IIT JEE , AIEEE Forum - Articles Category , Community shelf - Nice question... regarding constraint motion

magiclko

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

http://www.goiit.com/posts/list/3030.htm

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

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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