A uniform rod of mass m and length l is pivoted smoothly at point O as shown in figure. If a horizontal force F acts at the bottom of the rod and `omega` is the angular velocity of the rod which is a function of angle of rotation `theta`, then the maximum angular displacement of the rod is
(Acceleration due to gravity, g)

Options

A. `theta=2sin^(-1)((2F)/(mg))`
B. `theta=2cos^(-1)((2F)/(mg))`
C. `theta=2tan^(-1)((2F)/(mg))`
D. `theta=2cot^(-1)((2F)/(mg))`

Answer

C

Explanation

According to the question, we drawn the following figure,

As for maximum angular displacement, equilibrium of torque exists about the centre mass of rod.
So, `" "tau_(1)+tau_(2)=0 " …(i)"`
Torque due to the gravitational force,
`" "tau_(1)=mgsinphi times 1`
`rArr " "tau_(1)=mglsinphi " ...(ii)"`
Similarly, torque due to horizontal force,
`" "tau_(2)=-F2lcosphi`
`rArr " "tau_(2)=2Flcosphi " ...(iii)"`
From Eqs. (i), (ii) and (iii), we get
`tanphi=(2F)/(mg) rArr phi = tan^(-1)""(2F)/(mg)`
Since, the angular velocity of the rod is a function of angle of rotation `theta`,
As putting, `theta_("max")=2phi`
So, `phi =(theta_(max))/2`
Hence, `theta_(max)=2tan^(-1)""(2F)/(mg)`
The correct option is (c).