A particle `'P'` is moving on a circular path under the action of only one force action always toward the fixed point `'O'` on the circumference. Find the ratio of `(d^(2)theta)/(dt^(2))` & `(("d"theta)/(dt))^(2)`

Options

A. 2 tan `theta`
B. tan `theta`
C. `(tan theta)/(2)`
D. `(tan theta)/(3)`

Answer

A

Explanation

F cos `theta = m R omega^(2) , F "sin" theta = ma_(t)`
Angular velocity '`omega`' of line joining 'P' and 'C' is
`omega = (d)/(dt) (2 theta) = 2 (d theta)/(dt)`
`therefore " " (d theta )/(dt) = (omega)/(2)`
Tangential acceleration of particle about 'd' ,
`a_(t) = (F "sin" theta)/(m) = R alpha implies alpha = (F "sin" theta)/(mR)`
But `alpha = ( d omega)/(dt) = (d)/(dt) ( 2 (d theta)/(dt)) = 2 (d^(2) theta)/(dt^(2))`
`implies (d^(2) theta )/(dt^(2)) = (alpha)/(2)` , Hence `(d^(2) theta)/(dt^(2)) = (F "sin" theta)/(2 mR)`
`((d theta)/(dt))^(2) = (1)/(4) omega^(2) = (1)/(4) xx (F "cos" theta)/(mR)`
On solving , `((d^(2) theta)/(dt^(2)))/(((d theta)/(dt))^(2)) = 2 "tan" theta `