Two heavy metallic plates are joined together at `90^@` to each other. A laminar sheet of mass 30kg is hinged at the line AB joining the two heavy metallic plates. The hinges are frictionless. The moment of inertia of the laminar sheet about an axis parallel to AB and passing through its center of mass is 1.2 kg `m^2.` Two rubber obstacles P and Q are fixed, one on each metallic plate at a distance 0.5m from the line AB. This distance is chosen so that the reaction due to the hinges on the laminar sheet is zero during the impact. Initially the laminar sheet hits one of the obstacles with an angular velocity 1 rad/s and turns back. If the impulse on the sheet due to each obstacal is 6 N-s,
(a) Find the location of the center of mass of the laminar sheet from AB.
(b) At what angular velocity does the laminar sheet come back after the first impact?
(c) After how many impacts, does the laminar sheet come to rest?

Explanation

(i) 0.1, (ii) 1
Let x be the distance of centre of the mass from AB.
Moment of inertia of the laminar sheet about an axis passing through centre of mass and paralel to `AB =1.2 kgm^(2)`

`I_(AB) =I_(CM) + MX^(2)`.........(i)
Impulse `=MV_(2) + (-v_(1)) = M(v_(2) +v_(1))`
Where `omega_(1)`, and `omega_(2)` are the angular velocities of the sheet before and after the collision with obstacles.
Impulsive torque `=-6 xx 0.5 =L_(t) -L_(1)`
`L_(i) =L_(t) +3 rArr (L_(i) - L_(f))=3`
`I_(AB)(omega_(1)-omega_(2))=3 rArr I_(AB) (omega_(1) + omega_(2)) =3`........(iii)
From equations (ii) and (iii), `I_(AB) = (Mx)/2`
`I_(AB) = (Mx)/2 =1.2 + Mx^(2) rArr 30x^(2) -15x + 1.2=0, x=(15+-9)/60 =0.4` o `0.1m`
(i) Substituting `x=0.4 m` in equation (ii)
`6=30(0.4)(omega_(2)+1) rArr omega_(2) =-0.5` (this is not possible)
The laminar sheet turns back after collision.
(ii) Substituiting x=0.1 m in equation (ii)
`6=30(0.1) (omega_(1)+1)`
Angular velocity of lamination sheet after collision =1 rad/s.
Ths is possible if collision is elastic.