A 10 g dart traveling at 400 m/s sticks into a 10 kg, 1 m wide door at the edge opposite the hinge causing the door to swing open. What is the angular velocity of the door immediately after impact?

Answer 1

iniitial angular momentum about hinge = m v r = .010 *400 * 1 = 4 kg m^2/s
or
omega = v/r = 400/1
I = m r^2 = .010 (1)^2)
so I omega = .01 * 400 = 4 agains

now final angular momentum is the same
I = .010 *1^2 + (1/3)(10)1^2
= .01 + 10/3
= 3.34 kg m^2
so
I omega = 4 = 3.34 omega
omage = 1.2 radians/sec