Sol. Assume that the yo-yo is at rest before the application of the force F.

As there is no friction acting on the yo-yo the direction of rolling is only determined by the direction of the torque of the applied force F about its center. The direction of rolling is shown in fig. for θ = 0, π /2 or π .

The friction acting on the yo-yo is f = µN , where N is the normal reaction of the table, as shown in fig.(1). The yo-yo will slide without rolling if

The acceleration a of the center of mass of the yo-yo is given by

F cos θ – µN = Ma

Thus cos θ = + .

If this condition is satisfied, θ is independent of µ . If still depends on F unless a = 0 , i.e. no motion.

Let the acceleration of the center of mass of the yo-yo and its angular acceleration about the center be a and α respectively. We have Fig. (1)

F cos θ – f = Ma

F A – FB = MA 2 α .

For rolling without slipping, a = –A α . Eliminating α and a gives

f = .

As f < µN = µ (Mg – F = sin θ ),

For the yo-yo to roll without slipping irrespective of the smoothness of the table, i.e. independent of µ , we require

sin θ = cos θ = ,

or tan θ = .

Thus we require that, first on all, 2B < A, Mg < F. then two values of θ , one positive and one negative, with the same |sin θ | are possible.