Question

A 10 g bullet traveling at 440m/s strikes a 14kg , 1.2-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open.

What is the angular velocity of the door just after impact?

Answer 1

angular momentum of bullet just before impact = m*v*r = 0.01*440*1.2 = 5.28 kgm²s^-1

moment of inertia of door about hinge = 1/12ML²+ML²/4 where L is width

so I = 1/12*14*1.2*1.2+14*1.2*1.2/4= 6.72kgm²

moment of inertia of bullet = mr² = 0.01*1.2*1.2= 0.0144kgm²

total moment of inertia = 6.72+0.0144= 6.7344kgm²

angular velocity = angular momentum/moment of inertia = 5.28/6.7344 = 0.784 rad/sec