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Blazing goIITian

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5 Apr 2008 23:26:59 IST

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Challenge- A multimechanical question

None

I am posting a new type of ques here and its a multimechanical ques as it contains all the concepts of mechs viz, Kinematics,NLM,momentum,Collision,rotation.

In the figure, the 2 balls are are rolling without slipping. They undergo a perfectly inelastic collision at t=0. Find the time t taken by the total mass to stop after collision.

Given-Coefficient of friction=

Find the answer in terms of the given variable viz, M1,M2,R1,R2,V1,V2,

Yeh ques kar liya, samjho mechanics finish.

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Comments (18)

Glenn Jasper

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5 Apr 2008 23:30:53 IST

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nice ques dude.............

Kavya Sudha

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6 Apr 2008 10:29:34 IST

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first conserve momentum...find the velocity and then conserve angular momentum ....before they start pure rolling and friction will help it ...for starting pure rolling ...and in that time factor can b ecalculated ...by integrating it :)

its kinda lengthy question involiving all concepts....i shall giv eit a try...just post the answer just check it out ...if mine doesnt match thn i shall approach u again ..

The Werewolf

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8 Apr 2008 22:23:01 IST

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Tarin, can u post your solution because any solution you post I can definately find some fault in it , since your question carries a lot of bugs.

The Werewolf

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8 Apr 2008 22:56:00 IST

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Bug 1 : Is there friction between the two spheres ?

Bug 2 : What is the relation between M1 M2 (M1>M2 or M1<M2)and R1 R2(R1>R2 or R1<R2) ....its needed so that we can find which mass will stop first as these factors decide their angular accelerations.

Bug 3 : Atleast post the answer if not the solution.

Bug 4 : its high time now , and u should consider posting solution also.

The Werewolf

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8 Apr 2008 23:21:13 IST

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we can solve this problem easily if its given that their angular accelerations are also equal....if their angular acc are different they will need different times to stop and in that case both spheres will stop at different times and not simultaneously....i bet u must be thinking that they will stop at same time otherwise problem is easy .....they will stop at same time when their ang acc are also equal like their linear acc

anchit saini

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8 Apr 2008 23:23:05 IST

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i think tarin has made up this question on his own !!!!

Tarin Bansal

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8 Apr 2008 23:24:26 IST

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Bhai,
They are undergoing perfectly inelastic collision.
And as they are in same line, the whole mass wud not be able to roll now and it wud slide on the ground like a simple body given a velocity on a surface with friction. And will ultimately stop after some time. U just need to find that time to stop.

The Werewolf

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8 Apr 2008 23:28:53 IST

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EXPLAIN :And as they are in same line, the whole mass wud not be able to roll now

everyone knows that the spheres will still have their own independent rotational motion.

Tarin Bansal

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8 Apr 2008 23:34:30 IST

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Bhai,

After a perfectly inelastic collision, the 2 bodies stick together.

Just look at the figure-

after collision, the total mass will move together. Now I dont think it will roll.

If U still think it will roll, then take the condition that it is not rolling, it is sliding now.

But I am quite sure that such a mass cant roll now.

The Werewolf

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8 Apr 2008 23:39:51 IST

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What do u mean they stick together , they don't stick together , in technical terms they after the collision tend to have same velocity direction and magnitude....its that in general talk we say that they stick togather ...they dont get stick togather with a glue yaar so that they cant even rotate

Tarin Bansal

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8 Apr 2008 23:49:47 IST

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Buddy, maine ab tak jitne ques solve kiya hain and jahan bhi padha hai, I hav always seen and used that after PERFECTLY INELASTIC collision, the 2 bodies move as a single body with same velocity.

And I introduced this in this ques with this approach only.

The Werewolf

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8 Apr 2008 23:55:27 IST

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u r talking of linear collisions only where the acc of the two bodies after the collision are equal so there is no problem .....but here it is rotational collision where angular acc also cums into play and if that different for the two bodies they will need different times to stop separately

The Werewolf

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15 Jul 2008 14:13:06 IST

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tarin bro , pls provide answer to this question.

Tarin Bansal

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15 Jul 2008 14:16:48 IST

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Bhai,
Is thread ko kyon alive kar diya?? LOL

Its a self-made question and waise bhi I am afraid of mechanics, so I never tried to do this. I thought it wud be a nice workout for goiit's mechanical enigneers. ;) So posted it here. :D

VARUN RAJ

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15 Jul 2008 14:34:32 IST

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I DONT THINK ITS THAT A TUGH AS SUM NEVERTHLESS LETS CALCULATE THE ANGULAR MOMENTUM ABT THE PT OF CONTACT OF BOTH THE VEHICLES (MEANS THE BOTTOM MOST PT)
INITIAL MOMENTUM
ASSUMING THE RADIUS TO BE EQUAL
MV1R+1W1 AND THEN U CAN CALCULATE THE VELCOTY AND ANGULAR VELCOITYS RATIO
MEANS U WILL GET THAT MVR+1W1/V1 ....=W
SO THEN U CAN FIND OUT THE RELATIVE VELOCITY OF THE PT OF CONTACT AND THEN ITS SIMPLE

VARUN RAJ

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15 Jul 2008 16:21:20 IST

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SEE I GOT THE ANSWER ITS VERY COMPLEX AND THIS IS NOT AN IIT SUM THIS IS AN PHYSICS OYMAPAID SUM

SEE FIRST CALCULATE THE INERTIA OF THE COMBINED MASS (THIS IS THE TOUGHEST PART OF THE WHOLE SUM)

SEE FIRST WE NEED TO FIND THE POSITION OF THE COM IF THE SYSTEM AND THEN WE CAN FIND OUT THE THE INERTIA OF EACH SPHERE ABT THIS PT AND ADD IT GIVING THE INERTIA OF THE COMBINATION

THE Y AXIS OF THE COM WILL BE TAKING THE ORIGION AT THE GROUND COM=M1R1+M2R2/M1+M2

SIGNIFYING THIS AS Y1

TAKING THE RADIUS OF THE FIRST SPHERE AS THE ORIGION THE COM IS M2(R1+R2) /M1+M2

WE KNOW THE INERTIA ABT THE COM OF THE FIRST SPHERE (THAT IS THROUGH ITS RADIUS) SO D= M2(R1+R2) /M1+M2 SO D^2 CAN BE EASILY CALCULATED AND SO IST INERTIA IS EASILY CACULATED

INERTIA OF THE 2ND SPHERE ABT THE COM OF THE SYSTEM

D=R1+R2-M2(R1+R2) /M1+M2 SO D^2 CAN BE CALCULATED AND SO ANOTHE INERTIA IS ALSO CALCULATED

NOW LETS TAKE LINEAR MOMENTUM THIS CAN BE CONSERVED SINCE BEFORE COLLISION FRICION IS NOT ACTING AND IMMEDIATELY AFTER COLLISION IT IS ALSO NOT ACTING

SO M1V1+M2V2=(M1+M2)V3 (HERE V3 IS THE NEW VELOCITY)

V3=M1V1+M2V2/(M1+M2)

NOW CONSERVING ANGULAR MOMENTUM ABT THE BOTTOM MOST PT ANYWHERE U CAN TAKE SINCE INITIALLY NO FRICTION IS ACTING

V1=W1R1 W1=V1/R1 SIMILARLY W2=V2/R2

INERTIA OF MASS M1=2/5M1R1^2 AND INERTIA IF 2ND MASS =2/5M2R2^2

SO CONSERVING ANGUAR MOMENTUM

I1W1+M1VIR1+I2W2+M2V2R2=INERTIA CALCULATED ERALIER W3 +(M1+M2)V3AND R IS THE POSITION OF THE COM ON THE ORIGION

SO WE ET W

NOW SINCE THE DIRECTION OF ROLLING IS NOT CHANGING THE VEOCITY OF THE BOTTOMMOST PT OF CONTACT =V-W3Y1

AND SO WE FIND OUT THE VEOCITY OF PT OF CONTACT

BUT NOW WE KNOW THE MASS OF THE SYSTEM N=2MG

SO WE KNOW THE VALUE OF uN

TIME TAKEN TO STOP= U/A

AND SO THIS IS THE ANSWER

PLS RATE ME FOR MY EFFORTS

!!!!!!!CHEERS!!!!!!!!!!

Tarin Bansal

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15 Jul 2008 16:24:40 IST

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I dont know the answer. Some expert plz confirm the answer.

@varun, R U sure this is an olympiad question?? :O :O
I havent picked this question from anywhere. Its a self made question. :-|

VARUN RAJ

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15 Jul 2008 16:25:41 IST

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N I AM JUST TELLING THAT IT IS SO TOUGH THAT IT CANT BE AN IIT SUM

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