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  • A uniform rod of length is given an impulse <img alt="\mathrm{p}" src="https://entrancecorner.oncodec

A uniform rod of length \mathrm{l} is given an impulse \mathrm{p} at a distance \mathrm{x} from its \mathrm{CM} as shown in figure. The position of the IAOR from the centre of mass is -

Option: 1

\frac{l^{2}}{x}


Option: 2

\frac{l^{2}}{12x}


Option: 3

\frac{x^{2}}{12l}


Option: 4

\frac{l^{2}}{3x}


Answers (1)

best_answer

Using conservation of angular momentum about \mathrm{CM} of the rod. Assuming after impulse velocity of \mathrm{CM} becomes \mathrm{}\vartheta. Then

\mathrm{L =I \omega }
\mathrm{m \vartheta x =I_{c m} .\omega }
\mathrm{\omega =\frac{m \vartheta x}{I_{c m}} }
\mathrm{\omega =\frac{m \vartheta x}{\frac{m l^{2}}{12}}=\frac{12 \vartheta x}{l^{2}} }

If \mathrm{x_{0}} is the distance of IAOR

\mathrm{O=\vartheta-\omega x_{0} }
\mathrm{O=\vartheta-\left(\frac{12 \vartheta x}{l^{2}}\right) x_{0} }
\mathrm{\vartheta=\frac{12 \vartheta x}{l^{2}} \cdot x_{0} }
\mathrm{x_{0}=\frac{l^{2}}{12 x}} Ans.

Posted by

Divya Prakash Singh

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