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The velocity of two ends of rod of length $\mathrm{l=1 \mathrm{~m}}$ are given as $10 \mathrm{~cm} / \mathrm{s}$ and $\mathrm{v}$ , as shown in figure. The angular velocity of the rod is

Option: 1
$0.40 \mathrm{rad} / \mathrm{sec}$

Option: 2
$0.20\: \mathrm{rad/sec}$

Option: 3
$0.15 \mathrm{rad} / \mathrm{sec}$

Option: 4
$0.25 \mathrm{rad} / \mathrm{sec}$

Answers (1)

best_answer

Rod on being a rigid body,

$\mathrm{ v \cos \theta=10 \cos 30^{\circ} }$

$\mathrm{ v \cos 60^{\circ}=10 \cos 30^{\circ} }$

$\mathrm{ v \times \frac{1}{2}=10 \times \frac{\sqrt{3}}{2} }$

$\mathrm{ v=10 \sqrt{3} \mathrm{~cm} / \mathrm{s}}$
The angular velocity of the rod about $\mathrm{ ' A '.}$

$\mathrm{ \omega=\frac{v_{A B}}{r} \quad(\because v=r \omega) }$

$\mathrm{ \omega=\frac{(10 \sqrt{3}) \sin 60^{\circ}+10 \sin 30^{\circ}}{100} }$

$\mathrm{ \omega=\frac{10 \sqrt{3} \times \frac{\sqrt{3}}{2}+10 \times \frac{1}{2}}{100} }$

$\mathrm{ \omega=\frac{5 \times 3+5}{100}=\frac{20}{100} }$

$\mathrm{ \omega=0.20\: \mathrm{rad} / \mathrm{sec}}$

Hence option 2 is correct.

Posted by

Suraj Bhandari

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