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Kinematics!

I have been pondering over this question for a while and trying to derive a general formula for this.The problem is:

"In an \(N\)-sided polygon ,objects are placed at each of the vertices.They move with a velocity \(v\) such that they always move towards each other. Find the time when they will meet."

I perhaps feel it is an easy question,but still I have been struggling with it for quite some time.

Help would really be appreciated.

Thanks!

Note by Shivam Mishra
1 year, 5 months ago

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If the side of the polygon is of length \(a\), then: \[T=\frac{a}{v \left (1-\cos \frac{2\pi}{n} \right) } \]


Proof:

At any instant, joining the adjacent particles, we again get a regular \(n\) sided polygon.

We now conclude that the relative velocity (along the line joining them) between any two adjacent particles is constant throughout the motion. Also, all the particles meet when any two adjacent particles meet.

Note that the relative velocity along the line joining two adjacent particles is \(v(1-\cos \frac{2\pi}{n})\).

Hence, time required for all the particles to meet is given by the first equation.

Deeparaj Bhat - 1 year, 5 months ago

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Please can you present the derivation? @Deeparaj Bhat

Shivam Mishra - 1 year, 5 months ago

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I've added it now. Tell me if you any problem understanding. :)

Deeparaj Bhat - 1 year, 5 months ago

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@Deeparaj Bhat Thanks!!!

Shivam Mishra - 1 year, 5 months ago

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@Shivam Mishra You're welcome. :)

Deeparaj Bhat - 1 year, 5 months ago

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Can we somehow figure out equation of trajectory of motion of objects in case polygon under consideration is a square ? I tried to solve this problem last year but was unable to do so.

Aditya Sky - 1 year, 5 months ago

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Ok. I was able to find the trajectory in polar coordinates. We shall consider the trajectory of a single particle.

Let \(r\) be the distance of a particle from the centre of centre of the square (which is invariant wrt time). Let \(\theta\) denote the angle the radius vector makes with the initial line. Then, \[ \frac{d}{dt} (re^{i\theta})=ve^{i(\theta+\frac{3\pi}{4})} \]

Now, solve. :)


Edit:

If the side of the square is \(a\), then \[\sqrt2 r=a-vt \\ \theta- \theta_0 = \ln \left( 1- \frac{v}{a}t \right)\]

Deeparaj Bhat - 1 year, 5 months ago

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@Deeparaj Bhat @shivam mishra

Actually, the trajectory can be found explicitly for a n sided polygon.

Deeparaj Bhat - 1 year, 5 months ago

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You have to define "always move towards each other". You most likely mean "move towards the object clockwise from them".

Calvin Lin Staff - 1 year, 5 months ago

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Actually by that phrase I meant that their velocity vectors are always directed towards the particle that is in front of them,it could be clockwise or anticlockwise.

Shivam Mishra - 1 year, 5 months ago

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Right, so that should be clarified. "always move towards each other" is an ambiguous phrase.

Calvin Lin Staff - 1 year, 5 months ago

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@Harsh Shrivastava@Rajdeep Dhingra@Prakhar Bindal Please help!!!

Shivam Mishra - 1 year, 5 months ago

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Deeparaj bhaiya has already explained. Sorry I'm late.

Rajdeep Dhingra - 1 year, 5 months ago

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Don't call me 'bhaiya' yaar.

Deeparaj Bhat - 1 year, 5 months ago

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No problem ,well congrats on your ijso selection.

Shivam Mishra - 1 year, 5 months ago

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@Shivam Mishra Thanks.

Rajdeep Dhingra - 1 year, 5 months ago

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