JEE Main Physics QNo:3 2020 (Solution)

 A bead of mass m stays at point P(a,b) on a wire bent in the shape of a parabola y=4Cx²

and rotating with angular speed 𝝎.Find the value of 𝝎.

 

 

 

Solution: Since the path of the bead is a parabola it's velocity at P can be calculated using

`V=sqrt(2gh)`(in a projectile motion vertical motion is equivalent to free fall)

here h=b `therefore V=sqrt(2gb)`--------------(1)

since y=`4Cx^2`

here x=a and y=b `therefore``b=4Ca^2`

substituting in (1)

V=`sqrt(2g4ca^2)`

V=`2a sqrt(2gc)`

𝝎=V/r here r=a

𝝎=`2sqrt(2gc)` (Answer)