A simple pendulum of length L carries a bob of mass m. when the bob is at its lowest position, it is given that the monimum horizontal speed necessary for it to mvoe in a vertical circle about the point of suspension. When the string in horizontal, the net force on the bob is

Options

A. `sqrt(10)mg`
B. `sqrt(5)mg`
C. 4mg
D. 1mg

Answer

A

Explanation

`(1)/(2)xx mxx 5rg=(1)/(2)m xx v^2+mgr`
`(5)/(2) rg =(v^2)/(2)+gr`
`(3)/(2) rg =(v^2)/(2)`
`v^2= 3rg`.
Horizontal force, `f_x = f_c = (mv^2)/( r)`
`(mxx3rg)/( r) =3mg`
`f_(y)=mg`
`f_("net") =sqrt(f__(x)^(2) +f_(y)^(2))=sqrt((3mg)^2 +(mg)^2)`
`=sqrt(9m^2g^2+m^2g^2)`
`=mgsqrt(10)`.