Question 4d195

Jun 5, 2016

I found $\sqrt{\frac{\text{3g}}{L}} \to$Option(B)

Explanation:

Let

• $m \to \text{Mass of the rod}$
• $L \to \text{Length of the rod}$
• $g \to \text{Acceleration due to gravity}$
• $w \to \text{ Angular velocity of the rod when it comes at vertical position}$

The rod was initially at rest in its horizontal position. When it comes to vertical position after falling freely under gravity, it loses its PE due its shift of center of mass through a fall of height $\frac{L}{2}$

This loss of PE is equal to its gain in KE
So
$\frac{1}{2} \times I \times {w}^{2} = m g \times \frac{L}{2}$
Where I is the moment of inertia
about horizontal axis passing through the poit of suspension at the end.
"where I =1/3xxmxxL^2#

Inserting the value of I

$\frac{1}{2} \times \frac{1}{3} \times m \times {L}^{2} / 2 \times {w}^{2} = m g \times \frac{L}{2}$

$\text{Angular Velocity at vertical position of the rod} \left(w\right)$

$w = \sqrt{\frac{\text{3g}}{L}}$

Is it OK?