A uniform rod pivoted at its upper end hangs vertically. It is displaced through anangle of `60^@` and then released. Find the magnitude of the force acting on a particle of mass `dm` at the tip of the rod when the rod makes an angle of `37^@` with the vertical.

Answer

B, D

Explanation

Let l=length of the rod and m=mass of the rod.
Applying energy principle
`(1/2)Iomega^2-=mg1/2(cos37^0-cos60^0)`
`rarr 1/2.(momega)^2/3=mg1/2(4/5-1/2)t`
`rarr omega^2=(9g)/(10l)=0.9(g/l)`
Again `malpha=mg(1.2sin37^@)`
`=mg(1.2)xx3/5`
`:gt alpha=0.9(g/l) `
`=angular acceleration
So, to find out the force on the particle at the tip of the rod
`F_1=` centrifugal force
`=(dm)omega^2l=0.9(dm)g`
`F_t=` tangential force
`=(dm)alphal=0.9(dm)g`
`So total force ltbr. `F=sqrt(F_1^2+Ft^2)`
`=0.9sqrt2(dm)g`