Q: A uniform rod of mass m and length L is held vertically on a smooth horizontal surface. When the rod is released, choose the correct statement (s).
(a) The COM of the rod accelerates in vertical direction
(b) Initially the magnitude of normal reaction is mg
(c) When the rod becomes just horizontal, the magnitude of normal reaction is mg /2
(d) When the rod becomes just horizontal, the magnitude of normal reaction is mg/4
Click to See Answer :
Initially , N – m g = 0
N = m g
M g – N = m a …(i)
As , τ = I α
$\displaystyle m g (\frac{L}{2} cos \theta) = \frac{m L^2}{3} \alpha $
$\displaystyle \alpha = \frac{3 g}{2 L} cos\theta $ ….(ii)
By using , a = r α
$\displaystyle a = \frac{L}{2} \alpha $ …(iii)
From (ii) & (iii) we get,
$\displaystyle a = \frac{3 g}{4} cos\theta $
Fro (i) we get,
$\displaystyle m g – N = m \frac{3 g}{4} cos\theta $
When rod becomes horizontal θ = 0°
$\displaystyle m g – N = \frac{3 m g}{4} cos0^o $
$\displaystyle m g – N = \frac{3 m g}{4} $
$\displaystyle N = \frac{m g}{4} $