Question

A rod of length 50 cm is pivoted at one
end. It is raised such that it makes an
angle of 53° from the horizontal as
shown and released from rest. Its
angular speed when it completes
angular displacement 16° after realse
(in rads -1) will be (g = 10ms-2 and
\cos(53) = \frac{3}{5}

the diagram for the following que is in the pic 

explain the process in steps and mention the values clearly 

It's really very urgent pls ans it ASAP..
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Answer 1

Answer :-

**Consider attached for better understanding.

◾As per the question :-

▪️Length of rod = 50 cm or 0.5 m

▪️Pivoted angle = 53°

▪️Angular Displacement = 16°

▪️Gravitational pull = 10 m/s²

⏩ Solution :-

→As we can clearly see from the figure that after moving 16° the pivoted angle becomes 37°

→ Also we have these equation

▪️Rotatinal Kinetic Energy :-

» $\sf{K_r = \dfrac{1}{2} \it{I} \omega^2}$

➡️Also

» $\sf{K_r = mg\:\dfrac{\it{l}}{2}\:sin\theta}$

$\boxed{\sf{Note\::- \it{I} = \dfrac{ml^2}{3} }}$

→ Now equating and substituting values :-

$\implies \sf{\dfrac{1}{2}\it{I} \omega^2 = mg\:\dfrac{l}{2}\:sin\theta}$

$\implies \sf{\dfrac{1}{2} \dfrac{m\it{l^2}} \omega^2 = mg\:\dfrac{l}{2}\:(sin53^{\circ}-sin37^{\circ})}$

$\small{\implies \sf{\cancel{\dfrac{1}{2}} \dfrac{\cancel{m}\:\it{l^2}}{3} \omega^2 = \cancel{m}\:g\:\dfrac{\it{l}}{\cancel{2}}\:(\dfrac{4}{5} - \dfrac{3}{5})}}$

$\implies \sf{\dfrac{\it{l^2}}{3} \omega^2 = g\: \it{l}\: (\dfrac{1}{5})}$

$\implies \sf{\omega^2 = \dfrac{10 \times \it{l} \times \dfrac{3}{5}}{\it{l^2}}}$

$\implies \sf{\omega^2 = \dfrac{2 \times 3}{\it{l}}}$

$\implies \sf{\omega^2 = \dfrac{2\times3}{0.5}}$

$\implies \sf{\omega^2 = 4 \times 3}$

$\implies \sf{\omega = \sqrt{4\times3}}$

$\implies \sf{\omega = 2\sqrt{3}}$