As shown in the figure,a platform with a disk on it is rotating in the shown direction.The disk while rotating with it attains a speed of 5 m/s.The spring attached to it has a stiffness coefficient of 4 N/m,with an unstretched length of 1 m.The weight of the disk is 2 kgs.Determine at this instant the distance of the disk from the center of the platform where the spring is attached to a joint.The coefficient of kinetic friction between the platform and the disk is 0.1.

1) 2.276 m.
2) 1.456 m.
3) 4.707 m.
4) 5.234 m.

Free Body Diagram-
The FBD is shown in the adjoining figure.The frictional force opposes the relative motion between the disk and the platform.

Equations of motion-
ND - 19.62 = 0.-----------------------------(1)
0.1ND= 2at.----------------------------------(2)
4(x-1) = 2(25/x)-----------------------------(3)

Equation 1 gives-
ND = 19.62 N.

Equation 2 gives-
at = 0.981 m/s2.

Equation 3 is a quadratic equation-
2x2-2x-25 = 0
x = 4.707 m.