A circular table with smooth horizontal surface is rotating at an angular speed `omega` about its axis. A groove is made on the surface along a radius and a small particle is gently placed inside the groove at a distance l from the centre. Find the speed of the particle with respect to the table as its distance from the centre becomes L.

Answer

`w sqrt((b^(2) - a^(2)))`

Explanation

In a turning table the force along the radius of the table is the centrifugal force . Thus , the acceleration of a particle when it is at a distance x from the centre is

`a = omega^(2)x`
`" " v(dv)/(dx) = omega^(2) x`
or `" " v dv = omega^(2) x dx `
Integrating above equation , we get
`int_(0)^(v)v dv = int_(a)^(b) omega^(2) x dx `
` |(v^(2))/(2)|_(0)^(v) = omega^(2)|(x^(2))/(2)|_(a)^(b)`
or `" " v^(2) = omega^(2) (b^(2) - a^(2))`
`therefore " " v = omega sqrt(b^(2) - a^(2))`