A simple pendulum is constructed by attaching a bob of mas m to a string of length L fixed at its upper end. The bob oscillates in a vertical circle. It is found that the speed of the bob is v when the string makes an angle `theta` with the vertical. Find the tension in the string at this instant.

Options

A. `T=m ((v^(2))/(L)+g "cos" alpha)`,
`F= msqrt(g^(2)"sin"^(2)alpha + (v^(4))/(L^(2)))`
B. `T=m ((v^(2))/(L)+g "cos" alpha)`,
`F= msqrt(g^(2)"cos"^(2)alpha + (v^(4))/(L^(2)))`
C. `T=m ((v^(2))/(L)+g "cos" alpha)`,
`F= msqrt(g^(2)"sin"^(2)alpha - (v^(4))/(L^(2)))`
D. `T=m ((v^(2))/(L)+g "cos" alpha)`,
`F= msqrt(g^(2)"cos"^(2)alpha - (v^(4))/(L^(2)))`

Answer

A

Explanation

The bob moves in a circle of radius L. When it is at A, the centripetal force of magnitude `(mv^(2))/(L)` is provided by `T- mg "cos" alpha`
`therefore T- mg "cos" alpha=(mv^(2))/(L)`
`therefore T= (mv^(2))/(L)+mg "cos" alpha = m ((v^(2))/(L)+ g "cos" alpha)`
and the net force,
`F= sqrt((mg "sin" alpha)^(2)+((mv^(2))/(L))^(2))`
`= m sqrt(g^(2)"sin"^(2) alpha +(v^(4))/(L^(2)))`