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

Option: 1
$\mathrm{\frac{l^2}{x}}$

Option: 2
$\mathrm{\frac{l^2}{12 x}}$

Option: 3
$\mathrm{\frac{x^2}{12 l}}$

Option: 4
$\mathrm{\frac{l^2}{3 x}}$

Answers (1)

best_answer

Using conservation of angular momentum about CM of the rod .Assuming after impluse velocity of CM becomes $\mathrm{v}$ .Then

$\mathrm{L =I \omega }$

$\mathrm{m v x =I_{c m} \cdot \omega }$

$\mathrm{ \omega =\frac{m v_x}{I_{c m}} }$

$\mathrm{\omega=\frac{mvx}{\frac{ml^{2}}{12}}=\frac{12vx}{l^{2}}}$

If $\mathrm{ x_0 }$ is the distance of IAOR

$\mathrm{ 0=v-\omega x_0 }$

$\mathrm{ 0=v-\left(\frac{12 v_x}{l^2}\right) x_{0} }$

$\mathrm{ k=\frac{12 v x}{l^2} \cdot x_0 }$

$\mathrm{ x_0=\frac{l^2}{12 x} }$

Hence option 2 is correct.

Posted by

vinayak

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