11-17-2012, 12:04 PM |
#1 |

Junior Member Join Date: Nov 2012
Posts: 1
| angular momentum I am in college taking a physics class and for some reason i am having a problem with these equations so any help would be appreciated. here is the question and where i am with the solutions. a .005 kg bullet traveling horizontally at 1 X 10^3 m/s strikes an 18 kg door imbedding itself 10cm from the side opposite the hinge of this 1 meter door.the hinge is frictionless. (a) before the bullet hits does it have angular momentum--- this answer i got it does with reference to the hinge point. (b) evaluate the angular momentum--- so i started out with momentum equation mv(bullet) + mv(door) = ((m(bullet) + m(door) v(final) the mvdoor is 0 so i solved for initial momentum of bullet and got 5kgm/s according to our physics book we now multiply both sides by r to get angular momentum (r x P) where r is the perpendicular r. so r is .9 meters so i multiply that by 5kg(m)/s and get 4.5kg(m)/s and this is the correct answer. (c) is the mechanical energy of the bullet door system constant during collision ? answer no it loses some do to internal energy. (d) This question is where i got lost so i dont know if i am over thinking it or not. At what angular speed does the door swing open immediately after the collision. angular speed(w) = velocity(v)/radius(r) so my problem is angular momentum is Iw and in the previous momentum equation by me multiplying both sides by r the book says that makes it angular momentum so i have 4.5 kg(m)/s = Iw divide both sides by I(moment of inertia). So do i have to solve for moment of inertia? do I use the parallel axis theorem on the door and treat it like a bar? this is where i get confused? any help is extremely appreciated? |

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