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Unformatted text preview: Kinematics, Mechanics, and Frames of Reference Part 2 NonInertial frames If we are only interested in “kinematics” and not in the application of Newton’s laws of mo tion, then the use of reference frames accelerating with respect to the Earth is equally useful and straightforward. However, when Newton’s Laws are invoked, there are new features to explore, features which explain why such frames are known as noninertial frames. When we use such frames, we must also use the relationship between accelerations. Thus, if S is accelerating with value A relative to S , we have a = a A Kinematics A classic problem is the police sharpshooter and the falling bottle. For practice, a police sharpshooter, lying on the ground, is aiming at a target, hung from a tree at height h above the ground. The base of the tree is distant D from his gun. He points the muzzle of his gun directly at the target, and fires with a muzzle speed of V = V x ˆ x + V y ˆ y , but just at the moment that the bullet is fired, the target falls from the tree. Show that the bullet does hit the target. 6 ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡ ‡1 ? D h V x y t t From the diagram, we see that the initial direction of the bullet is such that V y /V x = h/D 1 Then, in the frame of the Earth, S , we would write for the bullet x b = V x t ; y b = V y t 1 2 gt 2 and for the target x t = D ; y t = h 1 2 gt 2 The origin of coordinates is at the muzzle of the gun. The bullet will hit the target if, at time t , x b = x t and y b = y t . These two conditions give V x t = D ; V y t 1 2 gt 2 = h 1 2 gt 2 and since the two terms 1 2 gt 2 cancel, we obtain V x t = D ; V y t = h This is exactly the required condition. The solution is perhaps more “elegant” in the frame fixed to the target, S . Funda mentally, however, it is the same solution. In this frame, the target is always at x t = D and y t = h , and the accelerations of both bullet and target are zero. Equally, the velocity of the target is zero. The velocity of the bullet, however, is V x = + V x ; V y = + V y so that its position at time t will be x b = V x t ; y b = V y t The bullet will hit the target if, at time t , x b = x t and y b = y t . Thus: V x t = D ; V y t = h which is exactly the same condition as before.which is exactly the same condition as before....
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 Fall '07
 Harris
 Force, Inertia, 4 g, Angular frequency, Fictitious force, 0  mg

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