A particle `'P'` is moving on a circular 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=mRomega^(2) , F sin theta=ma_(t)`
Angular velocity `'omega'` of line joining `'P'` and `'C'` is
`omega =d/(dt)(2theta)=2(d theta)/(dt),:.(d theta)/(dt)=omega/2`
Tangential acceleration of particle about `'d'`
`a_(t)=(F sin theta)/m=Ralpha rArralpha=(F sin theta)/(mR)`
But `alpha=(domega)/(dt)=d/(dt)(2(d theta)/(dt))=2(d^(2) theta)/(dt^(2))`
`rArr (d^(2)theta)/(dt^(2))=alpha/2`, Hence `(d^(2)theta)/(dt^(2))=(F sin theta)/(2mR)`
`((d theta)/(dt^(2)))^(2)=1/4omega^(2)=1/4xx(Fos theta)/(mR)`
On solving `(d^(2)theta//dt^(2))/(((d theta)/(dt))^(2))`