D6. Examples of 2D dynamics
[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{\ps}{\textrm{p}} \newcommand{\hs}{\textrm{h}} \newcommand{\cs}{\textrm{c}} \newcommand{\gs}{\textrm{g}} \newcommand{\Ns}{\textrm{N}} \newcommand{\Hs}{\textrm{H}} \newcommand{\Fs}{\textrm{F}} \newcommand{\ms}{\textrm{m}} \newcommand{\ts}{\textrm{t}} \newcommand{\us}{\textrm{u}} \newcommand{\vs}{\textrm{v}} \newcommand{\Rs}{\textrm{R}} \newcommand{\Ts}{\textrm{T}} \newcommand{\Ls}{\textrm{L}} \newcommand{\Bs}{\textrm{B}} \newcommand{\es}{\textrm{e}} \newcommand{\fs}{\textrm{f}} \newcommand{\is}{\textrm{i}} \newcommand{\Is}{\textrm{I}} \newcommand{\ks}{\textrm{k}} \newcommand{\js}{\textrm{j}} \newcommand{\rs}{\textrm{r}} \newcommand{\ss}{\textrm{s}} \newcommand{\Os}{\textbf{O}} \newcommand{\Gs}{\textbf{G}} \newcommand{\Cbf}{\textbf{C}} \newcommand{\Or}{\Os_\Rs} \newcommand{\Qs}{\textbf{Q}} \newcommand{\Cs}{\textbf{C}} \newcommand{\Ps}{\textbf{P}} \newcommand{\Ss}{\textbf{S}} \newcommand{\Js}{\textbf{J}} \newcommand{\P}{\textrm{P}} \newcommand{\Q}{\textrm{Q}} \newcommand{\Ms}{\textrm{M}} \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{\vecdot}[1]{\overline{\dot{#1}}} \newcommand{\OQvec}{\vec{\Os\Qs}} \newcommand{\QPvec}{\vec{\Qs\Ps}} \newcommand{\JQvec}{\vec{\Js\Qs}} \newcommand{\GJvec}{\vec{\Gs\Js}} \newcommand{\OPvec}{\vec{\Os\textbf{P}}} \newcommand{\OCvec}{\vec{\Os\Cs}} \newcommand{\OGvec}{\vec{\Os\Gs}} \newcommand{\GCvec}{\vec{\Gs\Cs}} \newcommand{\PGvec}{\vec{\Ps\Gs}} \newcommand{\abs}[1]{\left|{#1}\right|} \newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}} \newcommand{\matriz}[9]{ \begin{bmatrix} {#1} & {#2} & {#3}\\ {#4} & {#5} & {#6}\\ {#7} & {#8} & {#9} \end{bmatrix}} \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{\psio}{\dot{\psi}_0} \newcommand{\Pll}{\textbf{P}_\textrm{lliure}} \newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)} \newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)} \newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)} \newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}} \newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}} }[/math]
This unit proposes a systematic method for solving problems in dynamics of systems consisting of any number of rigid bodies. If this number is high, defining the strategy to achieve the desired results is essential to avoid excessive development.
The examples focus on problems in plane dynamics (2D). Unit D7 analyses examples of 3D dynamics.
D6.1 2D kinematics and 2D dynamics
The condition for a system to exhibit 2D dynamics is not only that its kinematics be plane: it is also necessary that the angular momentum of each rigid body [math]\displaystyle{ \ss_\is }[/math] in the system relative to its center of inertia [math]\displaystyle{ \Gs_\is }[/math] be orthogonal to the plane of motion (section D4.5). This is equivalent to saying that the direction orthogonal to the plane of motion is a principal direction of inertia for [math]\displaystyle{ \Gs_\is }[/math].
✏️ EXAMPLE D6.1: 2D kinematics and 3D dynamics
- The thin homogeneous bar is hinged at [math]\displaystyle{ \Os }[/math] to a support that rotates relative to the ground with constant vertical angular velocity [math]\displaystyle{ \Omega_0 }[/math]. The bar has a single-point contact with the smooth ground, so its motion is planar: all its points (except point [math]\displaystyle{ \Os }[/math]) describe horizontal circular trajectories.
- The fact that the kinematics of the bar is 2D may suggest that only the vertical component of the AMT is necessary to study its dynamics. But this is not the case.
- The angular momentum of the bar at point [math]\displaystyle{ \Os }[/math] is:
[math]\displaystyle{ \braq{\overline{\mathbf{H}}_{\text {RTO=E}}(\Os)}{}= \matriz{\Is}{0}{0}{0}{0}{0}{0}{0}{\Is}\vector{\Omega_0 \sin \beta}{\Omega_0 \cos \beta}{0}= \vector{\Is\Omega_0 \sin \beta}{0}{0} . }[/math]
- Only the horizontal component of this vector is variable (it has a variable direction because of [math]\displaystyle{ \Omega_0 }[/math]). Its geometrical time derivative yields:
[math]\displaystyle{ \dot{\overline{\mathbf{H}}}_{\text {RTO}} (\Os)=(\Uparrow \Omega_0) \times (\Rightarrow \Is \Omega_0 \sin \beta \cos \beta)= \otimes\Is \Omega_0^2 \sin\beta \cos\beta. }[/math]
This time derivative can also be obtained through the vector basis:
[math]\displaystyle{ \braq{\dot{\overline{\mathbf{H}}}_{\text {RTO}} (\Os)}{}= \vector{\Omega_0 \sin \beta}{\Omega_0 \cos \beta}{0} \times \vector{\Is \Omega_0 \sin \beta}{0}{0} =\vector{0}{0}{-\Is \Omega_0^2\sin\beta\cos\beta}. }[/math]
When applying the AMT at [math]\displaystyle{ \Os }[/math] to the bar, component 3, which is horizontal, cannot be ignored.
D6.2 Free-body diagram (FBD) and roadmap
When a system consisting of only one rigid body has 2D dynamics, the number of equations to solve the problem is 3: the two components of the LMT in the plane of motion and the component of the AMT perpendicular to this plane. Depending on what we want to calculate (equation of motion, constraint force...), it may not be necessary to apply both theorems.
Finding a quick way to achieve the desired result requires the development of a roadmap (which is a statement of the strategy to be followed). Exploring possible strategies and choosing an appropriate one requires a good understanding of the kinematics and dynamics of the rigid body. Therefore, it is always recommended to start by investigating its velocity distribution and the description of the interactions on it. The representation of these interactions constitutes the free-body diagram (FBD).
D6.3 Examples with just one rigid body
✏️ EXAMPLE D6.2: sliding limit condition
The homogeneous disk, with mass m and radius R, is on a smooth horizontal floor and has a single-point of contact with a rough wall. We want to investigate the minimum value of the friction coefficient between the disk and the wall, as a function of F, so that there is no sliding between these two elements.
The kinematic and dynamic descriptions of the disk are:
The unknowns of the problem are [math]\displaystyle{ \ddot{\theta}, \Ns, \Ts }[/math]. The tangential force T is the target of the calculation, since the limit condition for imminent sliding is formulated on this force: [math]\displaystyle{ \Ts_\mathrm{màx}=\mu_\mathrm{mín}\Ns. }[/math].
Roadmap
The force T appears both in the horizontal component of the LMT and in the component perpendicular to the plane of the AMT at G. However, the unknown [math]\displaystyle{ \ddot{\theta} }[/math] also appears in both equations. On the other hand, the value of N must also be known in order to impose the limit condition. Therefore: [math]\displaystyle{ \boxed{\text{Roadmap: LMT and AMT at } \Gs} }[/math]
[math]\displaystyle{
\left.\begin{array}{ll}
\text { LMT: } \hspace{1.2cm} \left\{\begin{array}{l}
(\rightarrow \mathrm{F} \cos \beta)+(\leftarrow \mathrm{T})=(\rightarrow \mathrm{mR} \ddot{\theta}) \\
(\downarrow \mathrm{F} \sin \beta)+(\uparrow \mathrm{N})=0
\end{array}\right. \\
\text { AMT at }\Gs: \quad (\otimes\Ts\Rs)=(\otimes \Is_\mathrm{G} \ddot{\theta})=\left(\otimes \frac{1}{2} \ms \Rs^2\ddot{\theta} \right)
\end{array}\right\}
\Rightarrow
\left\{\begin{array}{l}
\ddot{\theta}=\frac{2}{3} \frac{\mathrm{F}}{\mathrm{mR}} \cos \beta \\
\mathrm{T}=\frac{1}{3} \mathrm{~F} \cos \beta \\
\mathrm{N}=\mathrm{F} \sin \beta
\end{array}\right.
}[/math]
[math]\displaystyle{ \mathrm{T}_{\text {max }}=\mu_{\text {min }} \mathrm{N} \quad \Rightarrow \quad \frac{1}{3} \mathrm{F}_{\mathrm{max} } \cos \beta=\mu_{\text {min }} \mathrm{F}_{\mathrm{max}} \sin \beta \quad \Rightarrow \quad \mu_{\text{min} }=\frac{1}{3 \tan \beta} \\
}[/math]
✏️ EXAMPLE D6.3: bar with translational motion
Point P of the homogeneous bar, with mass m and length 2L, slides along a rough guide. We want to investigate for what value of the angle the angular velocity of the bar relative to the ground is zero.
The kinematic and dynamic descriptions of the bar are:
Roadmap
If the bar does not rotate, its angular momentum at any of its points is permanently zero, and therefore its derivative is also zero. This condition suggests the use of the AMT to solve the problem.
| If the AMT is applied at G, the condition for having a zero time derivative of the angular momentum with respect to this point implies that the direction of resultant of the two forces on P goes through G: [math]\displaystyle{ \boxed{\text{Roadmap: AMT at } \Gs} }[/math]
|
Since P is a point with nonzero acceleration relative to the ground, the unknown appears in the AMT, and that is why the LMT must be applied to calculate it.
[math]\displaystyle{ \text {LMT: } \quad\left\{\begin{array}{l}
(\nearrow \mathrm{N})+(\swarrow \mathrm{mg} \cos \delta)=0 \\
(\nwarrow \mu \mathrm{N})+(\searrow \mathrm{mg} \sin \delta)=(\searrow \mathrm{m} \dot{\mathrm{v}})
\end{array}\right\} \Rightarrow \dot{\mathrm{v}}=\mathrm{g}(\sin \delta-\mu \cos \delta) }[/math]
[math]\displaystyle{ \text{AMT at } \Ps : \quad \sum \overline{\Ms}_\mathrm{ext}(\Ps) - \PGvec \times \ms\acc{P}{Gal}=\overline{0} }[/math]
[math]\displaystyle{ \sum \overline{\Ms}_\mathrm{ext}(\Ps)=\PGvec \times \ms\overline{\gs} }[/math]
[math]\displaystyle{ \sum \overline{\Ms}_\mathrm{ext}(\Ps)=\left[ (\swarrow \Ls \cos \beta )+ (\searrow \Ls \sin\beta )\right]\times\left[ (\swarrow \ms \gs \cos \delta )+(\searrow \ms\gs\sin\delta)\right] = \left[ \odot \ms\gs\Ls (\sin\delta\cos\beta-\cos\delta\sin\beta)\right] }[/math]
[math]\displaystyle{ \PGvec \times \ms\acc{P}{Gal}=\left[ (\swarrow \Ls \cos \beta )+ (\searrow \Ls \sin\beta )\right] \times(\searrow\ms\dot{\vs})=(\odot\ms\Ls\dot{\vs}\cos\beta)=\left[ \odot \ms\gs\Ls (\sin\delta\cos\beta-\mu\cos\delta\cos\beta)\right] }[/math]
[math]\displaystyle{ \left[ \odot \ms\gs\Ls (\sin\delta\cos\beta-\cos\delta\sin\beta)\right]-\left[ \odot \ms\gs\Ls (\sin\delta\cos\beta-\mu\cos\delta\cos\beta)\right] =0 }[/math]
[math]\displaystyle{ \ms\gs\Ls\cos\delta(\sin\beta-\mu\cos\beta)=0 \quad \Rightarrow \quad \tan \beta =\mu }[/math]
✏️ EXAMPLE D6.4: ring on a rough ground
The homogeneous ring, with mass m and radius R, initially has a translational motion relative to the ground. We want to investigate how long it will take to stop sliding relative to the ground.
The kinematic description of the ring at three different time instants (initial, intermediate and final), and the dynamic description are:
Roadmap
While sliding, the ring has two DoF [math]\displaystyle{ (\vs,\Omega) }[/math], whereas at the end it only has one ([math]\displaystyle{ \vs_\fs }[/math]). Since we are trying to investigate how the motion changes from translation to rotation, we have to determine the equations of motion [math]\displaystyle{ (\dot{\vs},\dot{\Omega}) }[/math].
The LMT provides two equations, but both will contain N. Therefore, the AMT is also needed. The problem suggests two points for application: G and the contact point between the ring and the ground (point J). If the latter is chosen, it must be specified whether it is [math]\displaystyle{ \Js_\mathrm{ring} }[/math] , [math]\displaystyle{ \Js_\mathrm{ground} }[/math] or [math]\displaystyle{ \Js_\mathrm{geom} }[/math] section D4.5, and this is an added difficulty. For this reason, the roadmap will be:
[math]\displaystyle{ \boxed{\text{Roadmap: LMT and AMT at } \Gs} }[/math]
[math]\displaystyle{ \text {LMT: } \quad\left\{\begin{array}{l}
(\uparrow \mathrm{N})+(\downarrow \mathrm{mg})=0 \\
(\leftarrow \mu \mathrm{N})=(\rightarrow \mathrm{m} \dot{\mathrm{v}})
\end{array}\right\} \Rightarrow \dot{\mathrm{v}}=-\mu\mathrm{g} }[/math]
[math]\displaystyle{ \text{AMT at } \Gs : \quad \sum \overline{\Ms}_\mathrm{ext}(\Gs)=\dot{\overline{\Hs}}_\mathrm{RTG}(\Gs) \quad \Rightarrow \quad (\otimes \mu\ms\gs\Rs)=(\otimes\Is_\mathrm{G}\dot{\Omega})=(\otimes \ms\Rs^2\dot{\Omega}) \quad \Rightarrow \quad \dot{\Omega}=\mu\frac{\gs}{\Rs} }[/math]
Since [math]\displaystyle{ \dot{\vs} }[/math] and [math]\displaystyle{ \dot{\Omega} }[/math] are constant, the G motion and the rotation are uniformly accelerated. Therefore:
[math]\displaystyle{ \vs(\ts)=\vs_0 - \mu\gs\ts \quad,\quad \Omega(\ts)=\mu\frac{\gs}{\Rs}\ts }[/math]
[math]\displaystyle{ \text {For }\ts=\ts_\fs: \quad\left\{\begin{array}{l}
\vs(\ts_\fs)=\vs_0 - \mu\gs\ts_\fs \quad,\quad \Omega(\ts_\fs)=\mu\frac{\gs}{\Rs}\ts_\fs \\
\quad \\
\vs(\ts_\fs)=\Rs\Omega(\ts_\fs)
\end{array}\right\} \Rightarrow \ts_\fs=\frac{\vs_0}{2\mu\gs} }[/math]
✏️ EXAMPLE D6.5: articulated plate
The homogeneous plate of mass m is connected to the ground by two identical bars, of negligible mass hinged at their ends, and by an inextensible thread. We want to calculate the acceleration of the plate centre of inertia o just after the thread is cut.
Because of the hinges, the two bars describe an identical rotational motion relative to the ground, and this implies that the plate has a circular translational motion relative to the ground. The kinematic description (for any value of m) of velocities and accelerations is:
Although the system contains three bodies, it can be reduced to a single body if the bars are treated as CAE. The constraint torsor associated with the indirect constraint at any point on the plate is straightforwatrd from the kinematics described just above: it is a force in the direction of the bars and a moment orthogonal to the plane of the plate. The dynamic description of the plate is:
|
Roadmap [math]\displaystyle{ \mathrm{AMT}]_\mathrm{tangential} : (\searrow\ms\gs\sin\theta_0)=\ms\acc{G}{T} \quad \Rightarrow \quad \acc{G}{T}=(\searrow \gs\sin\theta_0) }[/math] |
D6.4 General diagram of interactions (GDI)
When solving multibody system dynamics problems, a very useful tool for defining an appropriate roadmap is the general diagram of interactions (GDI). It is a generic representation of the rigid bodies in the system, and the interactions between them and the reference frame (which is very often the ground). When those interactions are constraint interactions, a brief description must be given (for example, “articulation”, “cylindrical joint”, “spherical joint”, “single-point contact with sliding”...), and the number of unknowns associated to them must be indicated in parentheses (as has been done in the examples in unit D3).
The description of the system's velocity distribution and the DGI allow for an easy count of the number of equations available to solve the problem, and the number of unknowns (those associated with the DoF and those associated with the constraints), and to see whether the problem is determinate or indeterminate.
On the other hand, the DGI is an easy tool for visualizing the subsystems of rigid bodies in the system for which the application of the vector theorems would contain the unknown to be calculated. Once the subsystems that introduce fewer additional unknowns have been discovered, it is useful to accurately represent the external interactions on these subsystems in order to finally establish the appropriate strategy (which theorem to apply, at which point, which component to analyze...).
The systematic procedure proposed for solving dynamics problems is summarized in the following steps:
- description of the system’s velocity distribution
- general diagram of interactions
- exploration of the subsystems of rigid bodies for which the theorems include the unknowns to be calculated
- accurate description of the external interactions on these subsystems
- elaboration of a roadmap
D6.5 Examples of multibody systems
✏️ EXAMPLE D6.6: disk in a cylindrical cavity
| The homogeneous disk, with mass m and radius r, rotates and slides on a rough cylindrical ground driven by a motor that acts between the disk and the massless arm. There is a frictionless pin-slot constraint between the disk and the floor. The motor guarantees that the relative angular velocity between the disk and the floor is constant. We want to find the equation of motion for the [math]\displaystyle{ \theta }[/math] coordinate, the value of the motor torque that guarantees a constant [math]\displaystyle{ \omega_0 }[/math] and the force associated with the pin-slot constraint. |
The kinematic description of the system is:
AB:ground, REL: arm [math]\displaystyle{ \velang{disk}{ground}=\velang{disk}{AB}=\velang{disk}{REL}+\velang{}{tr}=(\otimes \omega_0)+(\odot\dot{\theta})=[\otimes(\omega_0-\dot{\theta})] }[/math]
Rigid body kinematics (rigid body: arm): [math]\displaystyle{ \vel{G}{T}=\vel{O}{T}+\OGvec\times\velang{arm}{T}=(\searrow \Rs)\times(\odot\dot{\theta})=(\nearrow\Rs\dot{\theta}) }[/math]
Rigid body kinematics (rigid body: disk): [math]\displaystyle{ \vel{J}{T}=\vel{G}{T}+\GJvec \times \velang{disk}{T}=(\nearrow\Rs\dot{\theta})+(\searrow \rs)\times[\otimes(\omega_0-\dot{\theta})]=(\swarrow [\rs\omega_0-(\Rs +\rs)\dot{\theta}] }[/math]
General diagram of interactions (s-p.c.w.s.=single-point contact with sliding).
It is a 2-DoF system ([math]\displaystyle{ \omega_0 }[/math] forced, [math]\displaystyle{ \dot{\theta} }[/math] free) with 4 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 3 equations per rigid body. It is a determinate problem:
equations: 2 rigid bodies[math]\displaystyle{ \times \frac{3 \mathrm{eqs.}}{\mathrm{r. body}}=6 }[/math]eqs.
unknowns: 2 associated with the DoF + 4 constraint unk. = 6 unk.
Roadmap for the equation of motion
Both the arm and the disk have a motion that depends on [math]\displaystyle{ \dot{\theta} }[/math]. The systems in which [math]\displaystyle{ \ddot{\theta} }[/math] will appear when applying the vector theorems are: disk, arm, disk + arm. These three options are indicated with dotted lines in the following GDI:
The intersection points between the dotted lines and the continuous lines indicate the external interactions on the system (which are the only ones to be taken into account in the vector theorems). The red ones correspond to unknowns. Therefore, the number of unknowns that will appear in the vector theorems for these three options are:
- system: disk, 3 constraint unk.[math]\displaystyle{ +\Gamma+\ddot{\theta}=5 }[/math] inc.
- system: arm, 3 constraint unk.[math]\displaystyle{ +\Gamma+\ddot{\theta}=5 }[/math] inc.
- system: disk + arm, 2 constraint unk.[math]\displaystyle{ +\ddot{\theta}=3 }[/math] inc.
The three equations generated by applying the vector theorems to the option (disk + arm) allow the calculation of the three unknowns, whereas, in the other two options, the number of unknowns exceeds the number of equations that can be generated. For the system (disk + arm), the external interactions are:
| Since the pin-slot constraint allows the translational motion of the arm in direction 2 and the rotation in direction 3 independently, the characterization of the torsor is straightforward, and yields on force component in direction 1 (which is the direction of translation prevented by the link).
|
Since O (fixed to the ground) does not belong to the disk, the angular momentum has to be calculated through the barycentric decomposition:
[math]\displaystyle{ \overline{\mathrm{H}}_\mathrm{RTO}(\Os)=\overline{\mathrm{H}}_\mathrm{RTG}(\Gs)+\OGvec \times \ms \vel{G}{RTO=T}= \left(\odot \frac{1}{2}\ms\rs^2\dot{\theta}\right) +(\searrow \Rs )\times \ms(\nearrow \Rs \dot{\theta})=\left[\odot \frac{1}{2}\ms(\rs^2 +2\Rs^2)\dot{\theta}\right] }[/math]
[math]\displaystyle{ \left.\begin{array}{l}
\text{LMT}]_2: (\nwarrow \Ns)+(\searrow \ms\gs\cos\theta)=\ms\acc{G}{T}]_{\nwarrow\searrow}=(\nwarrow \ms\Rs\dot{\theta}^2)\\
\text{AMT at }\Os]_3: [\odot\mu\Ns(\Rs+\rs)]+(\otimes \ms\gs\Rs\sin\theta)=\left[\odot \frac{1}{2}\ms(\rs^2+2\Rs^2)\ddot{\theta}\right]
\end{array}\right\} \Rightarrow
\left\{\begin{array}{l}
\Ns=\ms(\Rs\dot{\theta}^2+\gs\cos\theta)\\
(\rs^2+2\Rs^2)\ddot{\theta}-2\mu(\Rs+\rs)\Rs\dot{\theta}^2 + 2\gs\left[\Rs\sin\theta-\mu(\Rs+\rs)\cos\theta\right]=0
\end{array}\right. }[/math]
Roadmap for the motor torque
There are two system options for calculating the motor torque: arm, disk.
The normal force associated with the sliding single-point contact between the disk and the ground is no longer an unknown, and that is why the intersection with this line is indicated with a black dot.
If we choose the arm, the number of unknowns is 4 (3 constraint unknowns plus the motor torque). If we choose the disk opton, they can be reduced to 2: 2 constraint unk. + [math]\displaystyle{ \Gamma }[/math] = 3 unk. Therefore, that second option is more suitable. The description of external interactions on the disk is:
| If the AMT is applied to the center, the two components of the constraint force do not appear, and the motor torque is obtained directly:
[math]\displaystyle{ \boxed{\text{Roadmap: SYSTEM disk, AMT at }\Gs]_3} }[/math] [math]\displaystyle{ \text{AMT at }\Gs]_3: [\odot\mu\Ns(\Rs+\rs)]+(\otimes \Gamma)=\left(\odot \frac{1}{2}\ms\rs^2\ddot{\theta}\right) }[/math] [math]\displaystyle{ \hspace{2.5cm} \Gamma=\mu\Ns(\Rs+\rs)-\frac{1}{2}\ms\rs^2\ddot{\theta} }[/math] Replacing N and [math]\displaystyle{ \ddot{\theta} }[/math] by the results obtained before: |
[math]\displaystyle{ \Gamma=\mu\ms(\Rs+\rs)\frac{2\Rs^2}{\rs^2+2\Rs^2}(\Rs\dot{\theta}^2+\gs\cos\theta)+\ms\gs\frac{\Rs\rs^2}{\rs^2+2\Rs^2}\sin\theta }[/math]
Roadmap for the force associated with the pin-slot constraint
The systems in which this force will appear when the vector theorems are applied are: arm, disk + arm:
The number of unknowns in the option (disk + arm) is only that of the force to be calculated. This force is in direction 2 (according to the set of external interactions on this system shown in the previous figure).
Therefore: [math]\displaystyle{ \boxed{\text{Roadmap: SySTEMA(disk+arm), LMT}]_1} }[/math].
[math]\displaystyle{ \text{LMT}]_1: (\nearrow \Fs)+(\swarrow \ms\gs\sin\theta)=\ms\acc{G}{T}]_{\swarrow \nearrow}=(\nearrow \ms\Rs\ddot{\theta}) \quad \Rightarrow \quad \Fs=\ms(\Rs\ddot{\theta}+\gs\sin\theta) }[/math]
Replacing [math]\displaystyle{ \ddot{\theta} }[/math] by the equation of motion:
[math]\displaystyle{ \Fs=2\mu\ms\Rs\frac{\Rs(\Rs+\rs)}{\rs^2+2\Rs^2}\left(\Rs\dot{\theta}^2+\gs\cos\theta\right)+\ms\gs\frac{\rs^2}{\rs^2+2\Rs^2}\sin\theta }[/math].
✏️ EXAMPLE D6.7: boat dragged by a ship
A ship (represented simply as a support) moving at constant speed relative to the water (considered at rest relative to the ground) drags a boat of mass m and moment of inertia [math]\displaystyle{ \Is_\mathrm{G} }[/math]. The boat is articulated to a massless arm, which can slide along the smooth guide q-q'. Between the boat and the water there is viscous friction with constant c. Between the support and the arm there are two linear springs, and between the arm and the boat there is a torsional one. All springs have linear behavior. We want to find the equations of motion for the coordinates [math]\displaystyle{ \theta }[/math] and x.
The kinematics of the system relative to the water (which is the same as with respect to the ground) is:
[math]\displaystyle{ \left.\begin{array}{l}
\text{AB: water }\\
\text{REL: support}
\end{array}\right\} \vel{O}{water}=\vel{O}{support}+\vel{O}{ar}=(\uparrow \dot{\xs})+(\leftarrow\vs_0) }[/math]
[math]\displaystyle{ \text{Rigid body kinematics (rigid body: arm, }\velang{arm}{water}=\overline{0}): \vel{G}{water}=\vel{O}{water}=(\uparrow \dot{\xs})+(\leftarrow\vs_0) }[/math]
[math]\displaystyle{ \text{Rigid body kinematics (rigid body: boat, }\velang{bot}{water}=\otimes\dot{\theta}): \vel{C}{water}=\vel{G}{water}+ \GCvec\times\velang{boat}{water}=(\uparrow \dot{\xs})+(\leftarrow\vs_0)+(\nearrow\es\dot{\theta}) }[/math]
General diagram of interactions
The fact that the support has a predetermined translational motion [math]\displaystyle{ \vs_0 }[/math] indicates that it has lost 2 of the 3 DoF of the planar motion. This is equivalent to saying that there is a constraint on it that introduces 2 unknowns, and an actuator that guarantees the constant translation. That constraint and that actuator act may be considered to act between the boat and the water (or the ground, since they are at rest relative to each other).
| It is a system with three DoF ([math]\displaystyle{ \vs_0 }[/math] forced, [math]\displaystyle{ \dot{\xs} }[/math] and [math]\displaystyle{ \dot{\theta} }[/math] free) with 4 constraint unknowns. On the other hand, it contains 3 rigid bodies, and the two vector theorems generate 3 equations per rigid body. It is a determinate problem:
[math]\displaystyle{ \text{equations: 3 rigid bodies}\times\frac{\text{3 eqs.}}{\text{r. bodies}}=9 \text{ eqs.} }[/math] [math]\displaystyle{ \text{unkwons: 3 associated to the DoF + 6 constraint unk. = 9 unk.} }[/math] |
Roadmap for the equation fo motion of coordinate [math]\displaystyle{ \theta }[/math]
The only element whose motion depends on [math]\displaystyle{ \dot{\theta} }[/math] is the boat. Therefore, the systems in which [math]\displaystyle{ \ddot{\theta} }[/math] will appear when applying the vector theorems must necessarily include it: boat, boat + arm, boat + arm + support.
[math]\displaystyle{ \text{2 constraint unk.}+\ddot{\theta}+\ddot{\xs}=\text{4 unk.} \hspace{4.5cm}\text{2 contraint unk.}+\ddot{\theta}+\ddot{\xs}=\text{4 unk.} \hspace{3.5cm} \text{2 constraint unk.}+\Fs_\mathrm{act}+ \ddot{\theta}+\ddot{\xs}= \text{5 unk.} }[/math]
The first two options are the most appropriate. The descriptions of the external interactions in those cases are:
SYSTEM: boat
SYSTEM: boat + arm
For the boat system, the two constraint components can be avoided if the AMT is applied at G. On the other hand, applying the AMT at O to the system (boat + arm) only allows the elimination of the force F but not the moment M. Therefore:
[math]\displaystyle{ \boxed{\text{Roadmap: SYSTEMA boat, AMT at }\Gs]_3} }[/math]
[math]\displaystyle{ \text{AMT at }\Gs]_3: (\odot\ks'\dot{\theta})+(\odot\cs\dot{\xs}\es\cos\theta)+(\odot\cs\es^2\dot{\theta})+(\otimes\cs\vs_0\es\sin\theta)=(\otimes\Is_\mathrm{G}\ddot{\theta}) }[/math]
[math]\displaystyle{ \hspace{2.5cm} \Is_\mathrm{G}\ddot{\theta}+ (\ks'+ \cs\es^2)\dot{\theta}+\cs\es(\dot{\xs}\cos\theta-\vs_0\sin\theta)=0 }[/math]
Roadmap for the equation of motion of coordinate x
The two elements whose motion depends on are the boat and the arm. Therefore, the systems in which will appear when applying the vector theorems are: boat, boat + arm, arm + support, boat + arm + support.
The last two options include one more unknown (the actuator force), so the systems to be analyzed are reduced to the first two, for which the external interactions have already been represented above.
The [math]\displaystyle{ \ddot{\xs} }[/math] acceleration is orthogonal to the arm and will only appear in the LMT. For the boat system, the two constraint components associated with the joint between the boat and the arm have a nonzero projection in this direction. For the (boat+arm) system, however, no constraint unknown appears in the direction. Therefore:
[math]\displaystyle{ \boxed{\text{Roadmap: SYSTEMA (boat+arm), LMT}]_\updownarrow} }[/math]
[math]\displaystyle{ \text{LMT}]_\updownarrow:\left[\uparrow(\Fs_0-\ks\xs)\right]+ \left[\downarrow(\Fs_0+\ks\xs)\right]+(\downarrow\cs\dot{\xs})+(\downarrow \cs\es\dot{\theta}\cos\theta)=\ms\acc{G}{T}]_\updownarrow=(\uparrow \ms \ddot{\xs}) }[/math]
[math]\displaystyle{ \hspace{2.5cm} \ms\ddot{\xs}+\cs\dot{\theta}+2\ks\xs+\cs\es\dot{\theta} \cos\theta=0 }[/math]
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