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  circuler motion   4 Nickels awarded!
Tagged with:    [Post New]posted on 6 Jun 2007 20:55:16 IST    
I hav come across a gud question regarding constraint motion on the forum, dnt knw why others hadnt noticed this... so posting here, might help in understanding constraint motion... its here on this post
 
the situation is as shown in the figure, the left block is moving a constant velocity v, and the right one is fixed, find the speed of the sphere's centre, at the instant when x= r/(sqrrt2) and y= r(sqrrt), where r is the radius of the sphere
[Thumb - ou.JPG]

 
 
 
from the geometrical constraint we can get the relation between the downward velocity of the sphere, 'u' , and ' v '
let          l=y/2
 => (dl/dt)=(dy/dt) /2
              = v/2                         ........(i)
from the figure, x^2 + (l)^2=r^2, 
differentiating wrt time, we have,
     2x (dx/dt) + (2l) (dl/dt) = 0
=> x u + l (v/2) = 0
thrfore u = l v /2x
thrfore the downward velocity at the given instant, i.e. x=y,we have u= v/2
and now since the speed is asked we hav to consider the horizontal velocity of the sphere also, i.e. (dl/dt), which is v/2.
So vectorily adding both we hav v/sqrrt 2, n in vector form (v/2) (i-j)
 
Note: constraint relations are only used to find the relation between sppeds and not the velocities(thats y i hav neglected the sign, in deriving relation b/w v n u)
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a4asd (329)

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Olaaa!! Perrrfect answer. 57  bad job dude!! I dont approve of this answer! 1  [81 rates]

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pink_ele
pink_ele is offline comment by pink_ele    (posted on 6 Jun 2007 23:56:04 IST)
useful!!!!!!!!!!!
pirate1_from_jee
pirate1_from_jee is offline comment by pirate1_from_jee    (posted on 6 Jun 2007 23:57:41 IST)
Hard wrk!!!
magiclko
magiclko is offline comment by magiclko    (posted on 26 Jun 2008 13:50:27 IST)
this also was posted by me before....!!!!!!!!!!!
shivam999
shivam999 is offline comment by shivam999    (posted on 26 Jun 2008 15:27:02 IST)
bro first u correct the spelling of circular!!!!!!!!!!!
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