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.

A heavy partile slides under gravity download the inside of a smooth v

1.	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» Solution :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))`