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Ask iit jee aieee pet cbse icse state board community Discussion Response Post to: Nice question... regarding constraint motion
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Manasi (3976)

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Olaaa!! Perrrfect answer. 702  bad job dude!! I dont approve of this answer! 2  [939 rates]

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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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