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).

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).

Numerical

(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

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Ans: (a), (b) , (d)
Solution : As no force is acting in horizontal direction so center of mass of rod accelerates in vertical direction .

Initially , N – m g = 0

N = m g

Numerical

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} $