A rod of length l and mass M is fixed to one end of a string . The string passes over a massless , frictionless pulley . A small ball of mass m (lt M) is attached to the other side of the string . The ball has an opening through which the string passes and the ball can slide along the string with some friction . Initially , the ball is held at level with the lower end of the rod . Whenset free , both the force of friction , if the ball rises to the level of the upper end of the rod in time t ?

A rod of length l and mass M is fixed to one end of a string . The str

1.	A rod of length l and mass M is fixed to one end of a string . The string passes over a massless , frictionless pulley . A small ball of mass m (lt M) is attached to the other side of the string . The ball has an opening through which the string passes and the ball can slide along the string with some friction . Initially , the ball is held at level with the lower end of the rod . Whenset free , both the force of friction , if the ball rises to the level of the upper end of the rod in time t ?
Answer» Solution :When the system is set in motion , let `a_(M) , a_(m)` be the ACCELERATION of rod and ball respectively and F the force of friction applied by the string on the ball . The equation of MOTIONS are : `Mg - T = Ma_(M) ""……..(i)` ` F - mg = ma_(m) "" .......(ii)` The string is pulled up and the ball slides down due to its weight . Friction between the ball and string tries to hold it but ULTIMATELY it moves with acceleration `a_(m)` . As the string is massless , the net force on any portion of it must be zero . So , F = T Eqn . (i) can be written as : `Mg - F = Ma_(M) "" ....... (iii)` Adding eqns . (ii) and (iii) , `a_(m) + a_(M) = F((1)/(m) - (1)/(M)) "" ........ (iv)` As the rod moves DOWNWARD while ball upward , so , RELATIVE distance of the rod and ball in time t , `l = (1)/(2) (a_(m) + a_(M))t^(2)` Substituting for `(a_(m) + a_(M))` from eqn. (iv), `l = (1)/(2) F ((1)/(m) - (1)/(M))t^(2)` or `""F = (2 lmM)/(( M - m) t^(2))`