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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 May 2008 19:52:12 IST
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the assembly of two discs as shown in figure is placed on a rough horizontal surface and the front disc is given an initial angular velocity ' w '. determine the final linear and angular velcity when both the discs start rolling, it is given that friction is sufficeient to sustain rolling in the rear wheel from the starting of motion.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 May 2008 20:07:43 IST
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i am waiting for the reply..........................any one please...........................................
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 May 2008 20:41:22 IST
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hey i cant understand the diagram.. why will the second sphere move?
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 May 2008 20:47:58 IST
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the front disc is given an initial angular velocity w.
this question is from arihant..............................................
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 12 May 2008 21:09:23 IST
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CANT UNDERSTAND THE QUES IN THE END WHEN THEY WILL HAVE THE SAME VELOCITY THEN U WANT TO FIND OUT OR OTHERWISE
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 12:01:48 IST
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after both the discs start rolling , what is their final linear and angular velocity.
ans: angular velocity: w / 6 ,
linear velocity : w R / 6 .
any experts ...........please give the solution ..... .... ............ i am waiting !!!!!!!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 12:08:11 IST
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SEE I AM ASSUMING THAT THEY R CONNECTED BY STRINGS THEN LETS CONSIDER THE FIRST WHEEL SINCE IT IS MOVING BACKWARDS IT HAS A TENDENCY TO SLLIP SO FRICTION WILL ACT IN THE FORWARD DIRECTION f+T=MA FR=IA/R FOR THE 2 ND BLOCK TENSION WILL ACT IN THE FORWARD DIRECTION AND TO PREVENT RELATIVE MOTION FRICTION WILL ACT IN THE BACKWARDS DIRECTION AND IT WILL ALSO PRODUCE A TORQUE T-F=MA FR=IA
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
http://my.opera.com/vu2rje/blog/ |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 12:18:36 IST
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pls verify my methods
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
http://my.opera.com/vu2rje/blog/ |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 12:24:41 IST
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hey varun,
i solved this using ur equations but i am not getting the answer.
once u try urself,,,,
i didnot understand the first equation f+T=MA,
becos for the right disc which is rotating with an angular velocity w , the friction acts in the forward direction(i.e, rightward) and the tension is acting along the string (leftward)......and hence the equation comes to be
f - T= MA
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 13:00:33 IST
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YES IT IS F-T IN TH FIRST EQUATION
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From shadows a light shall spring
Renewed shall be blade that's broken
The crown less again shall be king.
MY BLOG
http://my.opera.com/vu2rje/blog/ |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 14:03:37 IST
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even if it is f-T=MA u wiill not get the answer. first u try it and then give the reply completely.......
any experts please answer to my question. i am not getting it.................
i am waiting only for the complete solution.................and not anything else......................
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 18:46:20 IST
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I SOLVED IT SEE SINCE THERE IS NO EXT FORCE ACTING ON THE SYSTEM SINCE THE FRICTIONAL FORCES CANCEL EACH OTHER AS IT IS SHOWN Y THE EARLIER POSTS. TENSION IS AN INTERNAL FORCE AND NEE DNOT BE CONSIDERED SO THE PRINCIPLE OF CONSERVATION OF ANGULAR MOMENTUM CAN BE APPLIED INITIAL ANGULAR MOMENTUM IS IW FINALLY THE ANGULAR AND THE LINEAR VELOCITIES OF BOTH THE SPHERE WILL BE EQUAL SINCE THEY R CONNECTED BY THE SAME STRING SO THE ANGULAR MOMENTUM OF ONE OF THEM IS IW2+MVR W2=VR SINCE THE SPHERE IS RLLING IVR+MVR=1/2MR^2*V*R+MVR =1/2MVR+MVR=3/2MVR SO THE ANGULAR MOMENTUM OF BOTH THE SPHERES R 3MVR IW=3MVR 1/2MR^2W=3MVR V=WR/6 COZ OF ROLLING V=RW W=V/R=WR/6R=WR/6 PLZ RATE ME COZ IT TOOK ME 5 HRS TO SOLVE IT 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.
MY BLOG
http://my.opera.com/vu2rje/blog/ |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 May 2008 19:04:51 IST
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mahnat hai teri,,, but u wasted time dude,, still i vote 4 u!!
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Jul 2008 14:09:39 IST
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The direction that you show for angular velocity is it correct?
For that direction (clockwise), the right sphere will never collide with the left one as the friction at its point of contact is rightwards.
However if the friction is leftwards, then friction will be rightwards and the right sphere can approach the left one. In that case we will need the distance between the spheres to find with what speed will the left sphere collide with the right one. If the collision be assumed elestic, each sphere gains equal speeds (equal to haf the impacting speed of the left sphere) in opposite directions, but the angular velocity of the left sphere remains at the value it was before collision.
Now, its a simple problem of the spheres being given a vertain configuration of linear and angular velocities and finding their speeds at pure rollning is a standard task.
Please check whether or not the distance between them is given
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