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A rod of mass $1kg$ is placed on a frictionless surface as shown in the figure. If its lower end starts sliding at a speed of $2 \mathrm{~m} / \mathrm{s}$ at an instant when its angle with the horizontal is $45^{\circ}$ . Find its total kinetic energy.

[Given, length of the rod is ' 1m']

Option: 1
$\frac{3}{4} Joule$

Option: 2
$\frac{4}{3} Joule$

Option: 3
$\frac{2}{19} Joule$

Option: 4
$\frac{19}{2} Joule$

Answers (1)

best_answer

In $\triangle QRS$ ,

$\tan \theta=\frac{R S}{Q S}, \quad \sin \theta=\frac{R S}{Q R} \\$

$Q S=R S \cot \theta, \quad Q R=R S \operatorname{cosec} \theta \\$

$Q S=\left(\frac{l}{2}\right) \cot \theta, \\$

$I_{Q}=I_{CM}+m(QS)^2$

$I_{Q}=\frac{m l^{2}}{12}+\frac{m l^{2}}{4} \cot ^{2} \theta \\$

$I_{Q}=\frac{m l^{2}}{12}\left[1+3 \cot ^{2} \theta\right] \\$

$K_{R}=\frac{1}{2} I_{Q} \omega^{2}=\frac{1}{2} \times \frac{m l^{2}}{12} \times\left(1+3 \cot ^{2} \theta\right)\left(\frac{V}{Q R}\right)^{2} \\$

$k_{R}=\frac{m l^{2}}{24}\left(1+3 \cot ^{2} \theta\right) \frac{v^{2}}{\left(\frac{l}{2}\right)^{2} \operatorname{cosec}^{2} \theta} \\$

$K_{R}=\frac{m v^{2}}{6} \frac{\left(1+3 \cot ^{2} \theta\right)}{\operatorname{cosec}^{2} \theta} \\$

$k_{R}=\frac{(1)(2)^{2}}{6} \frac{\left(1+3 \cot ^{2} 45\right)}{\operatorname{cosec}^{2} 45^{\circ}}=\frac{\not 2}{3} \times \frac{4}{\not 2} \\$

$K_{R}=\frac{4}{3} \text { Joule }$

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Rishi

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