Minimum distance between balls connected by rods

[etotheipi]

Homework Statement

Three balls (A,B,C) are connected in series by two light rods of length $l$, with B in the middle. All hinges are smooth. The three balls initially lie along a straight line, and an impulse delivers a speed $v$ to the ball A, perpendicular to the rods. Determine the minimum distance between the balls A and C.

Relevant Equations

N/A

I defined the angle $\beta$ as the angle from the right horizontal to the ball C, from B, and $\alpha$ as the angle from the left horizontal to the ball A, from B. I also work in the CM frame, which has a velocity downwards of magnitude $\frac{v}{3}$ w.r.t. the lab frame. The positions of the three balls in the CM frame are as follows $$\vec{r}_B = \begin{pmatrix}x\\y\end{pmatrix}$$ $$\vec{r}_A = \vec{r}_B - l\begin{pmatrix}\cos{\alpha}\\\sin{\alpha}\end{pmatrix}$$ $$\vec{r}_C = \vec{r}_B + l\begin{pmatrix}\cos{\beta}\\\sin{\beta}\end{pmatrix}$$From that I conserved momentum and angular momentum in the CM frame, respectively below:$$3m\vec{v}_B + ml\dot{\beta} \begin{pmatrix}-\sin{\beta}\\\cos{\beta}\end{pmatrix} + ml\dot{\alpha} \begin{pmatrix}\sin{\alpha}\\\ -\cos{\alpha}\end{pmatrix} = \vec{0}$$ $$3\vec{L}_B + ml \left[\begin{pmatrix}\cos{\beta} - \cos{\alpha}\\\sin{\beta} - \sin{\alpha} \\0\end{pmatrix} \right] \times \vec{v}_B + ml\vec{r}_B \times \left[ \dot{\beta} \begin{pmatrix}-\sin{\beta}\\\cos{\beta}\\0\end{pmatrix} + \dot{\alpha} \begin{pmatrix}-\sin{\alpha}\\\cos{\alpha}\\0\end{pmatrix} \right] + ml^2(\dot{\alpha} + \dot{\beta})\hat{z} = mlv \hat{z}$$ If we left multiply the COM equation by $\vec{r}_B$ then we can substitute the $\vec{0}$ into the COAM equation to obtain$$\begin{pmatrix}\cos{\beta} - \cos{\alpha}\\ \sin{\beta} - \sin{\alpha}\\0\end{pmatrix} \times \vec{v}_B + l(\dot{\alpha} + \dot{\beta})\hat{z} = v\hat{z}$$But I don't know what to do next. The distance between them is $s = l\sqrt{2+2\cos{(\alpha - \beta})}$, so it is required to determine the most negative value of $\cos{(\alpha - \beta)}$. Maybe it will also be necessary to use the centre of mass condition somewhere, which provides an extra constraint:$$3m\vec{r}_B + ml \begin{pmatrix}\cos{\beta} - \cos{\alpha}\\\sin{\beta}-\sin{\alpha}\\0\end{pmatrix} = \vec{0}$$I wondered if anyone could help out, thanks!

 

[etotheipi]

If I solve the very last equation for $\vec{r}_B$, differentiate for $\vec{v}_B$ and insert that into the second last equation, I can determine a further equation in terms of only the variables $\alpha$ and $\beta$.$$\frac{2l}{3}(\dot{\alpha} + \dot{\beta}) - \frac{l}{3}(\dot{\beta} \cos{(\alpha + \beta)} - \dot{\alpha} \cos{(\alpha - \beta)}) = v$$Assuming I have done the algebra right, now it is required to find the minimum value of $\cos{(\alpha - \beta)}$.

 

[anuttarasammyak]

Though I have not completed it yet, take a look at my calculation attached to check your try.

 

[etotheipi]

Though I have not completed it yet, take a look at my calculation attached to check your try.

Yes I think I get the same. We have defined $\alpha$ slightly differently (for instance mine takes $\alpha = 0$ at $t=0$ whilst you do $\alpha = \pi$ at $t=0$) however this is an arbitrary choice and as far as I could tell our calculations were the same .

 

[wrobel]

good task for the Lagrange–d'Alembert principle

 

[anuttarasammyak]

@etotheipi thanks for investigation. Energy conservation
\dot{\alpha}^2+\dot{\beta}^2-\dot{\alpha}\dot{\beta}cos(\alpha-\beta)=\frac{v^2}{l^2}
where $\alpha$ is my way in post #3. Is it helpful ?

 

[etotheipi]

@etotheipi thanks for investigation. Energy conservation
\dot{\alpha}^2+\dot{\beta}^2-\dot{\alpha}\dot{\beta}cos(\alpha-\beta)=\frac{v^2}{l^2}
where $\alpha$ is my way in post #3. Is it helpful ?

Thanks for your help, it could well be useful . I am going to sleep now but I'll have a closer look tomorrow morning!

 

[anuttarasammyak]

Formula of angular momentum conservation :
(2-cos(\alpha-\beta))(\dot{\alpha}+\dot{\beta})=3\frac{v}{l}
initial conditions at t=0 :
\alpha=\pi,\ \beta=0,\ \dot{\alpha}=\frac{v}{l}, \ \dot{\beta}=0

We expect
\dot{\alpha}=\dot{\beta}
when A and C take minimum distance. Otherwise the distance keeps increasing or decreasing. From energy and angular momentum conservation equations we get
\dot{\alpha}=\dot{\beta}=\frac{2v}{3l} and
cos(\alpha-\beta)=-\frac{1}{4}
which does not depend on v.

As for max, $\alpha-\beta=\pi$,
\dot{\alpha},\dot{\beta}=\frac{v}{l},0
For maximum two angular velocities do not have to coincident because $(cos\theta)'=sin\theta=0$ there.

I will appreciate your check and advice.

 

[etotheipi]

I think that looks good, the only thing I found was that I think the initial conditions in the CM frame will be $\dot{\alpha}(t=0) = \frac{2v}{3l}$ instead of $\frac{v}{l}$, although this doesn't affect the answer.

If the angle between the rods is $\phi = \alpha - \beta$, then for our closest distance case we will have $\dot{\phi} = 0 \implies \dot{\alpha} - \dot{\beta} = 0 \implies \dot{\alpha} = \dot{\beta}$, as you have used.

So yes, I think you are correct , thanks!

 

[etotheipi]

Actually no you are right, $\dot{\alpha}(t=0) = \frac{v}{l}$, like you said. My bad!