A uniform circular disc of mass 1.5 kg and raius 0.5 m is initially ar rest on a horiozntal frictonless surface. Three forces of equal matgnitude F = 0.5 N are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces the angular speed of the disc in `rad s^(-1)` is

Answer

2

Explanation

Refer to the question to understand that length of perpendicular drawn from the centre on the line of action of the force is `Rsin30^(@)` There are total three forces.
`tau_("net")=3tau_(1)=3FRsin30^(@)`
`=3(0.5)(0.5)(1)/(2)=(3)/(8)N-m`
`I=(1.5(0.5)^(2))/(2)=(3)/(16)`
`alpha=(tau)/(I)=2rad//s^(2)`
`omega=alphat=2` rad/s