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 o with vertical? [ g is the acceleration due to gravity]

Step-by-step solution

Using conservation of energy ∆ K + ∆ U = 0 1 2 I 0 ω 2 = - ∆ U I 0 = m o m e n t o f i n e r t i a a b o u t H i n g e 1 2 m l 2 3 ω 2 = - - m g l 4 ω = 3 g 2 l ⇒ C is correct ⇒ a r a d i a l = ω 2 . l 2 = 3 g 2 l l 2 = 3 g 4 ⇒ A is correct Now, τ = I . α ⇒ m g . l 2 sin ⁡ 60 = m l 2 3 . α ⇒ α = 3 3 g 4 l ⇒ B is incorrect Acceleration in vertical direction ⇒ a v = α l 2 sin ⁡ 60 o + ω 2 l 2 cos ⁡ 60 o = a v = 3 3 g 8 3 2 + 3 g 8 a v = 9 g 16 + 6 g 16 = 15 g 16 Now, using N L M ⇒ m g - N = m a v ⇒ N = m g - m a v =