9-12-08LectPHY2048

# 9-12-08LectPHY2048 - Change of coordinate reference frames...

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Change of coordinate reference frames (brief video) Random examples Cart demo

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Change of coordinate reference frames To motivate this development we first do a problem you had for homework without changing coordinate frames. Train T is moving at 45 m/s and catching up to locomotive L, 676 m ahead on the same track moving in the same direction at 8 m/s, when the train’s conductor wakes up and begins applying the brakes. T L v TO = 45 m/s v LO = 8 m/s Let t = 0 when T’s conductor begins applying the brakes and define positions in the coordinate frame, at that instant, as shown below: x LO = 676 m x TO = 0 m What deceleration must train T have to just avoid a collision ?
v T = v LO = 8 m/s Later, for an avoided collision, train P must have slowed to 8 m/s when the trains just kiss (some distance x T = x L down the track). x T = x L T L For the locomotive (moving at constant velocity): x L = x LO + v LO t For the train (decelerating) 2 1 TT OT O 2 x= x + v t+ a t Now the time we are interested in is when x T = x L so we equate these expressions, 2 1 LO LO TO 2 x+ vt = + a t 0

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2 1 LO LO TO 2 x+ vt a t = Giving us one equation, but with 2 unknowns: a and t . But we have another condition, which is that when the trains just touch their velocities must be the same (both equal to v LO ) (anything else is not a “just avoided collision”) so we also can also write for the time of the “collision”: LO T TO vv v a t = =+ LO TO t a = Solving this for the time when the velocities are equal We can now use this in the * equation above eliminating the unknown time. Rearranging and collecting terms, ( ) 2 1 LO LO TO 2 xv v t a t + −= *
() 2 LO TO LO TO 1 LO LO TO 2 vv xv v a aa −− ⎤⎡ +− = ⎥⎢ ⎦⎣ ( ) ( ) 22 LO TO LO TO 1 LO 2 x += ( ) 2 LO TO 1 LO 2 x a =− ( ) 2 LO TO 1 2 LO a x 2 1 2 2 mm 84 5 m ss a 1.01 676 m s ⎛⎞ ⎜⎟ ⎝⎠ = Note that the acceleration comes out negative as it should & that this was a fair amount of algebra.

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Coordinate Reference Frame Transformations S x S y Frame S (stationary) M x M y Frame M (moving) MS v G Frame M has constant velocity with respect to frame S.
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## This note was uploaded on 08/25/2011 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.

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9-12-08LectPHY2048 - Change of coordinate reference frames...

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