Show that if a rod held at angle `theta` to the horizontal and released, its lower end will not slip if the friction coefficient between rod and ground is greater than `(3sinthetacostheta)/(1+3sin^(2)theta)`

Explanation

Point A is momentarily at rest.
`alpha=(mg(l)/(2)costheta)/((ml^(2))/(3))=(3)/(2)(gcostheta)/(l)`
`thereforea_(C)=(l)/(2)alpha=(3)/(4)gcostheta`
Now `muN=ma_(x)`
or `muN=ma_(C)sintheta`
or `muN=(3)/(4)mgsinthetacostheta` ..(i)
further `mg-N=ma_(y)`
or `N=mg-ma_(C).costheta`
or `N=mg-(3)/(4)mgcos^(2)theta` .(ii)
dividing eq. (i) by eqs. (ii), we get
`mu=((3)/(4)sinthetacostheta)/(1-(3)/(4)cos^(2)theta)=(3sinthetacostheta)/(4-3cos^(2)theta)`
`=(3sinthetacostheta)/(1+3sin^(2)theta)`