Q7.24     A bullet of mass $10g$ and speed $500m/s$  is fired into a door and gets embedded               exactly at the centre of the door. The door is $1.0m$  wide and weighs $12kg$ . It is               hinged at one end and rotates about a vertical axis practically without friction. Find               the angular speed of the door just after the bullet embeds into it.            (Hint: The moment of inertia of the door about the vertical axis at one end is $ML^{2}/3$.)

The imparted angular momentum is given by :

$\alpha \ =\ mvr$

Putting all the given values in the above equation we get :

$=\ (10\times 10^{-3})\times500\times \frac{1}{2}$

$=\ 2.5\ Kg\ m^2\ s^{-1}$

Now, the moment of inertia of door is :

$I\ =\ \frac{1}{3}ML^2$

or                                                       $=\ \frac{1}{3}(12)1^2\ =\ 4\ Kg\ m^2$

Also,                                    $\alpha\ =\ I \omega$

or                                         $\omega\ =\ \frac{\alpha}{I}\ =\ \frac{2.5}{4}\ =\ 0.625\ rad\ s^{-1}$

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