1.

In the above question, the speed of each ball relative to ground just after they leave the disc is -

Answer»

`(Romega_(0))/(sqrt3)`
`(Romega_(0))/(sqrt2)`
`(2Romega_(0))/(3)`
None of these

Solution :The angular speed of the disc just after the BALLS leave the disc is `omega=(omega_(0))/(3)`
Let the speed of each ball just after they leave the disc be V.
From conservation of energy
`(1)/(2)[(1)/(2)mR^(2)]omega_(0)^(2)=(1)/(2)[(1)/(2)mR^(2)]omega^(2)+(1)/(2)[("m")/(2)]v^(2)+(1)/(2)[("m")/(2)]v^(2)`
solving we get `v=(2Romega_(0))/(3)`
NOTE : `v=sqrt((OMEGAR)^(2)+V_(r)^(2)) , V_(r)` = RADIAL velocity of tha ball


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