A uniform rod of length 4l and mass `m` is free to rotate about a horizontal axis passing through a point distant `l` from its one end. When the rod is horizontal its angular velocity is `omega` as shown in figure. calculate
(a). reaction of axis at this instant,
(b). Acceleration of centre of mass of the rod at this instant.
(c). reaction of axis and acceleration of centre mass of the rod when rod becomes vertical for the first time.
(d). minimum value of `omega`, so that centre of rod can complete circular motion.

Explanation

`alpha=(mgl)/((m(4l)^(2))/(12)+ml^(2))=(3)/(7).(g)/(l)`
`therefore (a_(C))_(v)=lalpha=(3)/(7)g`

Let `V` be the vertical reaction (upwards) at axis, then
`mg-V=ma_(C)=(3mg)/(7)`
`thereforeV=(4)/(7)mg` ..(i)
If H be the horizontal reaction (towards CO) at axis the
`H=m//omega^(2)` ..(ii)
`therefore` total reaction at axis
`N=sqrt(H^(2)+V^(2))=(4)/(7)mgsqrt(1+((7lomega^(2))/(4g))^(2))`
(b). `a_(C)=sqrt((a_(C))_(V)^(2)+(lomega^(2))^(2))`
`=sqrt(((3g)/(7))^(2)+(lomega^(2))^(2))`
(c) Let `omega'` be the angular speed of the rod when it becomes vertical for the first time. then from conservation of mechanical energy
`(1)/(2)I(omega^('2)-omega^(2))=mgl`
`thereforeomega^('2)=omega^(2)+(2mgl)/(I)`
`=omega^(2)+(2mgl)/((7)/(3)ml^(2))`
`=omega^(2)+(6)/(7)(g)/(l)`
Acceleration of centre of mass at this instance will be,
`a_(C)=lomega^('2)=lomega^('2)+(6g)/(7)`
Let V be the reaction (upwards) at axis at this instant, then
`V-mg=ma_(C)=mlomega^(2)+(6mg)/(6)`
`thereforeV=(13)/(7)mg+mlomega^(2)`
(d). from conservation of mechanical energy
`mgl=(1)/(2)Iomega_(min)^(2)`
`thereforeomega_(min)=sqrt((2mgl)/(I))=sqrt((2mgl)/((7)/(3)ml^(2)))`
`=sqrt((6)/(7)(g)/(l))`