Question

video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It is unlatched. A police officer fires a bullet with a mass of 10 g and a speed of 400 m/s into the exact center of the door, in a direction perpendicular to the plane of the door. Find the angular speed of the door just after the collision.

Answer 1

`\omega_f = 0.4\ rad/s`

Step-by-step explanation:

given,

width of door dimension = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

`I_(total) =I_(door) + I_(bullet)`

`I_(total) =(1)/(3)MW^2 + m((W)/(2))^2`

a) from conservation of angular momentum

`L_i = L_f`

`mv(W)/(2) = I_(total)\omega_f`

`mv(W)/(2)= ((1)/(3)MW^2 + m((W)/(2))^2)\omega_f`

`\omega_f= (mv(W)/(2))/((1)/(3)MW^2 + m((W)/(2))^2)`

`\omega_f= ((mv)/(2))/((MW )/(3)+((mW)/(4)))`

`\omega_f= ((0.01* 400)/(2))/((15* 1 )/(3)+((0.01* 1)/(4)))`

`\omega_f = 0.4\ rad/s`