Hey magicklo, last time I did a mistake and the correct answer is 4V according to me.this is a nice problem and thanks for posting it. The tip to solve it that one should not get stuck with the word 'variable force'. Even if the force is variable yet it must provide the necessary centripetal force mw²r in the radial direction otherwise the particle won't execute circular motion.

  Now, let us first derive the general method for solving the problem for any angle.  (diagram below) At a particular instant angle POA = . So, angle BPO = /2. So, if F be the value of the force at that instant then resolve it in radial and tangential direction.

               Thus, F cos /2 = mw²r

          So, tangential comp. F sin/2 = mw²r tan/2

     So, tangential acceleration  a_t =w²r tan /2

  So, angular acceleration   =w² tan/2

    Note that here the angular acceleration will not remain constant as w and  are not constant. This situation can be compared to (but not same as) the varying acceleration of S.H.M.

            Now,  = w² tan /2

             So, dw/dt = w²tan/2

         dw/d . d/dt = w²tan/2

          w dw/d = w²tan/2

              dw/d = w tan/2

     So,   1/w dw = tan /2 d

  Integrating

             

         1/w dw =  tan /2 d

 

       Now let w at A be w_a and at P be w_p. In this problem  at A = 0degree and  at P = 120 degree.

 

          So,    _{[ wa]}^{[wp ]} 1/w dw = _{[0 ]}^{[120 ]}  tan /2 d

 

           Or,  log w_p - log w_a = [ (log sec/2)/(1/2)]₀¹²⁰

 

           or,  log w_p/w_a = 2 log (sec60/sec0)

          or,     log w_p/w_a = log (sec60/sec0)²

 So, taking antilog

                   w_p/w_a  =   (sec60/sec0)² = (2/1)² = 4

       So,            w_p = 4w_a

                      rw_p = 4 rw_a

                        V_p = 4 V

 

           This is the answer.