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  • A uniform rod with a length '  ' and mass' <img alt="1 \mathrm{~kg}" src="h

A uniform rod with a length ' 1 \mathrm{~m} ' and mass' 1 \mathrm{~kg} is rotating with an angular velocity of 2 \mathrm{rad} / \mathrm{s} in a horizontal plane around an axis passing through its end. Find the distance of the point from its axis of rotation where tension is half of the maximum value.

Option: 1

\frac{1}{\sqrt{2}} m


Option: 2

\frac{1}{2} m


Option: 3

\frac{1}{\sqrt{3}} \mathrm{~m}


Option: 4

\frac{1}{3} m


Answers (1)

best_answer

\begin{aligned} T_x & =\frac{m \omega^2}{2}\left(l^2-x^2\right) \\ T_{\text {max }} & =\frac{m \omega^2}{2} l^2 \\ T_x & =\frac{T_{\text {max }}}{2}=\frac{1}{4} m \omega^2\left(l^2\right) \\ \frac{1}{2} m \omega^2\left(l^2-x^2\right) & =\frac{1}{4} m \omega^2 l^2 \\ l^2-x^2 & =\frac{l^2}{2} \\ & \frac{l^2}{2}=x^2 \Rightarrow \frac{l}{\sqrt{2}}=x \Rightarrow x=\frac{1}{\sqrt{2}} m \end{aligned}

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