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  • A bullet of mass 10 g and speed 500 m /r s is fired into a door and gets embedded exactly at the centre of the door.

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.)

Answers (1)

best_answer

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} 

Posted by

Devendra Khairwa

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