Let an crazy ball has to be projected on

Roughhorizontal surface such that after striking surface it comes back to initial position then itself repeat such cycle again and againwithout loss of energythen Find all Required conditions For This Event ?

NOTE:

\(\bullet\) Here in this situation i take ideal conditions .So try this it is really funny and interesting !

But This is not practically possible because it's loses energy because of air resistance and many other factor.

See practical example of this :

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## Comments

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TopNewest@Deepanshu Gupta @Ronak Agarwal @Mvs Saketh @Krishna Sharma @megh choksi @satvik pandey @jatin yadav @Sudeep Salgia @Karthik Kannan @Pratik Shastri what do you think about this question ??

I think that we should project it in that way such that it follows the angle \(\theta\) and then \(90-\(\theta\)) and crazy ball should repeat this condition So that Range is same , Am I correct ?

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I would like to see what other guys think about that. I think I don't have sufficient knowledge to comment on this question. :(

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I explored about it and I came up with this.

I think the ball retraces its path after each collision. For that, after each collision the magnitude of X and Y components of velocity should be equal.

Let the initial X and Y component of the velocity of the ball be Vx and Vy. So at the time of 1st collision X and Y component of velocity will be -Vy and Vx.

As

Relative velocity of separation =e(Relative velocity of approach)So \({ V }_{ y' }=e{ V }_{ y }\) but in order to retrace its path magnitude of \({ V }_{ y' }\) and \({ V }_{ y }\) should be equal. So \(e=1\)

Let \({ J }_{ N }\) be impulse due to Normal force and \({ J }_{ f }\) be impulse due to frictional force.

So \({ J }_{ f }=\mu { J }_{ N }\)

and \({ J }_{ N }=2m{ V }_{ y }\)

also \({ J }_{ f }=-2m{ V }_{ x }\) where (-)ve sign indicates the direction of \(J_{f}\)

So \(\mu =\frac { { V }_{ x } }{ { V }_{ y } } \)

Now Angular impulse about CoM= \(-RJ_{f}\) (taking anti clockwise direction +ve)

So \(-RJ_{f}=I({ \omega }_{ f }-{ \omega }_{ i })\) I think that \({ \omega }_{ f }=-{ \omega }_{ i }\)

So \(\omega =\frac { 5{ V }_{ x } }{ 2R } \).

What do think guys. Please reply. @Deepanshu Gupta @Mvs Saketh @Ronak Agarwal @Mardokay Mosazghi @KARAN SHEKHAWAT @Karan Siwach @Nathanael Case @Pratik Shastri .

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\[{ J }_{ F }\quad =\quad \mu { J }_{ N }\].

\(\bullet\) You Have Taken an Assumption Try To Find what was it ?

\(\bullet\) Also By doing all calculation comment on the Nature of Ball , i.e is it Possible That this Phenomena Can occur if we Throw an Rigid Ball ( Say Cock Ball) ?

\(\bullet\) Also Find what Should Be The Minimum and Maximum required \(\mu \) For this Phenomena ?

Try To Answer These question So that You Understand Many things from This question !

And Your Mathematical Calculation are Correct !

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1) I think the normal force that acts on the ball for an infinitely small time period over which the collision occurs should be constant .

2) The given conditions will take place only if the ball does not loose any kinetic energy after each collision. So I think it should be Rigid.

3) It's minimum value can't be zero. :D I don't know how to find minimum and maximum value of \(\mu\). Please give some hint. :)

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And Minimum Coffecient of friction is \({ \mu }_{ min }\quad =\quad \cot { \theta } \). where \(\theta \) is inclination angle( angle of projection ) , Think why or why not?

Maximum \(\mu \) can be infinitely Large , Think why or why not ?

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Flexible, Because when The Collision b/w ball and surface occurs then it takes some time ( may be fraction of second) But during this fraction of Time Friction does work and due to which if body is not rigid Then there should be some energy Loss (may be in too small amount )Takes Place So This Phenomena Can't Happens ! And If we consider ball as perfectly Flexible ( I don't want say that elastic , which is loose talk) then due to it's Flexibility it Can Stored Potential energy in it during The time interval in which work is done by friction , So that when Ball get's detached from surface it regains it's original Energy and also regain it's original Shape and Size (In ideal Conditions, which is already Mentioned above in Note ) ( even thesmallamount of energy which is lost due to work done by friction is regained by it)Hence I Still Says That Ball must be

Flexible! (and of course it must be elastic also)Log in to reply

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Rigidity' means ? From My Views Cock ball or an Lather Type Ball is Rigid , According to me , I mean that Rigidity means Very Hard mass which Can't be compressed and Expand by applying a force i.e we can say and Solid Ball , Solid Rigid Rod etc.I'am gotta Confused with this Term '

Rigidity'what is Precise Meaning of this Term ?

thanks ! My good Friend :)

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