Two particles of mass `m` each are kept on a horizontal circular platfrom on two mutually perpendicular radii at equal distance `r` from the center of the table. The particles are connected with a string, which is just taught when the platfrom is not rotating. The coefficient of static friction between the platfrom and block is `mu` (now if angular speed of platfrom `m` is slowly increased). Find the maximum angular speed `(omega)` of platfrom about it center so that the blocks remain stationery relative to platform. (If `mu=(1)/(sqrt(2))`, `r=2.5m` and `g=10m//s^(2)`)

Explanation

For small values of `omega` friction will be directed radially inwards as the tension in the spring is zero. The string will develop tension only if the centinfugal force `F_(C )` exceeds the limiting friction `f_(c )`. i.e. when
`momega^(2) r gt mu mg` `(f_(c )=mumg)`
or `omega=sqrt((mug)/(r))`
in this case direction of friction will be as shown in the figure.
For equilibrium
`F_(c )=Tcos45^(@)+f_(c )costheta`
And `Tsin45^(@)=f_(c )sintheta`
Eliminiating `T`, we get
`F_(c )=f_(c )(sintheta+costheta)`
i.e. `momega^(2)r=mumg[sqrt(2)sin(theta+45^(@))]`
or `omega^(2)=(sqrt2 mug)/(r )sin(theta+45^(@))`
maximum value of `sin(theta+45^(@))` is `1`
`:. ` Maximum value of `omega=sqrt((sqrt(2)mug)/(r ))=2`