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(10/4)` mg along horizontal
B. `mg` along vertical
C. `(sqrt(10))/4mg` alonng 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 oif the rod about the pivot when is passes through the horizontal position is given by

`mgxxL/2sin30^@=(mL^(2))/3xxomega^(2)/2`
`omega=((3g)/(2L))`
Radial acceleration of the centre of mass (as centre of mass is moving in a circle of radius `L//2`) is given
`a_(r)=omega^(2)L/2=(3a)/4`
Torque about pivot, in the horizontal position, is `tau=mgL/2=Ialpha`
`alpha=(mgL//2)/(mL^(2)//3=(3g)/(2L)`
Tangential acceleration of the centre of mass `a_(t)=L/2 alpha=(3g)/4`
Draw the `FBD` of the rod at an instant when it passes through the horizontal position. Use Newton's second law of equation.

`R_(1)=ma_(r)=(3mg)/4`
`mg-R_(2)=mxxa_(t)=(3mg)/4`
`R_(2)=(mg)/4`
So, reaction force by the pivot on the rod `R=R_(1)^(2)+R_(2)^(2)=sqrt(10) mg/4` at angle of `tan^(-1)(R_(2)//R_(1))=[=tan^(-1)(1/3)]` with the horizontal.