IIT JEE , AIEEE Forum - Physics , Mechanics

gonecrazy

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

magiclko

the answer is v(sqrrt 2)...isnt it???

adityak

the answer is v(sqrrt 2)
thats correct

gonecrazy

hey manasi and aditya,i want the solution

magiclko

Hiiiiiiiiiii again crazyguy

well m xtremely srry cudnt give the solution then, as i was in a hurry, n also that the answer is not vsqrt2, but v/ sqrt2, i hav mistaken somewhere that tym, newayz here's the full solution to the problem.....

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)

gonecrazy

hey manasi, thanx once again

, a salute for u

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