Question

1. Find angular speed of door after collision

A 1.5m wide door (mass 15kg) 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 10g and a speed of 400m/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 if the bullet embeds itself in the door. The moment of inertia of a door rotating about its edge is equal to ML where M is the mass of the door and L is its width. Right After Right Before

2. What is the change in kinetic friction?

Answer 1

by angular momentum conservation

Li = Lf

mv*L/2 = ( 1/3*ML² +m(L/2)² ) ω

10*10^-3* 400*1.5/2 = ( 1/3*15*1.5^2+10*10^-3*(1.5/2)^2 ) ω

10.10-9 _400-101+ ( < 15x1.52 + 10 109 (5) Result: 3 = 11.2556 w

from here solve for ω

ω≈0.266533 rad/s answer

2)

change in kinetic energy

KE f - KE i

= 1/2*I*ω² - 1/2*m*v²

=0.5*( 1/3*15*1.5^2+10*10^-3*(1.5/2)^2 )*0.266533^2 - 1/2*10*10^-3*400^2 =

05(ž 15x152 +10-10° (* 0.266533 - 10-10-* 4002 Result: -799.6002010999491246875

change in kinetic energy = -799.6 J answer