A stationary horizontal disk has moment of inertia I = 0.09 kg m² about a fixed frictionless vertical axle. A pellet of mass 0.04 kg moves tangentially at 50 m s⁻¹ and embeds at radius 0.3 m. Treat the pellet as a point mass. During the short collision the external torque about the axle is negligible. Find the final angular speed and kinetic energy lost. Explain which conservation law applies and why linear momentum of disk plus pellet need not be conserved.

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Angular speed 6.41 rad s⁻¹; kinetic energy lost 48.08 J. Angular momentum about the axle is conserved. The pellet initially has angular momentum mur about the axle because its velocity is tangential. The stationary disk contributes no initial angular momentum. The axle may exert an impulse but provides no torque about itself. After embedding, the total rotational inertia is I + mr² = 0.0936 kg m². Angular momentum conservation gives mur = (I + mr²)ω. Therefore ω = 6.41 rad s⁻¹. The final kinetic energy is ½(I + mr²)ω² = 1.923 J. Initial kinetic energy is ½mu² = 50 J, so the loss is 48.08 J. Embedding is inelastic; this energy becomes internal energy, deformation and sound. The fixed axle supplies an external horizontal impulse, so total linear momentum need not remain constant. The positive energy loss and unchanged angular momentum are consistent, not contradictory.

A pellet embeds in a rotating disk · set 10

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