A disc is freely rotating with an angular speed ω ω on a smooth horizontal plane. It is hooked at a rigid peg P & rotates about P without bouncing. Its angular speed after the impact will be equal to.
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During the impact, the impact forces passes through the point P. Therefore the torque produced by it about P is equal to zero. Consequently the angular momentum of the disc about P, just before & after the impact remains the same
⇒ ⇒ L 2 = L 1 . . .
Where L 1 = Angular momentum of the disc about P just before the impact = I 0 ω ω = 
mr 2 ω ω .
⇒ ⇒ L 2 = Angular momentum of the disc about P just after the impact
= I ω ω = ( 
mr 2 + mr 2 ) ω ω = 
∴ ∴ 
⇒ ⇒ ω ω ′ ′ = 
.
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