Three identical thin rods, each of mass `m` and length `l`, are joined to form an equilateral triangular frame. Find the moment of inertia of the frame about an axis parallel to its one side and passing through the opposite vertex. Also find its radius of gyration about the given axis.

Options

A. `v_("MAX")^(2)=(4ga)/(sqrt(3))`
B. `v_("MAX")^(2)=sqrt(3)" ga"`
C. `v_("MAX")^(2)=(2ga)/(sqrt(3))`
D. `v_("MAX")^(2)=(ga)/(sqrt(3))`

Answer

C

Explanation

Using parallel axis theorem for the side BC, the moment of inertia of the triangle about an axis passing through A and perpendicular to the plane of the triangle,
`I_(A)=2((1)/(3)ma^(2))+((1)/(12)ma^(2)+m((sqrt(3)a)/(2))^(2))=(3)/(2)ma^(2)`
If the angular velocity of the triangle at any instant is `omega`, the velocity of the vertex B at that instant is `a omega`
Therefore, the velocity of B is maximum at the instant the angular velocity is maximum, i.e. when the side BC becomes horizontal
Let the angular velocity at this instant be `omega_("MAX")`
Then, by conservation of energy
Gain in kinetic energy = Loss in potential energy
`(1)/(2)I_(A)omega_("MAX")^(2)=3mg` (Loss in height of CM of triangle)
`(1)/(2)((3)/(2)ma^(2))omega_("MAX")^(2)=3mg((a)/(sqrt(3))-(a)/(2sqrt(3)))=(sqrt(3))/(2)mga" "implies" "omega_("MAX")^(2)=(2g)/(sqrt(3)a)`
Therefore, the maximum velocity of the vertex B is given by `v_("MAX")^(2)=omega_("MAX")^(2)a^(2)=(2ga)/(sqrt(3))`