Two particles of masses m1 and m2 are joined by a light rigid rod of length r. The system rotates at an angular speed ω about an axis through the centre of mass of the system and perpendicular to the rod. Show that the angular momentum of the system is L=μ r2ω where μ is the reduced mass of the system defined as μ=m1+m2m1+m2.
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Hence proved m1m2 / (m1+m2)r^2ω = Lur^2ω
Explanation:
Angular momentum due to the mass m1 at the centre of system
L1 = m1 r1 ω
= m1 x (m2r/m1+m2)^2 x ω
r2 = r - m2r / m1+m2 = m1r / m1+m2
L2 = m2 r2 ω
m2 x (m1r/m1+m2)^2 x ω
Therefore net angular momentum = L1 + L2
= m1 m2^2r^2ω/ (m1+m2)^2 ÷ m2 m1^2 r^2ω / (m1+m2)^2
= m1 m2 (m1+m2) r^2ω /(m1+m2)^2
m1m2 / (m1+m2)r^2ω = Lur^2ω
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