Minimum distance between balls connected by rods

So, for the closest distance, the initial conditions are ##\alpha(t=0) = \pi##, ##\beta(t=0) = 0##, ##\dot{\alpha}(t=0) = \frac{v}{l}##, ##\dot{\beta}(t=0) = 0##.
  • #1
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!
 
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  • #2
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)}##.
 
  • #3
Though I have not completed it yet, take a look at my calculation attached to check your try.
 

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  • #4
anuttarasammyak said:
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 :smile:.
 
  • #5
good task for the Lagrange–d'Alembert principle
 
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  • #6
@etotheipi thanks for investigation. Energy conservation
[tex]\dot{\alpha}^2+\dot{\beta}^2-\dot{\alpha}\dot{\beta}cos(\alpha-\beta)=\frac{v^2}{l^2}[/tex]
where ##\alpha## is my way in post #3. Is it helpful ?
 
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  • #7
anuttarasammyak said:
@etotheipi thanks for investigation. Energy conservation
[tex]\dot{\alpha}^2+\dot{\beta}^2-\dot{\alpha}\dot{\beta}cos(\alpha-\beta)=\frac{v^2}{l^2}[/tex]
where ##\alpha## is my way in post #3. Is it helpful ?

Thanks for your help, it could well be useful :smile:. I am going to sleep now but I'll have a closer look tomorrow morning!
 
  • #8
Formula of angular momentum conservation :
[tex](2-cos(\alpha-\beta))(\dot{\alpha}+\dot{\beta})=3\frac{v}{l}[/tex]
initial conditions at t=0 :
[tex]\alpha=\pi,\ \beta=0,\ \dot{\alpha}=\frac{v}{l}, \ \dot{\beta}=0[/tex]

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

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

I will appreciate your check and advice.
 
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  • #9
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 :smile:, thanks!
 
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  • #10
Actually no you are right, ##\dot{\alpha}(t=0) = \frac{v}{l}##, like you said. My bad!
 
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Related to Minimum distance between balls connected by rods

1. What is the minimum distance between two balls connected by a rod?

The minimum distance between two balls connected by a rod is equal to the sum of the radii of the two balls.

2. How is the minimum distance between balls connected by rods calculated?

The minimum distance between balls connected by rods can be calculated using the Pythagorean theorem, where the hypotenuse is equal to the sum of the radii of the two balls and the other two sides are the lengths of the rods.

3. Why is it important to consider the minimum distance between balls connected by rods?

It is important to consider the minimum distance between balls connected by rods in order to prevent the balls from colliding or overlapping, which could affect the accuracy and reliability of the system.

4. Can the minimum distance between balls connected by rods be adjusted?

Yes, the minimum distance between balls connected by rods can be adjusted by changing the length of the rods or the size of the balls.

5. Are there any other factors that can affect the minimum distance between balls connected by rods?

Yes, other factors such as the material and shape of the balls and rods, as well as external forces, can also affect the minimum distance between balls connected by rods.

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