A slender rod of mass `m` and length `L` is pivoted about a horizontal axis through one end and released from rest at an angle of `30^@` above the horizontal. The force exerted by the pivot on the rod at the instant when the rod passes through a horizontal position is

Options

A. `sqrt(10)/g mg` along horizontal
B. mg along vertical
C. `sqrt(10/4)`mg along a line making an angle of `tan^(-1) (1/3)` with the horizontal.
D. `sqrt(10)/4=mg` along a line making an angle of `tan^(-1) (3)` with the horizontal.

Answer

C

Explanation

The angular velocity of the rod about pivot when it passes through horizontal position is given by
`mg xx L/2 sin 30^(@) = (mL^(2))/3 xx omega^(2)/2 , omega = sqrt((3g)/(2L))`

Radial accleration of the cenre of mass (as centre of mass is moving in a circle of radius `L//2`) is given by `a_(r) = omega^(2) L/2 = (3g)/4`
Torque about pivot, in horizontal position, is,
`tau = mg L/2 = I alpha , alpha = (mgL//2)/(mL^(2)//3) = (3g)/(2L)`
Tangential acceleration of the centre of mass, `a_(t) = L/2 alpha (3g)/4`

`mg-N_(1) = ma_(t) rArr N_(1) =mg-(3mg)/4 = (mg)/4`
`N_(2) = ma_(r)=(3mg)/4` `therefore N=sqrt(N_(1)^(2) + N_(2)^(2)) = sqrt(10/4) = mg`
Angle with horizontal is `tan^(-1)(N_(1)/N_(2)) = sqrt(10)/4 mg`
Angle with horizontal is `tan^(-1)(N_(1)/N_(2)) = tan^(-1)(1/3)`