A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60° with vertical?

Correct option is D. The angular speed of the rod will be √(3g/2L)
ΔK + ΔU = 0
1/2 I₀ω² = -ΔU
1/2 mL²/3 ω² = -(-mgL/4)
ω = √(3g/2L)
→ aradial = ω²L/2 = √(3g/2L) L/2 = 3g/4
→ τ = I₀α
α = mgL/2 sin 60°/mL²/3 = 3√3g/4L
→ av = (αL/2)sin 60° + ω²L/2cos 60° where av is the vertical acceleration.
→ av = (3√3g/8)(√3/2) + (3g/4)(1/2)
→ av = 9g/16 + 3g/8
→ av = 15g/16
net vertical forces = mav
→ mg - N = mav
→ N = mg - mav = mg - m(15g/16) = mg/16