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
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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