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?

Answer

(a) `1 = 0.1 m` ;
(b) `w' = 1 rad//s` ;
( c) laminar sheet will never come to rest.

Explanation

Let `r` be the perpendicular distance of `CM` from the line `AB` and `omega` is the angular velocity of the sheet just after colliding with rubber obstacle for the first time.
Obviosly the linear velocity of `CM` before and after collision will be
`v_(f) = ( r) (1 rad//s) = r` & `v_(f) = r omega`
`vec v_(i)` and `vec v_(f)` will be in opposite directions.
linear impulse of `CM` = change in linear momentum of `CM`
`6 = m(vec v_(f) - vec v_(i))`
`6 = 30 (r + r omega)`
`r (1 + omega) = (1)/(5)`
Similarly, angulasr impulse about `AB` = change in angular momentum about `AB`.
Angular impulse = linear impulse xx perpendicular distance of impulse from `AB`.
Hence, `6 xx 0.5 m = I_(AB) (omega + 1)`
[initial angular velocity `= 1 rad//sec`,]
`3 = [I_(CM) + Mr^(2)] )(1 + omega)`
`3 = [1.2 + 30 r^(2)] [ 1 + omega]` ....(2)
Solving Eq. (1) and (2) for `r`, we get
`r = 0.4 m` and `r = 0.1 m`
but at `r = 0.4 m, omega` comes out to be negative `(-0.5 rad//s)(s)` which is not acceptable. Therefore
(a) `r` = distance of `CM` from `AB = 0.1`
(b) Substuting `r = 0.1 m` in Eq. (1), we get `omega = 1 rad//s` i.e., the angular velocity with which sheet comes back after the first impact is `1 rad//s`.
( c) Since, the sheet returns with same angular velocity of `1 rad//s`, the sheet will never come to rest.