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. 4 mg
D. 1 mg

Answer

a

Explanation

According to the question, the bob is moving in a vertical circle, so the velocity at lowest point `V = sqrt(5rg)` Now, applying conservation of mechanical energy, we get
`(1)/(2) m xx 5rg = (1)/(2) mv^(2) + mgr`
or `(5)/(2) rg = (v^(2))/(2) + rg` or `(3)/(2) rg = (v^(2))/(2)`
or `v^(2) = 3rg`
Now, horizontal force on the bob, `Fx = F_(C ) = (mv^(2))/(r )`
`= (m xx 3rg)/(r ) = 3 mg` [from Eq. (i)]
vertical force on the bob, `F_(y) = mg`
`:. F_("net") = sqrt(F_(x)^(2) + F_(y)^(2))`
`= sqrt((3mg)^(2) + (mg)^(2)) = sqrt(10) mg`