A uniform rod pivoted at its upper end hangs vertically. It is displaced through

Question

A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60

^{\circ}

and then released. Find the magnitude of the force acting on a particle of mass 'dm' at the tip of the rod, when rod makes an angle of 37

^{\circ}

with the vertical.

Options:

A . 0.9(dm)g

B . 0.9√3(dm)g

C . 0.9√2(dm)g

D . (910)(√5)(dm)g

Answer: Option C
:
C
Let I = length of the rod, and m = mass of the rod.
Applying energy principle

(\frac{1}{2}) l ω^{2} - O = m g (\frac{1}{2}) (c o s 37^{\circ} - c o s 60^{\circ})

\Rightarrow \frac{1}{2} \times \frac{m l^{2}}{3} ω^{2} = m g \times \frac{1}{2} (\frac{4}{5} - \frac{1}{2})

\Rightarrow ω^{2} = \frac{9 g}{10 l} = 0.9 (\frac{g}{l})

Again

(\frac{\frac{m}{2}}{3}) \propto= m g (\frac{1}{2}) s i n 37^{\circ} = \frac{m g l}{2} \times \frac{3}{5}

∴ α = 0.9 (\frac{g}{l}) =

angular acceleration.
So, to find out the force on the particle at the tip of the rod

F_{1} -

centrifugal force=(dm)

ω 2 l = 0.9 (d m) g

F_{t} -

tangential force=(dm)

ω l = 0.9 (d m) g

So, total force

F = \sqrt{(F_{1}^{2} + F_{t}^{2})} = 0.9 \sqrt{2} (d m) g

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