A uniform circular disc of radius `R` is placed on a smooth horizontal surface with its plane horizontal and hinged at circumference through point O as shown . An impulse P is applied at a perpendicular distance `h` from its centre C. The value of h so that the impulse due to hinge is zero, is

Options

A. `R`
B. `R//2`
C. `R//3`
D. `R//4`

Answer

B

Explanation

`omega=("angular impulse")/(I)` (about O) ltbr `=(P(R+h))/((3)/(2)mR^(2))`
Now, `V_(COM)=Romega=(PR(R+h))/((3)/(2)mR^(2))`
`P+P_("Hinge")=mv_(COM)=(mpR(R+h))/((3)/(2)mR^(2))`
solving, this equation we get `h=(R)/(2)`