C5. Rigid body kinematics: planar motion

[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{\Es}{\textrm{E}} \newcommand{\Ts}{\textrm{T}} \newcommand{\Ls}{\textrm{L}} \newcommand{\Bs}{\textrm{B}} \newcommand{\es}{\textrm{e}} \newcommand{\is}{\textrm{i}} \newcommand{\rs}{\textrm{r}} \newcommand{\Os}{\textbf{O}} \newcommand{\Js}{\textbf{J}} \newcommand{\Is}{\textbf{I}} \newcommand{\Or}{\Os_\Rs} \newcommand{\Qs}{\textbf{Q}} \newcommand{\Cs}{\textbf{C}} \newcommand{\Ps}{\textbf{P}} \newcommand{\Ss}{\textbf{S}} \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{\CPvec}{\vec{\Cs\Ps}} \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}[2]{\vec{\Omega}^{\textrm{REL}}_{\textrm{AB}}} \newcommand{\velQab}{\vel{Q}{AB}} \newcommand{\velQrel}{\vel{Q}{REL}} \newcommand{\velQar}{\vel{Q}{ar}} \newcommand{\cir}[2]{\Is\Cs\textbf{R}^{#1}_{#2}} \newcommand{\dth}{\dot{\theta}} \newcommand{\omeg}[2]{\vec{\Omega}^{\textrm{#1}}_{\textrm{#2}}} }[/math]

A rigid body is said to have a planar motion when all its points describe planar trajectories. This is the case when two conditions are fulfilled: the velocity along the ISA is zero [math]\displaystyle{ \vvec_{\textrm{ISA}}=0 }[/math], and the direction of the angular velocity is constant (and so [math]\displaystyle{ \Alfavec }[/math] is parellel to [math]\displaystyle{ \Omegavec }[/math] as it is only associated with a change in value of [math]\displaystyle{ \Omegavec }[/math] ).

In this unit, the results obtained for the velocity distribution in a general motion of a rigid body in space are particularized to the case of planar motion.

---

---

C5.1 Instantaneous Center of Rotation (ICR)

When a rigid body has a planar motion, the velocity distribution on all sections perpendicular to its angular is the same. Thus, just one section has to be analysed, and only the intersection of the ISA with that section is needed. That point is the Instantaneous Center of Rotation ([math]\displaystyle{ \textrm{ICR}\equiv\textrm{I} }[/math], Figure C5.1), and it is the only point (on that section) having a zero velocity. The velocity of any other point is associated to the rotation about the ICR:

[math]\displaystyle{ \vel{P}{R} = \vel{I}{R}+\omeg{S}{R} \times \vecbf{IP} =\omeg{S}{R} \times \vecbf{IP} }[/math]  ; [math]\displaystyle{ |\vel{P}{R}| = \Omega^{\textrm{S}}_{\textrm{R}} · |\overline{\textbf{IP}}| }[/math]

Figure C5.1 Planar motion of a rigid body, and Instantaneous Center of Rotation

Knowing the ICR is always useful. There two methods to localize it:

• Taking into account that the velocity of a point is always orthogonal to the straight line through that point and the ICR: if we know the direction of the velocity of two points of the rigid body (and those two directions are not parallel), the intersection of the straight lines orthogonal to those velocities is the ICR (Figure C5.2a).
• Taking into account that the velocity of a point is proportional to its distance to the ICR: if we know quantitatively (that is, both in direction and value) the velocity of two points of the rigid body (whose velocities are parallel), the triangle of velocities defined by those velocities has the ICR in one of its vertices (Figure C5.2b).

Figure C5.2 Two methods to determine the position of the ICR.

Sometimes, the ICR is always the same point of the rigid body. In that case, it is a permanent center of rotation, and it is also the center of curvature of all points in the rigid body.

C5.2 Examples

FALTEN TITOLS ALS EXEMPLES

✏️ EXAMPLE C5-2.1

---

The pulley does not slide on the inextensible cable. Its ICR relative to the ground can be found in a straightforward way from that constraint condition. The two cable spans that are not in contact with the pulley can be treated as rigid bodies with ICR located at their endpoint fixed to the ground (they both have zero velocity).

The velocity of all other points in those spans cannot have a component in the direction of the cable (otherwise it would not be inextensible!), but they may have a component perpendicular to it. The nonsliding condition between cable and pulley implies that the velocity of the contact points tangent to the pulley have also a velocity direction perpendicular to the cable. The intersection of the two directions orthogonal to those points is the ICR of the pulley.

✏️ EXAMPLE C5-2.2

---

The bar [math]\displaystyle{ \Os\Js }[/math] has a sliding single-point contact with the ground, and it is linked to the [math]\displaystyle{ \Os\Qs }[/math] bar through a revolute joint. Moreover, the bar [math]\displaystyle{ \Os\Qs }[/math] is linked through revolute joints to two wheels moving on inclined surfaces.

The position of the ICR of bar [math]\displaystyle{ \Os\Js }[/math] relative to the ground can be deduced from all those constraint conditions.

The velocity of [math]\displaystyle{ \Js }[/math] relative to the ground is horizontal, hence the ICR of the [math]\displaystyle{ \Os\Js }[/math] bar has to be on the vertical line through [math]\displaystyle{ \Js }[/math]. On the other hand, the velocity of the wheels centers has to be strictly parallel to the inclined surfaces. As those centers are also point ¡s of the[math]\displaystyle{ \Os\Qs }[/math] bar, the intersection of the directions through those centers and orthogonal to the inclines is the ICR of the [math]\displaystyle{ \Os\Qs }[/math] bar.

From that ICR, we can obtain the direction of the [math]\displaystyle{ \Os }[/math] velocity relative to the ground. The intersection of the direction orthogonal to that velocity and the vertical line through [math]\displaystyle{ \Js }[/math] is the ICR of the [math]\displaystyle{ \Os\Js }[/math] bar.[math]\displaystyle{ \Os\Js }[/math].

✏️ EXAMPLE C5-2.3

---

Element [math]\displaystyle{ \Os\Qs }[/math] is linked to the ground and to the bar through revolute joints. Between the bar and the support, there is a prismatic joint and the support does not slide on the ground. The ICR of the bar relative to the ground can be obtained from all these constraint conditions. The ICR relative to the ground of the [math]\displaystyle{ \Os\Qs }[/math] element and the support are straightforward ([math]\displaystyle{ \Os }[/math] and [math]\displaystyle{ \Js }[/math], respectively). Regarding that of the bar, it is located on the [math]\displaystyle{ \Os\Qs }[/math], direction, as that direction is perpendicular to the velocity of point [math]\displaystyle{ \Qs }[/math].

Finding out the direction of the velocity of another point of the bar is not straightforward. The prismatic joint between the bar and the support restricts the direction of the velocity of points in the bar relative to the support: it has to be that of the bar. From that information, the velocity of any point in the bar relative to the ground can be obtained through a composition of movements. If ground=AB and support=REL, the transportation motion is a rotation about [math]\displaystyle{ \Js }[/math].

Though the directions of the relative and the transportation motions are qualitatively accurate, the direction of its composition (which is that of the absolute motion) is not univocally defined in general because of the lack of a quantitative knowledge of those two velocities. However, if we discover a point in the bar where one of those two velocities is zero or parallel, the precise direction of the absolute velocity can be obtained. As the direction of the transportation velocity changes from one point in the bar to another, but that of the relative velocity does not, we may move along the bar until we discover a point where [math]\displaystyle{ \vec{\textbf{v}}_{REL}||\vec{\textbf{v}}_{tr} }[/math]. From that point, [math]\displaystyle{ \vec{\textbf{v}}_{AB} }[/math] is parallel to the bar. The intersection of the line perpendicular to the bar through that point and the [math]\displaystyle{ \Os\Qs }[/math] direction is the ICR of the bar relative to the ground.

✏️ EXAMPLE C5-2.4

---

The center of the carrier is linked to the ground and to the three identical wheels through revolute joints. A central wheel is articulated to the ground and in contact with those three wheels. All those elements are inside a ring.

If there is no sliding at any contact point, and the angular velocity of the carrier and the rind relative to the ground are known (in the present case, they have the same value [math]\displaystyle{ \omega }[/math] but opposite sign), the angular velocity of the central wheel is univocally defined.

The ICR relative to the ground of the carrier, the ring and the central wheel are the same point: the mechanism center (fixed to the ground, therefore with zero velocity). From the ICR and the angular velocities of the carrier and the ring angular velocities, it is possible to calculate the velocity of the wheels centers and that of their contact point with the ring.

As the velocities of those two points (for a same wheel) have opposite directions, the wheel ICR has to be located somewhere between them. The value of the velocity of a point is proportional to the distance between that point and the ICR. Hence, the ICR is at a distance (2/5)r from the center.

FALTA PARAGRAF QUE SI EST[A A LA VERSIO EN CATALA

From that ICR, the velocity of the contact point between the wheels and the central one is straightforward, and the angular velocity of the latter as well.

Video C5.1 Aplicació d'una transmissió epicicloidal: canvi de marxes de les bicicletes nòrdiques

✏️ EXAMPLE C5-2.5

---

The pulleys do not slide on the inextensible cable. The two upper ones are articulated to the ground.

The velocity of the block can be calculated from the velocity of the center of the lower pulleys and the constraint conditions on the system.

As in example C5-2.1, the cable spans that are not in contact with the pulleys can be treated as rigid bodies. As there is no pendulum motion, all points on the left span have zero velocity relative to the ground (as the upper one is linked to the ground).

The nonsliding condition between the left span and the first pulley (starting from the left) yields the ICR of the latter, and from that, the velocity of the point diametrically opposed to the ICR: it is twice that of the center, as its distance to the ICR is twice that of the center to the ICR.

That velocity [math]\displaystyle{ \downarrow 2\vs }[/math] is transmitted to all points in the next cable span. As the ICR of the second pulley is its central point, the point on the horizontal diameter tangent to the third cable span has a velocity [math]\displaystyle{ \uparrow 2\vs }[/math], and that velocity is transmitted to the third pulley through the cable.

As the left endpoint of the horizontal diameter of the third pulley goes up with [math]\displaystyle{ \uparrow 2\vs }[/math] but the pulley center goes down with [math]\displaystyle{ \downarrow 2\vs }[/math], the ICR of that pulley will be the midpoint between those two points. From that information, it is easy to prove that the block goes up with speed [math]\displaystyle{ 6\vs }[/math].

✏️ EXAMPLE C5-2.6

---

The pulley fixed to the ground and the moving pulley are Connected through an arm articulated to their centers, and an inextensible belt whose endpoints are fixed to the fixed pulley and does not slide on the moving pulley. The ICR of the arm relative to the ground is the center [math]\displaystyle{ \Os }[/math] of the fixed pulley. However, the location of the ICR of the moving pulley is not straightforward, but it can be obtained from the constraint conditions.

If we assume that the angular velocity of the arm relative to the ground is counterclockwise with value [math]\displaystyle{ \omega }[/math], we can calculate the velocity of the center of the moving pulley, but that is not enough to locate its ICR relative to the ground.

As the arm rotates counterclockwise relative to the ground, the fixed pulley rotation relative to the arm is clockwise with value [math]\displaystyle{ \omega }[/math]. Taking into account the velocity transmission through the belt (as in example C5-2.5), the angular velocity of the moving pulley relative to the arm is readily obtained.

Now we may analyse the kinematics of the moving pulley relative to the ground through a composition of movements: [math]\displaystyle{ \velang{pulley}{AB}=\velang{pulley}{REL}+\velang{pulley$\in$ REL}{AB}=\otimes\; 2\omega+\odot\;\omega=\otimes\;\omega }[/math] The ICR of the pulley relative to the ground can be obtained from that angular velocity and the velocity of the pulley center. The result is independent from the velocity assumed for the arm.

✏️ EXAMPLE C5-2.7

---

The wheel moves without sliding on a semicylindrical support fixed to the ground. The geometry of the trajectory of [math]\displaystyle{ \Ps }[/math] relative to the ground and, in particular, its curvature radius, is totally determined by the kinematical restrictions imposed by tht nonsliding constraint [math]\displaystyle{ \re{P}{T}=\frac{\vs^2_T(\textbf{P})}{|\accn{P}{T}|} }[/math]

Let’s assume that the angular velocity of the wheel relative to the ground is clockwise with value [math]\displaystyle{ \omega }[/math]. As [math]\displaystyle{ \Js }[/math] is its ICR, the instantaneous velocity of [math]\displaystyle{ \Cs }[/math] and [math]\displaystyle{ \Ps }[/math] is straightforward..

On the other hand, [math]\displaystyle{ \Cs }[/math] is always located at a distance r from the ground (otherwise the wheel would not touch the ground or would be embedded in the ground). Hence, its trajectory is circular with radius 3r and center [math]\displaystyle{ \Os }[/math], and the intrinsic components of [math]\displaystyle{ \acc{C}{T} }[/math] are straightforward.

The acceleration of [math]\displaystyle{ \Ps }[/math] can be obtained from the equation of acceleration distribution of a rigid body:

[math]\displaystyle{ \acc{P}{T}=\acc{C}{T}+\velang{wheel}{T}\times\left(\velang{wheel}{T}\times\CPvec\right)+\accang{roda}{T}\times\CPvec=(\rightarrow r\dot\omega)+\left(\downarrow\frac{1}{3}r\omega^2\right)+(\otimes\;\omega)\times[(\otimes\;\omega)\times(\uparrow r)]+(\otimes\;\dot\omega)\times(\uparrow r) }[/math]

The normal component is:

[math]\displaystyle{ \accn{P}{T}=\left(\downarrow\frac{1}{3}r\omega^2\right)+(\otimes\;\omega)\times[(\otimes\;\omega)\times(\uparrow r)]=\left(\downarrow\frac{1}{3}r\omega^2\right)+(\otimes\;\omega)\times(\rightarrow r\omega)=\left(\frac{4}{3}r\omega^2\right) }[/math]

and the radius of curvature is [math]\displaystyle{ \re{P}{R}=\frac{\vs^2_T(\textbf{P})}{|\accn{P}{T}|}\frac{(2r\omega)^2}{(4/3)r\omega^2}=3r }[/math]. The center of curvature is at a distance 3r from [math]\displaystyle{ \Ps }[/math] downwards (as the normal acceleration of [math]\displaystyle{ \Ps }[/math] points downwards).

---

Video C5.2 Translació: la roda de l'extrem es trasllada

Video C5.3 Mecanismes equivalents

Animació interactiva C5.1 Cinemàtica plana d'una bicicleta Animació interactiva C5.2 Corró

---

---

C5.3 Introduction to vehicle kinematics

An interesting example of planar motion is that of a vehicle without suspensions moving on horizontal ground (Figure C5.3): the chassis has a planar motion ([math]\displaystyle{ \velang{chassis}{T}=\vec{\dot\psi} }[/math], and [math]\displaystyle{ \vs_{\textrm{ISA}}=0 }[/math] because its distance to the ground is constant) though the wheels move in 3D (as their rotation relative to the ground is the superposition of two Euler rotations).

Figure C5.3 Movement of a vehicle without suspensions on horizontal ground. The chassis has a planar motion, but the wheels move in 3D
El xassís té moviment pla, però les rodes tenen moviment a l’espai

The angular velocity of the rear wheels is the superposition of the chassis precession [math]\displaystyle{ \vec{\dot\psi} }[/math] and the spin [math]\displaystyle{ \vec{\dot\varphi} }[/math]: [math]\displaystyle{ \velang{rear left/right}{T}=\vec{\dot\psi}+\vec{\dot\varphi}^{\textrm{rear left/right}} }[/math]. For the front wheels, the rotation associated with steering has to be added: [math]\displaystyle{ \velang{front left/right}{T}=\vec{\dot\psi}+\vec{\dot\varphi}^{\textrm{front left/right}}+\vec{\dot\delta}^{\textrm{left/right}} }[/math]. Anyway, the angular velocity of any wheel relative to the ground has a vertical component (that includes the precession [math]\displaystyle{ \vec{\dot\psi} }[/math] and another one parallel to its axis (different for each wheel). If the wheels do not slide on the ground, the velocity of the contact point [math]\displaystyle{ \Js }[/math] is instantaneously zero, and the direction of the velocity of the center is that of the horizontal diameter (example C4-1.2). Remembering this is useful to analyse in a very efficient way the kinematics of ground vehicles (provided that the simplifying hypotheses mentioned at the beginning of this section hold).

✏️ EXAMPLE C5-3.1

---

The wheels of the tricycle do not slide on the ground. At a certain time, the steering angle of the front wheel is [math]\displaystyle{ 45\deg }[/math] and the speed of its center is [math]\displaystyle{ \vs_0 }[/math]. From these data and what has been explained about the direction of the center of nonsliding wheels relative to the ground, it is possible to determine the location of the chassis relative to the ground, and the value and direction of the precession [math]\displaystyle{ \vec{\dot\psi} }[/math].

The velocity of the center of the left rear wheel can be obtained as rotation about that ICR (as the wheel center belongs both to the wheel and the chassis). As the result has to be equal to [math]\displaystyle{ r\dot\varphi }[/math], the wheel spin is straightforward: [math]\displaystyle{ \dot\varphi=\frac{\vs_0}{\sqrt{2L}}\frac{L-s}{r} }[/math].

✏️ EXAMPLE C5-3.2

---

The vehicle with articulated chassis has no steered wheels, the two chassis [math]\displaystyle{ \Ss_1 }[/math] and [math]\displaystyle{ \Ss_2 }[/math] are identical, and the four wheels as well The rotation of the two wheels of [math]\displaystyle{ \Ss_1 }[/math] determines the motion of the vehicle, hence the rotaion of the wheels of [math]\displaystyle{ \Ss_2 }[/math]. If the wheels of chassis [math]\displaystyle{ \Ss_1 }[/math] rotate with a same spin value ([math]\displaystyle{ \dot\varphi=\omega }[/math]), all points on the rear axle have the same speed [math]\displaystyle{ (r\omega) }[/math] and [math]\displaystyle{ \Ss_1 }[/math] has a translational motion (it does not rotate). Hence, the velocity of the articulation point [math]\displaystyle{ \Os }[/math] is also [math]\displaystyle{ r\omega }[/math] and its direction is the longitudinal one of chassis [math]\displaystyle{ \Ss_1 }[/math].

The velocity of the center of the right wheel of chassis [math]\displaystyle{ \Ss_2 }[/math] is proportional to its spin through the radius: [math]\displaystyle{ r\omega' }[/math]. In the other hand, the equiprojectivity yields immediately the velocity of the axle midpoint (point [math]\displaystyle{ \Ps }[/math]): [math]\displaystyle{ r\omega'/\sqrt{2} }[/math]. The ICR of chassis [math]\displaystyle{ \Ss_2 }[/math] is the intersection of the direction perpendicular to the velocity of point [math]\displaystyle{ \Ps }[/math] and the chassis axle. From the velocity of [math]\displaystyle{ \Ps }[/math] and the ICR, we may determine the angular velocity of [math]\displaystyle{ \Ss_2 }[/math] relative to the ground.

Finally, equating the speed [math]\displaystyle{ r\omega' }[/math] to that obtained as rotation about the ICR of chassis [math]\displaystyle{ \Ss_2 }[/math], we obtain the spin [math]\displaystyle{ \omega' }[/math]: [math]\displaystyle{ \omega'=\frac{\omega}{2\sqrt{2}} }[/math].

---

A very complete analysis of the kinematics of different ground vehicles (always under the simplifying hypotheses mentioned at the beginning of this section) can be found at [Batlle, J.A., Barjau, A. (2020) chapter 3, Appendix 3B a Rigid body kinematics. Cambridge University Press].

---

---

C5.E General examples

🔎 EXAMPLE C5-E.1: wheel in a cylindrical cavity

---

The two wheels, with radii R and 2R, are mutually fixed and rotate without sliding in a horizontal cylindrical cavity, with radius 3R and fixed to the ground.

1. Find the angular velocity of the wheel relative to the ground

The [math]\displaystyle{ \theta }[/math] angle in the figure orientates the straight line that goes through the cavity centre (point [math]\displaystyle{ \Os }[/math] fixed to the ground) and the wheels centre [math]\displaystyle{ \Cs }[/math]. Therefore [math]\displaystyle{ \velang{$\Os_\Es\Cs$}{E} = \vec{\dot{\theta}} = \odot\dth }[/math]. This angular velocity is not that of the wheels! Point [math]\displaystyle{ \Os }[/math] of the wheels goes through the cavity centre only instantaneously, as t becomes the contact point with the cavity later on, which is the wheels ICR relative to the ground.

As point [math]\displaystyle{ \Cs }[/math] belongs to both the [math]\displaystyle{ \vec{\Os_\Es\Cs} }[/math] line and the wheels, its velocity can be calculated from the ICR of [math]\displaystyle{ \vec{\Os_\Es\Cs} }[/math] and the wheels ICR:

[math]\displaystyle{ \cir{\Os_\Es\Cs}{\Es} = \Os_\Es }[/math]

[math]\displaystyle{ \vel{C}{E} = \velang{$\Os_\Es\Cs$}{E}\times\vec{\Os_\Es\Cs} = (\odot\dth)\times(\nearrow 2\Rs)^* = (\nwarrow 2\Rs\dth)^* }[/math]

[math]\displaystyle{ \cir{\Os_\Es\Cs}{\Es} = \Js }[/math]

[math]\displaystyle{ \vel{C}{E} = \velang{wheels}{E}\times\vec{\Js\Cs} = \velang{wheels}{E}\times(\swarrow\Rs)^* = (\nwarrow 2\Rs\dth)^*\Rightarrow\velang{wheels}{R} = (\otimes 2\dth) }[/math]

NOTE*: though the drawings [math]\displaystyle{ \nearrow }[/math], [math]\displaystyle{ \nwarrow }[/math] i [math]\displaystyle{ \searrow }[/math] suggest that the vectors define a 45° angle with the horizontal black line in the figure (origin of the [math]\displaystyle{ \theta }[/math] angle), they have to be understood only qualitatively. From now on, the asterisk indicates that the vector direction is only qualitative.

2. Find the curvature radius of the P trajectory relative to the ground

The calculation of the curvature radius calls for the knowledge of the velocity and the normal acceleration:

[math]\displaystyle{ \mathfrak{\textrm{R}}_{\mathrm{E}}(\mathbf{P})=\frac{\mathbf{v}_{\mathrm{E}}^2(\mathbf{P})}{\left|\mathrm{a}_{\mathrm{E}}^n(\mathbf{P})\right|}=\frac{\left(\bar{\mathbf{\mathbf{\Omega}}}_{\mathrm{E}}^{\text {wheel }} \times \overline{\mathbf{J P}}\right)^2}{\left|\mathrm{a}_{\mathrm{E}}^n(\mathbf{P})\right|}=\frac{(2 \mathrm{\textrm{R}} \dot{\theta})^2}{\left|\mathrm{a}_{\mathrm{E}}^n(\mathbf{P})\right|} }[/math]

The acceleration can be obtained through RBD from that of point [math]\displaystyle{ \Cs }[/math], which describes a circular trajectory about the ground-fixed point [math]\displaystyle{ \mathbf{O} }[/math] ([math]\displaystyle{ \Os_\Es }[/math]), with radius 2R and associated angular velocity [math]\displaystyle{ \overline{\dot{\theta}} }[/math]:

[math]\displaystyle{ \overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{C})=\overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{O}_\mathrm{E})+\overline{\boldsymbol{\mathbf{\mathbf{\Omega}}}}_{\mathrm{E}}^{\mathbf{O}_{\mathrm{E}}\mathrm{C}} \times\left[\overline{\boldsymbol{\mathbf{\mathbf{\Omega}}}}_{\mathrm{E}}^{\mathbf{O}_{\mathrm{E}}\mathrm{C}} \times\left(\nearrow 2\textrm{R}\right)^*\right]+\overline{\boldsymbol{\alpha}}_{\mathrm{E}}^{\mathbf{O}_{\mathrm{E}}\mathrm{C}}\times\left(\nearrow 2\textrm{R}\right)^*=(\odot \dot{\theta}) \times\left[(\odot \dot{\theta}) \times\left(\nearrow 2\textrm{R}\right)^*\right]+(\odot \ddot{\theta}) \times\left(\nearrow 2\textrm{R}\right)^* =(\nwarrow 2 \textrm{R} \ddot{\theta})^*+\left(\swarrow 2 \textrm{R} \dot{\theta}^2\right)^* }[/math]

[math]\displaystyle{ \overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{P})=\overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{C})+\overline{\boldsymbol{\mathbf{\mathbf{\Omega}}}}_{\mathrm{E}}^{\text {wheel}} \times\left(\bar{\mathbf{\mathbf{\Omega}}}_{\mathrm{E}}^{\text {wheel }} \times \overline{\mathbf{C P}}\right)+ \overline{\boldsymbol{\alpha}}_{\mathrm{E}}^{\text {wheeñ }} \times \overline{\mathbf{C P}}= (\nwarrow 2 \textrm{R} \ddot{\theta})^*+\left(\swarrow 2 \textrm{R} \dot{\theta}^2\right)^*+(\otimes 2 \dot{\theta}) \times\left[(\otimes 2 \dot{\theta}) \times(\swarrow \textrm{R})^*\right]+(\otimes 2 \ddot{\theta}) \times(\swarrow \textrm{R})^* }[/math]

[math]\displaystyle{ \overline{\mathbf{a}}_{\mathrm{E}}^{\mathrm{n}}(\mathbf{P})=\left(\swarrow 2 \mathrm{\textrm{R}} \dot{\theta}^2\right)^*+(\otimes 2 \dot{\theta}) \times\left[(\otimes 2 \dot{\theta}) \times(\swarrow \mathrm{\textrm{R}})^*\right] =\left(\swarrow 2 \textrm{R} \dot{\theta}^2\right)^*+(\otimes 2 \dot{\theta}) \times(\searrow 2 \textrm{R} \dot{\theta})^*=\left(\swarrow 6 \textrm{R} \dot{\theta}^2\right)^* }[/math]

[math]\displaystyle{ \Re_{\mathrm{E}}(\mathbf{P})=\frac{(2 \mathrm{\textrm{R}} \dot{\theta})^2}{6 \mathrm{\textrm{R}} \dot{\theta}^2}=\frac{2}{3} \mathrm{\textrm{R}} }[/math]

🔎 Exercici C5-E.2

---

EN CONSTRUCCIÓ

Resolució ➕

EN CONSTRUCCIÓ

🔎 Exercici C5-E.3

---

EN CONSTRUCCIÓ

Resolució ➕

EN CONSTRUCCIÓ

© Universitat Politècnica de Catalunya. Tots els drets reservats

---

---

<<< C4. Cinemàtica del sòlid rígid

D.1 EN CONSTRUCCIÓ >>>