Please help me sove this physical pendulum problem?

Two things you need to know here:

1) period T = 2π√(I / mgx), and

2) I = Icm + mx² = mL²/12 + mx² = m(L²/12 + x²)

(by the parallel axis theorem)

Then T = 2π√( m(L²/12 + x²) / mgx) = 2π√((L²/12 + x²) / gx)

Find where T is max/minimized:

d((L²/12 + x²) / x)/dx = 1 - L²/12x² = 0

x² = L²/12

x = L/√12 = L*√12 / 12 = L√3 / 6 ← distance from center

Since L = 1.85 m, x = 0.534 m

and T = 2π√( (1.85²/12 + 0.534²) / 0.534*9.8) s = 2.07 s ← resulting period

We should check the endpoints.

By observation, as x → 0, T → infinity, so that's not a minimum.

When x = L/2, T = 2π√( (1.85²/12 + (1.85/2)² / (9.8*1.85/2) ) s = 2.23 s -- not a minumum.

(a)

The weight of the stick is solely responsible for its torque.

Iθ" = mgx sinθ

Find the moment of inertia of the stick at different pivot points via the parallel axis theorem.

(mL²/12 + mx²)θ" = mgx sinθ

(L²/12 + x²)θ" = gx sinθ

For small oscillations sinθ ≈ θ.

θ" = [gx / (L²/12 + x²)] θ

ω² = gx / (L²/12 + x²)

d(ω²)/dx = [(L²/12 + x²)(g) - (gx)(2x)] / (L²/12 + x²)² = 0

g(L²/12 - x²) = 0

x = L/sqrt(12)

(b)

ω² = [gL/sqrt(12)] / (L²/12 + L²/12)

ω² = [g/sqrt(12)] / (L/6)

(2π/T)² = g sqrt(3) / L

T = 2π sqrt{L / [g sqrt(3)]}