C5. Rigid body kinematics: planar motion
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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:
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).
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
✏️ 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).
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✏️ 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}}_\text{REL}||\vec{\textbf{v}}_\text{tr} }[/math]. From that point, [math]\displaystyle{ \vec{\textbf{v}}_\text{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.
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✏️ 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:
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).
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).
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
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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 {wheel }} \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]
3. Find the curvature radius of the trajectory of point O of the wheel relative to the ground
- The steps are exactly the same as in the previous section:
- [math]\displaystyle{ \mathfrak{R}_{\mathrm{E}}\left(\mathbf{O}_{\text {wheel }}\right)=\frac{\mathbf{v}_{\mathrm{E}}^2\left(\mathbf{O}_{\text {wheel }}\right)}{\left|\mathrm{a}_{\mathrm{E}}^{\mathrm{n}}\left(\mathbf{O}_{\text {wheel}}\right)\right|}=\frac{\left(\overline{\boldsymbol{\mathbf{\Omega}}}_{\mathrm{E}}^{\text {wheel }} \times \overline{\mathbf{J} \mathbf{O}_{\text {wheel}}}\right)^2}{\left|\mathrm{a}_{\mathrm{E}}^{\mathrm{n}}\left(\mathbf{O}_{\text {wheel }}\right)\right|}=\frac{(3 \mathrm{R} \dot{\theta})^2}{\left|\mathrm{a}_{\mathrm{E}}^{\mathrm{n}}\left(\mathbf{O}_{\text {wheel }}\right)\right|} }[/math]
- [math]\displaystyle{ \overline{\mathbf{a}}_{\mathrm{E}}\left(\mathbf{O}_{\text {wheel }}\right)=\overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{C})+\bar{\mathbf{\mathbf{\Omega}}}_{\mathrm{E}}^{\text {wheel }} \times\left(\bar{\mathbf{\mathbf{\Omega}}}_{\mathrm{E}}^{\text {wheel }} \times \overline{\mathbf{C O}}\right)+\overline{\boldsymbol{\alpha}}_{\mathrm{E}}^{\text {wheel }} \times \overline{\mathbf{C O}} =(\nwarrow 2 \mathrm{R} \ddot{\theta})^*+\left(\swarrow 2 \mathrm{R} \dot{\theta}^2\right)^*+(\otimes 2 \dot{\theta}) \times\left[(\otimes 2 \dot{\theta}) \times(\swarrow 2 \mathrm{R})^*\right]+(\otimes 2 \ddot{\theta}) \times(\swarrow 2 \mathrm{R})^* \\ }[/math]
- [math]\displaystyle{ \overline{\mathbf{a}}_{\mathrm{E}}^{\mathrm{n}}\left(\mathbf{O}_{\text {wheel }}\right)=\left(\swarrow 2 \mathrm{R} \dot{\theta}^2\right)^*+(\otimes 2 \dot{\theta}) \times\left[(\otimes 2 \dot{\theta}) \times(\swarrow 2 \mathrm{R})^*\right] =\left(\swarrow 2 \mathrm{R} \dot{\theta}^2\right)^*+(\otimes 2 \dot{\theta}) \times(\nwarrow 4 \mathrm{R} \dot{\theta})^*=\left(\swarrow 10 \mathrm{R} \dot{\theta}^2\right)^* }[/math]
[math]\displaystyle{ \mathfrak{R}_{\mathrm{E}}\left(\mathbf{O}_{\text {wheel }}\right)=\frac{(3 \mathrm{R} \dot{\theta})^2}{10 \mathrm{R} \dot{\theta}^2}=\frac{9}{10} \mathrm{R} }[/math]
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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.
🔎 EXAMPLE C5-E.2: gear change
The three wheels, with radius r, are articulated to an arm whose centre is articulated to the ground (E), and that rotates with angular velocity [math]\displaystyle{ \omega }[/math] relative to E. The central wheel is fixed to the ground. There is no sliding at the contact points between wheels, and between the wheels and the external ring.
1. Find the ICR of all elements relative to E
- The ICR of the arm relative to the ground is the arm midpoint as it is directly articulated to the ground. For geometrical reasons, the ring ICR is located at that same point. In both cases, they are permanent rotation centres.
- The two internal small wheels have a non sliding contact with the central wheel, which is fixed to the ground. Therefore, instantaneously those points have zero velocity relative to the ground and they are the ICR of the small wheels.
2. Find the angular velocities of the wheels and the ring relative to E.
- The velocity of the wheels centres can be obtained from the ring angular velocity and its ICR. On the other hand, those velocities come from the wheels rotation velocities about their ICR. Therefore:
- [math]\displaystyle{ 2 \mathrm{r} \mathbf{\mathbf{\Omega}}_{\mathrm{E}}^{\text {arm}}=\mathrm{r} \mathbf{\mathbf{\Omega}}_{\mathrm{E}}^{\text {small wheel }} \Rightarrow \mathbf{\mathbf{\Omega}}_{\mathrm{E}}^{\text {small wheel }}=2 \mathbf{\mathbf{\Omega}}_{\mathrm{E}}^{\text {arm }}=2 \mathbf{\mathbf{\omega}} . }[/math]
- From $\mathbf{\mathbf{\Omega}}_{\mathrm{E}}^{\text {small wheel }}$ , we can calculate the velocity of the contact point between the small wheel and the ring. Again, that velocity comes from the rotation of the ring about its ICR. Therefore:
- [math]\displaystyle{ 2 \mathrm{r} \mathbf{\mathbf{\Omega}}_{\mathrm{E}}^{\text {small wheel }}=3 \mathrm{r} \mathbf{\mathbf{\Omega}}_{\mathrm{E}}^{\text {ring}} \Rightarrow \mathbf{\mathbf{\Omega}}_{\mathrm{E}}^{\text {ring }}=\frac{2}{3} \mathbf{\mathbf{\Omega}}_{\mathrm{E}}^{\text {small wheel }}=\frac{4}{3} \mathbf{\mathbf{\omega}} . }[/math]
🔎 EXAMPLE C5-E.3: wheels with connecting rod
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The two wheels have the same radius and are connected through a rod articulated to two points of their periphery.
1. Find the OCR of the connecting rod relative to the ground.
- The wheels ICR are the contact points with the ground. From them, we can obtain the velocity direction of two points in the rod. The intersection of the two lines perpendicular to those velocities yields the rod ICR.
2. Find the angular velocities of the three elements relative to the ground.
- The angular velocity of each element can be obtained from the speed of a point of that element and its distance to the corresponding ICR:
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🔎 EXAMPLE C5-E.4: vehicle with an upper wheel
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- The vehicle consists of a T-shaped chassis and three identical wheels articulated to it. The two lower wheels have nonsliding contact points with the ground and the upper wheel. The chassis has a translational motion relative to the ground with speed v to the left.
1. Find the ICR and the angular velocities of th wheels relative to the ground
- The lower wheels have exactly the same motion. The speed of their centres is v, and the ICR is the contact point with the ground. Therefore, their angular velocity is counterclockwise with value [math]\displaystyle{ \vs/\rs }[/math].
- The upper wheel kinematics is not so straightforward as its ICR relative to the ground is not known. However, it can be obtained geometrically from the non sliding contacts with the lower wheels: the ICR has to be located at the intersection of the straight lines perpendicular to the velocity direction of those points. Therefore, [math]\displaystyle{ \textrm{ICR}^{\mathrm{upper wheel}}_{\mathrm{E}} = \mathbf{P} }[/math].
- Alternatively, the upper wheel ICR can be found from the analysis of the vehicle relative to the chassis. In that reference frame, the wheels ICR are located at their centres. As the chassis has a translational motion relative to the ground, the angular velocity of any wheel relative to the chassis is the same as that relative to the ground:
- The description of the upper wheel kinematics relative to the ground can be obtained from a composition of movements:
- [math]\displaystyle{ \left.\begin{array}{l}\text { AB: ground } \\ \text { REL : chassis }\end{array}\right\} \overline{\mathbf{v}}_{\mathrm{AB}}(\mathbf{P})=\overline{\mathbf{v}}_{\mathrm{REL}}(\mathbf{P})+\overline{\mathbf{v}}_{\mathrm{tr}}(\mathbf{P})=(\rightarrow \mathrm{v})+(\leftarrow \mathrm{v})=\overline{0} \Rightarrow \mathbf{P}=\mathrm{ICR}_{\mathrm{E}}^{\text {upper wheel }} }[/math]
2. Find the acceleration of point P relative to the ground.
The acceleration of point [math]\displaystyle{ \mathbf{P} }[/math] relative to the ground can be obtained from that of the upper wheel centre ([math]\displaystyle{ \mathbf{C} }[/math]) through the equation of accelerations for the rigid body (RBK) applied to that wheel:
- [math]\displaystyle{ \begin{aligned} & \overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{P})=\overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{C})+\bar{\mathbf{\Omega}}_{\mathrm{E}}^{\text {upper wheel }} \times\left(\bar{\mathbf{\Omega}}_{\mathrm{E}}^{\text {upper wheel }} \times \overline{\mathbf{C P}}\right)+\overline{\boldsymbol{\alpha}}_{\mathrm{E}}^{\text {upper wheel}} \times \overline{\mathbf{C P}}=(\leftarrow \dot{\mathrm{v}})+\left(\otimes \frac{\mathrm{v}}{\mathrm{R}}\right) \times\left[\left(\otimes \frac{\mathrm{v}}{\mathrm{R}}\right) \times(\uparrow \mathrm{R})\right]+\left(\otimes \frac{\dot{\mathrm{v}}}{\mathrm{R}}\right) \times(\uparrow \mathrm{R})= \\ & =(\leftarrow \dot{\mathrm{v}})+\left(\otimes \frac{\mathrm{v}}{\mathrm{R}}\right) \times(\rightarrow \mathrm{v})+(\rightarrow \dot{\mathrm{v}})=\left(\downarrow \frac{\mathrm{v}^2}{\mathrm{R}}\right) \end{aligned} }[/math]
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🔎 EXAMPLE C5-E.5: vehicle with differently sized wheels
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The vehicle consists of a chassis and three wheels articulated to it. The wheels with radii R and [math]\displaystyle{ \rs_2 }[/math] do not slide on the ground, and that with radius [math]\displaystyle{ \rs_1 }[/math] has a non sliding contact with the R wheel. The chassis has a rightwards translational motion relative to the ground with speed v.
1. Find the ICR and the angular velocity of wheel 1 relative to the ground
- The ICR of the R wheel and wheel 2 relative to the ground are their contact points with the ground. As their centres have a v velocity to the right, their rotations are clockwise: [math]\displaystyle{ \bar{\mathbf{\mathbf{\Omega}}}_{\mathrm{E}}^{\text {wheel } 2}=\otimes \frac{\mathrm{v}}{\mathrm{r}_2}, \bar{\mathbf{\mathbf{\Omega}}}_{\mathrm{E}}^{\text {Rwheel }}=\otimes \frac{\mathrm{v}}{\mathrm{R}} }[/math]. Those angular velocities coincide with the angular velocities relative to the chassis, as the latter does not rotate relative to the ground.
- The kinematic analysis of the small wheel relative to the chassis is straightforward, and a composition of movements leads to the kinematics relative to the ground.
- [math]\displaystyle{ \overline{\mathbf{v}}_{\mathrm{E}}(\mathrm{ICR} \equiv \mathrm{I})=\overline{\mathbf{v}}_{\mathrm{E}}(\mathbf{C})+\bar{\mathbf{\mathbf{\Omega}}}_{\mathrm{E}}^{\text {wheel1 }} \times \overline{\mathrm{CI}}, \:\:\:\:\:\: \overline{0}=(\rightarrow \mathrm{v})+\left(\odot \frac{\mathrm{v}}{\mathrm{r}_1}\right) \times(\uparrow \mathrm{d}) \Rightarrow \mathrm{d}=\mathrm{r}_1 }[/math]
- The ICR coincides with the point located in the highest position
2. Find the centre of curvature of the [math]\displaystyle{ \mathbf{P} }[/math] trajectory relative to the ground
- The curvature centre is obtained from the curvature radius: [math]\displaystyle{ \mathfrak{R}_{\mathrm{E}}(\mathbf{P})=\frac{\mathrm{v}_{\mathrm{E}}^2(\mathbf{P})}{\left|a_{\mathrm{E}}^{\mathrm{n}}(\mathbf{P})\right|} }[/math].
- The velocity of point [math]\displaystyle{ \mathbf{P} }[/math] relative to the ground can be calculated form the ICR of wheel [math]\displaystyle{ 2(\equiv \mathrm{I}) }[/math]:
- [math]\displaystyle{ \overline{\mathbf{v}}_{\mathrm{E}}(\mathbf{P})=\bar{\mathbf{\mathbf{\Omega}}}_{\mathrm{E}}^{\text {wheel2 }} \times \overline{\mathbf{I} \mathbf{P}}=\left(\otimes \frac{\mathbf{v}}{\mathrm{r}_2}\right) \times\left[\uparrow\left(\mathrm{r}_2-\mathrm{r}_1\right)\right]=\left[\rightarrow \frac{\mathbf{v}}{\mathrm{r}_2}\left(\mathrm{r}_2-\mathrm{r}_1\right)\right] }[/math]
- The acceleration of point [math]\displaystyle{ \mathbf{P} }[/math] relative to the ground can be calculated from that of the wheel2 centre([math]\displaystyle{ \mathbf{C} }[/math]):
- [math]\displaystyle{ \begin{aligned} & \overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{P})=\overline{\mathbf{a}}_{\mathrm{T}}(\mathbf{C})+\bar{\mathbf{\mathbf{\Omega}}}_{\mathrm{E}}^{\mathrm{wheel} 2} \times\left(\overline{\boldsymbol{\mathbf{\mathbf{\Omega}}}}_{\mathrm{E}}^{\mathrm{wheel} 2} \times \overline{\mathbf{C P}}\right)+\overline{\boldsymbol{\alpha}}_{\mathrm{E}}^{\mathrm{wheel} 2} \times \overline{\mathbf{C P}}=(\rightarrow \dot{\mathrm{v}})+\left(\otimes \frac{\mathrm{v}}{\mathrm{r}_2}\right) \times\left[\left(\otimes \frac{\mathrm{v}}{\mathrm{r}_2}\right) \times\left(\downarrow \mathrm{r}_2\right)\right]+\left(\otimes \frac{\dot{\mathrm{v}}}{\mathrm{r}_2}\right) \times\left(\downarrow \mathrm{r}_2\right)=\\ &=(\rightarrow \dot{\mathrm{v}})+\left(\otimes \frac{\mathrm{v}}{\mathrm{r}_2}\right) \times(\leftarrow \mathrm{v})+(\leftarrow \dot{\mathrm{v}}) \\ & \overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{P})=\overline{\mathbf{a}}_{\mathrm{E}}^{\mathrm{n}}(\mathbf{P})=\left(\uparrow \frac{\mathrm{v}^2}{\mathrm{r}_2}\right), \:\:\:\:\:\:\: \Re_{\mathrm{E}}(\mathbf{P})=\frac{\mathrm{v}^2 \mathrm{r}_2^2 /\left(\mathrm{r}_2-\mathrm{r}_1\right)^2}{\mathbf{v}^2 / \mathrm{r}_2}=\frac{\mathrm{r}_2}{\left(\mathrm{r}_2-\mathrm{r}_1\right)^2} \end{aligned} }[/math]
- The curvature centre of the [math]\displaystyle{ \mathbf{P} }[/math] trajectory is located above [math]\displaystyle{ \mathbf{P} }[/math] (as the normal acceleration points upwards) at a distance higher than [math]\displaystyle{ \mathrm{r}_2 }[/math], and therefore above the wheel centre.
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🔎 Exercici C5-E.6: bicicleta
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The bicycle moves without sliding on a flat ground. The sprocket and the chainring have different radii, r y R respectively, and the pedals length is L.
1. Find the ICR and the angular velocity of the pedals relative to the ground
- The wheels ICR relative to the ground are their contact points with the ground. The sprocket is fixed to the rear wheel, therefore its ICR coincides with that of the wheel. If the bicycle moves forward with speed v relative to the ground, the angular velocity of those elements is straightforward:
- [math]\displaystyle{ \begin{aligned} & \overline{\mathbf{v}}_{\mathrm{E}}\left(\mathbf{C}_{\text {wheel }}\right)=(\rightarrow \mathrm{v})=\bar{\mathbf{\Omega}}_{\mathrm{E}}^{\text {wheel }} \times \overline{\mathbf{ICR}_{\mathrm{E}}^{\text {wheel }} \mathbf{C}_{\text {wheel }}}=\bar{\mathbf{\Omega}}_{\mathrm{E}}^{\text {wheel }} \times\left(\uparrow \mathrm{R}_{\text {wheel }}\right) \\ & \bar{\mathbf{\Omega}}_{\mathrm{E}}^{\text {wheel }}=\left(\otimes \frac{\mathrm{v}}{\mathrm{R}_{\text {wheeñ }}}\right) \end{aligned} }[/math]
- That angular velocity is the same as that of the wheel relative to the bicycle frame, as the frame has a translational motion relative to the ground:
- [math]\displaystyle{ \left.\begin{array}{l}\text { AB: } \mathrm{E} \\ \text { REL : frame }\end{array}\right\} \bar{\mathbf{\Omega}}_{\mathrm{AB}}^{\text {wheel }}=\bar{\mathbf{\Omega}}_{\mathrm{REL}}^{\text {wheel }}+\bar{\mathbf{\Omega}}_{\mathrm{ar}}=\bar{\mathbf{\Omega}}_{\mathrm{REL}}^{\text {wheel }}+\bar{\mathbf{\Omega}}_{\mathrm{AB}}^{\text {wheel}\in\text{REL}}=\bar{\mathbf{\Omega}}_{\mathrm{AB}}^{\text {wheel }}+\overline{0}=\left(\otimes \frac{\mathrm{v}}{\mathrm{R}_{\text {wheel }}}\right) }[/math].
- The ICR of the pedals relative to the frame is straightforward (though that relative to the ground is not): it is its centre (which is also the chainring centre). From that information, the angular velocity of the pedals relative to the frame can be obtained taking into account that the chainring rotation (which is the same as that of the pedals) is transmitted to the rear wheel through an inextensible chain:
- [math]\displaystyle{ \textrm{r} \mathbf{\Omega}_{\text {frame }}^{\text {sprocket }}=\textrm{r} \frac{\textrm{v}}{\textrm{R}_{\text {wheel }}}=\textrm{R} \mathbf{\Omega}_{\text {frame }}^{\text {chainring }} \Rightarrow \mathbf{\Omega}_{\text {frame }}^{\text {chainring }}=\mathbf{\Omega}_{\text {frame }}^{\text {pedals }}=\frac{\textrm{r}}{\textrm{R}} \frac{\textrm{v}}{\textrm{R}_{\text {wheel }}} }[/math]
- The ICR of the pedals relative to the ground can be calculated through the velocity distribution equation for the pedals:
- [math]\displaystyle{ \begin{aligned} & \overline{\mathbf{v}}_{\mathrm{E}}\left(\mathrm{ICR}_{\mathrm{E}}^{\text {pedals }} \equiv \mathrm{I}\right)=\overline{0}=\overline{\mathbf{v}}_{\mathrm{E}}(\mathrm{C})+\bar{\mathbf{\Omega}}_{\mathrm{E}}^{\text {pedals }} \times \overline{\mathrm{CI}}=(\rightarrow \mathrm{v})+\left(\otimes \frac{\mathrm{r}}{\mathrm{R}} \frac{\mathrm{v}}{\mathrm{R}_{\text {wheel }}}\right) \times(\downarrow \mathrm{d}) \\ & \overline{0}=(\rightarrow \mathrm{v})+\left(\leftarrow \mathrm{d} \frac{\mathrm{r}}{\mathrm{R}} \frac{\mathrm{v}}{\mathrm{R}_{\text {wheel }}}\right) \Rightarrow \mathrm{d}=\mathrm{R}_{\text {wheel }} \frac{\mathrm{R}}{\mathrm{r}} \\ & \end{aligned} }[/math]
- For [math]\displaystyle{ \mathrm{R}\gt \mathrm{r} }[/math], the pedals ICR relative to the ground is under ground level.
2. Find the curvature centre of the P trajectory relative to the ground
- The curvature centre is obtained from the curvature radius: [math]\displaystyle{ \mathfrak{R}_{\mathrm{E}}(\mathbf{P})=\frac{\mathrm{v}_{\mathrm{E}}^2(\mathbf{P})}{\left|\mathrm{a}_{\mathrm{E}}^{\mathrm{n}}(\mathbf{P})\right|} }[/math].
- The velocity and acceleration of point P [math]\displaystyle{ \mathbf{P} }[/math] relative to the ground can be calculated from those of the pedals centre ([math]\displaystyle{ \textbf{C} }[/math]):
- [math]\displaystyle{ \overline{\mathbf{v}}_{\mathrm{E}}(\mathbf{P})=\overline{\mathbf{v}}_{\mathrm{E}}(\mathbf{C})+\bar{\mathbf{\Omega}}_{\mathrm{E}}^{\text {pedals }} \times \overline{\mathrm{IP}}=(\rightarrow \mathrm{v})+\left(\otimes \frac{\mathrm{r}}{\mathrm{R}} \frac{\mathrm{v}}{\mathrm{R}_{\text {wheel }}}\right) \times(\downarrow \mathrm{L})=(\rightarrow \mathrm{v})+\left(\leftarrow \mathrm{L} \frac{\mathrm{r}}{\mathrm{R}} \frac{\mathrm{v}}{\mathrm{R}_{\text {wheel }}}\right) }[/math]
- [math]\displaystyle{ \overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{P})=\overline{\mathbf{a}}_{\mathrm{E}}(\mathbf{C})+\bar{\mathbf{\Omega}}_{\mathrm{E}}^{\text {pedals }} \times\left(\bar{\mathbf{\Omega}}_{\mathrm{E}}^{\text {pedals }} \times \overline{\mathbf{C P}}\right)+\bar{\mathbf{\alpha}}_{\mathrm{E}}^{\text {pedals }} \times \overline{\mathbf{C P}}= (\rightarrow \dot{\textrm{v}})+\left(\otimes \frac{\textrm{r}}{\textrm{R}} \frac{\textrm{v}}{\textrm{R}_{\text{wheel}}}\right) \times\left[\left(\otimes \frac{\textrm{r}}{\textrm{R}} \frac{\mathrm{v}}{\textrm{R}_{\text {wheel }}}\right) \times(\downarrow \textrm{L})\right]+\left(\otimes \frac{\mathrm{r}}{\mathrm{R}} \frac{\dot{\textrm{v}}}{\textrm{R}_{\text {wheel }}}\right) \times(\downarrow \textrm{L}) }[/math]
- [math]\displaystyle{ \overline{\mathbf{a}}_{\mathrm{E}}^{\mathrm{n}}(\mathbf{P})=\left(\otimes \frac{\textrm{r}}{\textrm{R}} \frac{\textrm{v}}{\textrm{R}_{\text {wheel }}}\right) \times\left(\leftarrow \textrm{L} \frac{\textrm{r}}{\textrm{R}} \frac{\textrm{v}}{\textrm{R}_{\text {wheel }}}\right)=\left[\uparrow \textrm{L}\left(\frac{\textrm{r}}{\textrm{R}}\right)^2 \frac{\textrm{v}^2}{\textrm{R}_{\text {wheel }}^2}\right],\:\:\:\:\:\: \Re_E(P)=\frac{\textrm{L} \frac{\textrm{r}}{\textrm{R}} \frac{\textrm{v}}{\textrm{R}_{\text {wheel }}}}{\textrm{L}\left(\frac{\textrm{r}}{\textrm{R}}\right)^2 \frac{\textrm{v}^2}{\textrm{R}_{\text {wheel }}^2}}=\left(\frac{\textrm{R}}{\textrm{r}}\right)^2 \textrm{R}_{\text {wheel }} }[/math]
- The curvature centre of the [math]\displaystyle{ \mathbf{P} }[/math] trajectory is located above [math]\displaystyle{ \mathbf{P} }[/math] (as the normal acceleration points upwards).
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