A weightless rod of length l with a small load of mass m at the end is hinged at point A as shown and occupies a strictly vertical position, touching a body of mass M. A light jerk sets the system in motion. a. For what mass ratio M//m will the rod form an angle alpha=pi//6 with the horizontal at the moment of the separation from the body? b. What will he the velocity u of the body at this moment? Friction should be neglected.

A weightless rod of length l with a small load of mass m at the end is

1.	A weightless rod of length l with a small load of mass m at the end is hinged at point A as shown and occupies a strictly vertical position, touching a body of mass M. A light jerk sets the system in motion. a. For what mass ratio M//m will the rod form an angle alpha=pi//6 with the horizontal at the moment of the separation from the body? b. What will he the velocity u of the body at this moment? Friction should be neglected.
Answer» Solution :`a=a_(t)sinalpha-a_(c)cos ALPHA` At the TIME of separation For `M, SigmaF_(x)=0` and `a=0` Acceleratin of the load in horizontal direction `a=0=a_(1)sinalpha-v^(2)/lcosalpha` `a_(t)sinalpha=(v^(2))/lcosalpha` But `a_(t)=gcosalpha` HENCE `v=sqrt(GL sin alpha)` `u=vsinalpha=sinalpha sqrt(gl sin alpha)` `mgl=mgl sin alpha+(mv^(2))/2+1/2Mv^(2)sinalpha` `M/m=(2-3sinalpha)/(sin^(3)alpha)=4` `u=vsinalpha=1/2sqrt((gl)/2)`