A heavy partile slides under gravity download the inside of a smooth vertical tube dheld in vertical plane. It starts from the highest point with velocity `sqrt(2ag)` , where a is the radius of the circle. Find the angular position `theta` (as shown in figure) at which the vertical acceleration of the particle is maximum.

Answer

A, B, C

Explanation

At position `theta` ,
`v^(2)=v_(0)^(2)+2gh`
where, `h=a(1-costheta)`
:. `v^(2)=(sqrt(2ag))^(2)+2ag(1-costheta)`
or `v^(2)=2ag(2-costheta)` ...(i)
`N+mgcostheta=(mv^(2))/(a)`
or `N+mgcostheta=2mg(2-costheta)`
or `N=mg(4-3costheta)`
Net vertical force,
`F=Ncostheta+mg`
`=mg(4costheta-3cos^(2)theta+1)`
This force (or acceleration) will maximum when `(dF)/(dtheta)=0`
or `-4sintheta+6sinthetacos=0`
So, either
`sintheta=0` ,
`theta=0^(@)` ,
or `costheta=(2)/(3)` ,
`theta=cos^(-1)((2)/(3))`
`theta=0^(@)` is unacceptable
Therefore, the desired position is at
`thetacos^(-1)((2)/(3))`