Question

A disc is freely rotating with an angular speed 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.

• A. ω
• B. $\frac { \omega } { 3 }$
• C. $\frac { \omega } { 2 }$
• D. None of these

Answer 1

Answer: (B)

$\frac { \omega } { 3 }$

Answer 2

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

⇒ L₂ = L₁ . . . (1)

Where L₁ = Angular momentum of the disc about P just before the impact = I₀ω = $\frac { 1 } { 2 }$ mr² ω.

⇒ L₂ = Angular momentum of the disc about P just after the impact
= I ω = ($\frac { 1 } { 2 }$mr² + mr²)ω= ∴ ⇒ ω′ = $\frac { 1 } { 3 } \omega$.