I found this exercise on a textbook which provided solely the description of what is happening and the answers with little explanation. The solution being provided, I am mostly interested in understanding what is happening. Consider the following system composed of three particles, each of mass $m$, connected by rigid massless rods in the form of an equilateral triangle. The system is initially at rest (particle $A$ is above particle $B$ in the initial state). An impulse $\hat{F}$ is applied to $A$, causing the particle $B$ to slide without friction on a supporting base.

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The idea is to solve for the values of $\dot x$ and $\dot \theta$ velocities immediately after the impulse; and evaluate the constraint impulse $\hat{N}$.

My initial idea was to use the linear impulse, focusing in the $x$ direction first. Since particle B will be sliding, the linear acceleration of the system would be $3m\ddot x$. By taking particle B as the reference point, and since the system center of mass would be engaged in an arc motion, its acceleration $x$ component (tangential) directly after the impulse would be $ml\sqrt{3}\ddot\theta$ (the radius would be $l/\sqrt{3}$).

So the impulse would be $\hat{F}=3m\dot x+ml\sqrt{3}\dot \theta$.

But the textbook solution indicates that $\hat{F}=3m\dot x+\frac{3}{2}ml\dot \theta$.

So my first problem is: where does the $\frac{3}{2}$ comes from? Or am I working this completely wrong?

Upon seeing this solution I wondered if perhaps this $\frac{3}{2}ml\ddot \theta$ tangential component of angular acceleration would be the result of the addition of an $l/2$ radius (along the vertical axis) which would be the vertical distance towards particle C from B; plus the $l$ radius which would be the vertical distance towards particle $A$; multiplied by $\ddot \theta$. Which would result in a "tangential" component along the $x$ axis of $\frac{3}{2}ml\ddot \theta$. Would this be an acceptable proposition?

My idea was then to state the angular impulse as the variation of the angular momentum, resulting in the multiplication of $\hat{F}$ by the distance between particle $B$ and the center of mass: $l/\sqrt{3}$. This way I would end up with a system of two equations and two variables and solve for $\dot x$ and $\dot \theta$.

Only, again the solution provided is rather far from my idea because it states that $\hat{F}l = \frac{3}{2}ml\dot x + 2ml^2\dot \theta$. And I see little relationship with the linear impulse equation.


This question did not get up votes nor answers, maybe because as it turns out, this was not that complicated. In any case, I still propose a detailed answer just in case.

Solving for $\dot x$ and $\dot \theta$

The impulse applied on top would be defined as the variation in the linear momentum of the particles system. For more convenience, we will be using particle $B$ as a reference point.

\begin{equation} \hat{\mathbf{J}} = \sum\limits_{i=1}^{3}\Delta \mathbf{p_i} \end{equation}

where $\hat{\mathbf{J}}$ is the total impulse vector affecting the system, comprised of the horizontal $\hat{F}$ and the vertical constraint impulse $\hat{N}$; and $\sum\limits_{i=1}^{3}\Delta \mathbf{p_i}$ is the sum of the linear momentum variation of the system particles. Since the system is initially at rest, its particles initial velocities are null, so the linear momentum variation is simply the sum of their individual final momenta.

Because particles $A$ and $C$ are subject to a linear velocity $\dot x$ resulting from the sliding of particle $B$; and to an angular velocity $\dot\theta$ resulting from the rotation about particle $B$; the above equation can then be developed as:

\begin{equation} \left(\hat{F}\,\vec{i}+\hat{N}\,\vec{j}\right) = m\left( \left( -l\frac{\sqrt{3}}{2}\vec{i} + \frac{l}{2}\vec{j} \right) \times \dot\theta\vec{k} + \dot x\vec{i} \right) + m\left( l\vec{j} \times \dot\theta\vec{k} + \dot x\vec{i} \right) + m\dot x\vec{i} \end{equation}

where $\vec{i}$, $\vec{j}$ and $\vec{k}$ are unit vectors of the frame and the $\times$ operator denotes the cross product between vectors. On the right hand side of the above equation, the first term corresponds to the linear momentum of particle $C$, the middle one corresponds to the linear momentum of particle $A$ and the last term is the linear momentum of particle $B$ which is only subject to a sliding motion.

By developing the above equation, the $\hat{F}$ and $\hat{N}$ components of the applied linear impulse vector $\hat{\mathbf{J}}$ are identified as the horizontal linear impulse applied on particle $A$:

\begin{equation} \hat{F} = 3m\dot x + \frac{3}{2}m l \dot\theta \end{equation}

and the constraint linear impulse applied by the supporting base on particle $B$:

\begin{equation} \hat{N} = \frac{\sqrt{3}}{2}m l \dot\theta \end{equation}

Since the constraint linear impulse $\hat{N}$ is considered a consequence of the linear impulse $\hat{F}$ and the value of $\hat{N}$ is unknown, we still need one equation to solve for the $\dot x$ and $\dot \theta$ values. Keeping particle $B$ as the reference point for the angular momentum of each particle, we can find $\dot x$ and $\dot \theta$ without having to solve for $\hat{N}$. Thus, we can state a relationship between the angular impulse at $B$ resulting from the linear impulse on particle $A$ and the variation of the angular momentum of each particle:

\begin{equation} \hat{\mathbf{M}} = \sum\limits_{i=1}^{3}\Delta \mathbf{H_i} = \sum\limits_{i=1}^{3}\mathbf{r}_i \times \Delta \mathbf{p_i} \end{equation}

where $\hat{\mathbf{M}}$ stands for the angular impulse around particle $B$ of the linear impulse $\hat{F}$ applied on $A$; $\sum\limits_{i=1}^{3}\Delta \mathbf{H_i}$ represents the sum of the individual angular momenta around $B$ of particles $A$ and $C$; and $\mathbf{r}_i$ is the vector going from $B$ to the other particles.

On the left hand side of the above equation we have

\begin{equation} \hat{\mathbf{M}} = \hat{F}\,\vec{i} \times l\vec{j} = \hat{F}l\vec{k} \end{equation}

and on the right hand side, since the initial velocities we have:

\begin{equation} \sum\limits_{i=1}^{3}\mathbf{r}_i \times \Delta \mathbf{p_i} = \left( -l\frac{\sqrt{3}}{2}\vec{i} + \frac{l}{2}\vec{j} \right) \times \left( m\left( \frac{l}{2}\dot\theta + \dot x \right)\vec{i} + ml\theta\frac{\sqrt{3}}{2}\vec{j} \right) + l\vec{j} \times m(l\dot\theta + \dot x)\vec{i} \end{equation}

where the first cross product stands for the angular momentum of particle $C$ around $B$; and the second cross product stands for the angular momentum of particle $A$ around $B$. By developing this last equation with that of $\hat{M}$ just before it, we find that

\begin{equation} \hat{F}l = \frac{3}{2}ml\dot x + 2ml^2\dot\theta \end{equation}

Thus, by recalling the horizontal linear impulse equation while forming a system with the above equation,

\begin{equation} \hat{F} = 3m\dot x + \frac{3}{2}m l \dot\theta \end{equation}

we can solve for the values of $\dot x$ and $\dot\theta$:

\begin{equation} \dot\theta = \frac{2\hat{F}}{5ml} \end{equation}


\begin{equation} \dot x = \frac{2\hat{F}}{15ml} \end{equation}

Solving for the value of the constraint linear impulse $\hat{N}$

By recalling the vertical linear impulse equation previously obtained

\begin{equation} \hat{N} = \frac{\sqrt{3}}{2}m l \dot\theta \end{equation}

and by injecting the value of the angular velocity $\dot\theta$ just above:

\begin{equation} \hat{N} = \frac{\sqrt{3}}{5}\hat{F} \end{equation}


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