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![[Post New]](/templates/default/images/icon_minipost_new.gif) 8 May 2008 22:02:30 IST
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A cylinder of mass m is kept on the edge of a plank of mass 2m and length 12m, which in turn is kept on smooth ground.coefficient of friction between the plank and the cylinder is 0.1. The cylinder is given an impulse , which imparts it a velocity 7m/s but no angular velocity.Find the time after which the cylinder falls off the plank.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 May 2008 17:14:56 IST
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please any one , give the solution to my question.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 May 2008 18:07:17 IST
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IS THE PLANK FRICTIONLESS
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 May 2008 18:22:18 IST
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yes the plank is frictionless.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 May 2008 18:41:54 IST
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FOR HOW MUCH TIME IS THE IMPULSE IMPARTED
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 May 2008 21:19:33 IST
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the question is as given above. no more doubts please.
i want the solution.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 May 2008 21:35:36 IST
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The time taken would depend on the radius of the cylinder so i think it should be given.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 May 2008 21:41:12 IST
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Is the answer 4 ? Friction is an internal force.. So no movement of centre of mass of plank + cylinder system.. So when it rolls off, the the position of the plank, will be at (2,0), considering initial position of ball to be at origin.. This you get by conserving position of centre of mass.. Now only force acting on plank is friction due to cylinder, which is umg, where u is coeff of friction. Acceleration is ug/2
(1/2)(ug/2)t^2=4
Putting values you get t=4.. Only thing is, I didn't make use of the initial velocity.. Am I missing something here?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 9 May 2008 22:12:17 IST
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the answer is 2.25 sec. i want the solution.
why no body is ready?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 May 2008 12:34:16 IST
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hey goiitians whats this ? i am waiting for the solution from 2 days . why no one helping me? try the
problem once. its from arihanth actually.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 May 2008 12:40:50 IST
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i think its just a waste of time ......... yaar............
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 May 2008 12:51:25 IST
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ok.. as a rule... i never solve questions first time unless my help is not enough the first time.
so try this and see if you get it. otherwise i will solve it.
for any body to roll, it requires a torque. this torque is usually provided by friction. so the friction b/w the plank and cylinder will casue the cylinder to
1) decelerate
2) roll in a clocwise sense.
friction will act to the left (all this in reference to your figure).
this friction stops acting when the cylinder stops rolling. so what will happen is that the VELOCITY in the right direction will decrease, and at the same time the CLOCKWISE angular velocity will increase, until v =rw.
now the real problem is : will this v = rw before the cylinder falls off the plank or not?
the problem is very very long, and we will have to do this in 2 cases.
in the first case, we say that the cylinder continues to decelerate till the end. this is the easier case, so try it first. frictional force can be found, and use newtons second law and write torque equations. if the answer is absurd, like imaginary roots etc, then take second case.
assume that it decelerates for t seconds and from then v = rw..
well its long.. but this is the only method i know.
shit this question is a very long one. i guess you will be disappointed that i did not solve it.. and now that i see it i dont even want to.. lol.
shreyas
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 May 2008 12:57:34 IST
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SEE I WILL JUST GIVE U A BRIEF METHOD OF MY WORKING SUPPOSE AN FORCE IS IMPARTED THEN THE FRICTION WILL ACT IN THE FORWARD DIRECTION BUT HERE THE IMPULSE IS IMPARTED WHIC IS EQUAL TO THE CHANGE IN MOMENTUM OF THE SYSTEM BUT SINCE THE SPHERE DOES NOT ROLL THAT MEANS THE ANGULAR MOMENTUM DOES NOT CHANGE. THIS MEANS THAT THE IMPULSE OF THE FRICTION IS EQUAL TO THAT OF THE APPLIED FORCE. SO 2*FDT=CHANGE IN THE MOMENTUM OF THE SYSTEM AFTER SOME TIME THE FORCE STPS ACTING BUT THE FRICTION DOESNT SO THE BODY WILL ALSO ROTATE BUT THIS DOESNT NEED TO BE CONSIDERED. SO IN OTHER WORDS THIS SUM IS SIMPLE BUT IT IS MADE COMPLEX 7 MS IS THE INITIAL VELOCITY AND FRICTION WILL ACT IN THE OPP DIRECTION TO SLOW RELATIVE MOTION AND IT WILL ACT IN THE FORWARD DIRECRTION FOR THE PLANK SO U CAN CALCULATE THE NASWER SIMPLY PLZ RATE ME (I WONT SAY IF I AM USEFUL BCOZ THIS WAS OF COURSE USEFUL) !!!!!!CHEERS!!!!!!!!
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
url=http://sig.graphicsfactory.com/]
                           
[/url]
Glitter Graphics
[url=http://sig.graphicsfactory.com/]
                   
[/url]
Glitter Graphics
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