IIT JEE , AIEEE Forum - Physics , Mechanics

magiclko

here's the solution

Using conservation of energy between the initial position n at the position at angle Q, we have

( M_p w² / 2) = [mg (1-cosQ)] (l/2)

where M_p is the moment of inertia wrt axis passing thru P , having value

M_p = [m(l/2)²] /3

frome this we have w² = 3g(1-cosQ)/l ............(1)

Also M_p (alpha) = mg(l/2) sin Q

where alpha is the angular acceleration,

so, alpha = (3gsinQ) / 2l ...........(2)

Now as in the given figure, F_y = mg + ma_r cosQ + ma_t sinQ

and F_x = ma_t cos Q- ma_r sinQ

Where a_r = radial acceleration

= w² (l/2)

and a_t = tangential acceleration

= alpha (l/2)

thus using equation 1 and 2, we can solve for F_x and F_y