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Bar pendulum minimum period condition

Short answer

For a bar pendulum, T = 2π √((k²+l²)/(gl)). Minimizing T² gives d/dl(k²/l + l)=0 → l=k. Thus minimum period when distance from CG equals radius of gyration. Then the centre of oscillation is also at distance k, so suspension and oscillation points coincide.

Show that in a bar pendulum, the minimum time period is achieved if the radius of gyration is equal to the distance of the point of suspension or point of oscillation from the center of gravity.

For a bar (compound) pendulum, let G be the centre of gravity, k the radius of gyration about G, and l the distance of the point of suspension O from G.

Time period

I_O = Mk² + Ml²
T = 2π √(I_O/(Mgl)) = 2π √((k² + l²)/(g l))

Condition for minimum

Write T² = (4π²/g) (k²/l + l). For T to be minimum, d(T²)/dl = 0:

d/dl (k²/l + l) = −k²/l² + 1 = 0
l² = k²   →   l = k

Second derivative: 2k²/l³ > 0 at l = k, so it is a minimum.

Hence the time period is minimum when the distance of the point of suspension from the centre of gravity equals the radius of gyration about G.

Since the centre of oscillation lies at l′ = k²/l from G on the other side, for l = k we get l′ = k²/k = k. So the distance of the point of oscillation from G is also equal to k.

 O ----- G ----- O'
   <-k->   <-k->
   L = 2k

Minimum period: T_min = 2π √(2k/g), with equivalent simple-pendulum length L = 2k. At this point the suspension and oscillation points coincide in their distances from G, so the four collinear points merge into two.

✓ Answer checked by Claude Opus 5.5 (high effort)

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