Question

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 1

Correct Answer - `sqrt((gl)/2)`
`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)`