Question

Problem 12.79

A 10 g bullet traveling at 380 m/s strikes a 8.0 kg, 1.0-m-wide door at the edge opposite the hinge. The bullet embeds itself in the door, causing the door to swing open.

Part A

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

Answer 1

let

m = 10 g = 0.01 kg

v = 380 m/s

L = 1 m

M = 8 kg

let w is the angular velocity of dorr after impact.

Apply conservation of anguar momentum

initial angular momentum of bullet = final angular mkimentum if door and bullet

m*v*L = I*w (here I is moment of Inertia of door and bullet about hinge)

m*v*L = (M*L^2/3 + m*L^2)*w

==> w = m*v*L/(M*L^2/3 + m*L^2)

= 0.01*380*1/(8*1^2/3 + 0.01*1^2)

= 1.42 rad/s <<<<<<<<<<---------------Answer