Asked by Yoyo

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?

Answers

Answered by Damon
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
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