert{\velang{ring}{E}}{E}=(\otimes\ddth)+(\Rightarrow\psio\dth) }[/math]
Calculation of theG velocity relative to the ground
[math]\displaystyle{ \vel{G}{E}=\vel{O}{E}+\velang{ring}{E}\times\OGvec=(\downarrow\dot{x})+[(\Uparrow\psio)+(\odot\dth)]\times(\searrow\Ls)= }[/math]
[math]\displaystyle{ =(\downarrow\dot{x})+(\Uparrow\psio)+(\rightarrow\Ls\ctheta)+(\odot\dth)\times(\searrow\Ls)= (\downarrow\dot{x})(\otimes\Ls\dpsi\ctheta)+(\nearrow\Ls\dth) }[/math]
Alternatively, the same operation can be done through the vector basis fixed to the ring:
[math]\displaystyle{ \braq{\vel{G}{E}}{B}=\vector{0}{-\dot{x}}{0}+\vector{0}{\psio}{\dth}\times\vector{\Ls\stheta}{-\Ls\ctheta}{0}=\vector{\dth\Ls\ctheta}{-\dot{x}+\dth\Ls\stheta}{-\psio\Ls\stheta} }[/math]
C4-E-Ex3-3-neut.png
Calculation of the G acceleration relative to the ground
[math]\displaystyle{ \acc{G}{E}=\acc{O}{E}+\velang{ring}{E}\times(\velang{ring}{E}\times\OGvec)+\accang{ring}{E}\times\OGvec= }[/math]
[math]\displaystyle{ =(\downarrow\ddot{x})+\left[\left(\Uparrow\psio\right)+\left(\odot\dth\right)\right]\times\left(\left[(\Uparrow\psio)+(\odot\dth)\right]\times(\searrow\Ls)\right)+(\Rightarrow\psio\dth)\times(\searrow\Ls) }[/math]
The number of required operations is rather high, therefore it is advisable to perform them through the vector basis:
[math]\displaystyle{ \braq{\acc{G}{E}}{B}=\vector{0}{-\ddot{x}}{0}+\vector{0}{\psio}{\dth}\times\left(\vector{0}{\psio}{\dth}\times\vector{\Ls\stheta}{-\Ls\ctheta}{0}\right)+\vector{\psio\dth}{0}{\ddth}\times\vector{\Ls\stheta}{-\Ls\ctheta}{0}= }[/math]
[math]\displaystyle{ =\vector{-(\psio^2+\dth^2)\Ls\stheta+\ddth\Ls\ctheta}{-\ddot{x}+\dth^2\Ls\ctheta+\ddth\Ls\stheta}{-2\psio\dth\Ls\ctheta} }[/math]


*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 C4-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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