First slide

Rotational motion

Question

One fourth length of a uniform rod of length 2l and mass m is placed on a horizontal table and the rod is held horizontally. The rod is released from rest. The normal reaction on the rod as soon as the rod is released will be

Difficult

A

$\frac{5 m g}{27}$
B

$\frac{4 m g}{31}$
C

$\frac{m g}{9}$
D

$\frac{4 m g}{19}$

Solution

The torque applied by gravity about the edge
$\begin{array}{l} O = τ = m g (3 ℓ / 4) \\ \Rightarrow I_{0} α = \frac{3}{4} m g . . . . . . (a) \\ where I_{0} = I_{g} + m {(\frac{3 ℓ}{4})}^{2} \\ = \frac{m ℓ^{2}}{12} + \frac{9}{16} m ℓ^{2} = \frac{m ℓ^{2}}{4} (\frac{1}{3} + \frac{9}{4}) \end{array}$
$I_{0} = \frac{31 m ℓ^{2}}{48} . . . . . . (b)$
$(a) and (b) yield, α = \frac{3 m g ℓ / 4}{31 m ℓ^{2} / 48}$
$\Rightarrow α = \frac{144 g}{124 ℓ} = \frac{36 g}{31 ℓ} . . . . . . (c)$
Newton's second law of motion:
$m g - N = m a . . . . . (d)$
$K i n e m a t i c s : a = (\frac{3 ℓ}{4}) α = \frac{36 g}{31}$
Using (d) and (e) we obtain
$N = m g - m (\frac{27 g}{31}) \Rightarrow N = \frac{4 m g}{31}$

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