Three particles A, B and C each of mass m, are connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side l. This body is placed on a horizontal frictionless table (x-y plane) and is hinged to it at the point A so that it can move without friction about the vertical axis through A . the body is set into rotational motion on the table about A with a constant angular velocity `omega`.

(a) Find the magnitude of the horizontal force exerted by the hinge on the body.
(b) At time T, when the side BC is parallel to the x-axis, a force F is applied on B along BC (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time T.

Explanation

a. The centre of mass of the system is at the centroid of a triangular assembly. The `CM` moves along a cirular path with constant angular velocity. Therefore, there must be a horizontal centripetal fore directed towards the axis at the hinge.
From figure a and we find
`AD=lsin60^@((lsqrt(3))/(2))`
`AO=2/3AD=l^(2)/(sqrt(3))=r`
The centripetal acceleration `a_(c)=omega^(2)r=omega^(2)(l/(sqrt(3)))`
Tangential acceleration `a_(1)=alphar=alpha(l/(sqrt(3)))`

b. Let `F_(x)` and `F_(y)` be the forces applied by the hinges along `x`-axis and `y`-axis, respectively. The system is in non centroidal rotation. The three equations of motion are
`SigmaF_(x)=F_(x)+F=(3m)a_(t)=3m(l/sqrt3alpha)`.........i
`SigmF_(y)=F_(y)=3m(l/(sqrt(3)))omega^(2)`............ii
`Sigmatau=Fxx((sqrt(3))/2l)=2ml^(2)alpha` .............iii
From eqn ii `alpha=(sqrt(3)F)/(4ml)`
From eqn i `F_(x)+F=3m(1/sqrt(3)alpha)xx(sqrt(3)F)/(4ml)=(3F)/4`
`implies F_(x)=F/4`
From eqn ii `F_(y)=sqrt(3)mlomega^(2)`