A uniform circular sector has radius R and total angle 2θ, where θ is measured in radians. Its centre of mass lies on the angle bisector at a distance d from the centre. Then d is:
Options

Question

A uniform circular sector has radius R and total angle 2θ, where θ is measured in radians. Its centre of mass lies on the angle bisector at a distance d from the centre. Then d is:

G1 Physics · Accepted Answer

Take the angle bisector as the x-axis and describe the sector by polar coordinates: −θ ≤ φ ≤ θ and 0 ≤ r ≤ R. With uniform surface density σ, dm = σr dr dφ. The total mass is M = σ∫_{−θ}^{θ}∫₀ᴿ r dr dφ = σR²θ. The first moment along the bisector is ∫x dm = σ∫_{−θ}^{θ}∫₀ᴿ (r cosφ)r dr dφ = σ(R³/3)(2 sinθ). Therefore d = x(CM) = [2σR³ sinθ/3]/[σR²θ] = 2R sinθ/(3θ).