4.3 Reference Frames
4.3 Reference Frames
MIT 8.01 Classical Mechanics, Fall 2016
View the complete course: http://ocw.mit.edu/8-01F16
Instructor: Prof. Deepto Chakrabarty
License: Creative Commons BY-NC-SA
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Content
2.87 -> we can consider a given coordinate
5.58 -> system as a reference frame within which
7.56 -> we can describe the kinematics of an
9.54 -> object by the kinematics I mean the
11.49 -> position the velocity and the
13.349 -> acceleration as a function of time
14.88 -> basically a geometric description of the
17.099 -> motion some aspects of these kinematics
19.77 -> will look different in different
20.939 -> reference frames and I'd like to examine
22.529 -> that now first I want to define what I
25.59 -> mean by an inertial reference frame an
27.779 -> inertial reference frame is one in which
29.34 -> an isolated body one with no net force
32.129 -> acting on it moves at constant velocity
34.23 -> where that constant velocity might be
36.18 -> zero another way of saying this is that
38.729 -> an inertial reference frame is one in
40.53 -> which Newton's laws of motion apply
42.69 -> recall that Newton's first law of motion
45.15 -> states that an isolated object with no
47.129 -> forces acting on it moves at constant
48.989 -> velocity so let's begin by considering a
52.68 -> bit observer in a particular reference
54.059 -> frame we'll call that reference frame s
55.739 -> and denoted by coordinate axes x and y
59.85 -> and let's consider an object that in
64.35 -> that court in that reference frame is at
66.84 -> a position vector small R we can then
69.96 -> consider a second reference frame which
71.73 -> I'll call the frame s Prime and I'll
73.95 -> denote that with coordinate axes X Prime
76.8 -> and y Prime in my example gear I'm going
79.56 -> to assume that the coordinate axes in
81.72 -> frame s prime are parallel to but
84.99 -> displaced away from the coordinate axes
88.71 -> in frame s more generally we could have
92.01 -> the S prime coordinate axes rotated with
94.71 -> respect to the frame s axes that's a
97.08 -> complication I'm not going to add now
98.46 -> but conceptually it's not really
99.99 -> different for simplicity we'll stick two
103.08 -> parallel axes in this example so imagine
106.65 -> we have two observers one in frame s at
109.11 -> the origin and one of the origin of
110.85 -> frame s Prime both looking at the same
112.89 -> object the observer s will measure a
116.4 -> position vector little R observer s
119.88 -> prime will measure a position vector
122.58 -> little R Prime the two observers have a
126.81 -> relative position vector capital R which
129.869 -> is the position of s Prime relative to
132.81 -> the origin of frame s so what we would
135.54 -> like to see
136.11 -> is how are these different position
138.12 -> vectors related well from the geometry
140.55 -> of the diagram we can see that the
143.88 -> position measured by observer s which is
146.85 -> just little R is equal to the position
153.57 -> of observer s prime relative to observer
156.3 -> s which is capital R plus the position
161.46 -> vector measured by the observer at s
163.83 -> prime which is little R prime I can
168.42 -> rewrite that if I'd like to write the
171.45 -> position measured by the observer s
173.22 -> Prime in terms of what's measured by the
175.68 -> observer at s I could just rearrange
177.75 -> this and write that little R prime is
180.15 -> equal to little R - capital R now let's
188.64 -> add a further complication and assume
190.95 -> that the observer s prime is not just at
195 -> a different location from the observer
196.62 -> of s but is moving at constant velocity
199.17 -> relative to frame s so we'll assume that
202.019 -> frame s prime is moving at constant
205.53 -> velocity with respect to frame s at a
208.32 -> constant velocity vector V ok so V
214.2 -> vector
219.37 -> is a constant and in that case my offset
224.019 -> of observer s prime relative to observer
227.349 -> s which is the vector capital R is a
229.659 -> function of time so capital R is a
234.31 -> function of time and it's given by the
237.28 -> offset at time zero which I will call
240.06 -> capital or not
243.9 -> plus the elapsed motion due to the
248.349 -> constant velocity which is capital V
251.79 -> times time so since capital R is a
256.12 -> function of time that tells me that in
258.639 -> this equation little R Prime the
261.76 -> position vector measured by the observer
264.669 -> in frame s prime is also going to be a
267.52 -> function of time even if capital sorry
271.72 -> even if little R is a constant so notice
274.15 -> what that means if the object is at rest
276.639 -> in frame s the object will appear to be
282.82 -> moving its position vector will be time
284.95 -> dependent in frame s prime because
287.65 -> capital R the location of s prime
290.289 -> relative to s is changing so this
293.83 -> relation tells us how the position
295.96 -> vectors in the two frames are related
298.33 -> what about the velocities well to
302.349 -> compute how the velocities are related
304.389 -> we can just take the time derivative of
306.039 -> the relation of the position vectors so
308.83 -> in this particular case we have that the
311.11 -> time derivative of the s prime position
315.4 -> D little R prime DT is equal to the time
318.82 -> derivative of the position in frame s
320.88 -> which is d little R DT minus the time
326.65 -> derivative of the offset of s prime
329.59 -> relative to s so that's minus D capital
333.699 -> R DT and I can rewrite that in terms of
338.32 -> symbols for the velocity so I have here
341.849 -> the velocity little V prime which is the
344.889 -> velocity measured by an observer in
346.539 -> frame s Prime and that's equal to the
349.93 -> velocity little V which is the velocity
351.37 -> measured by frame s
353.61 -> - d capital r dt which we see is just
360.09 -> capital v vector which is the velocity
363.16 -> of s prime relative to s so this is how
366.79 -> the velocities are related and again
368.5 -> notice that if the object is stationary
371.11 -> in one frame it will still have a will
374.89 -> be so if it's zero in one frame it will
378.37 -> be nonzero in the other frame so in
380.56 -> general you will measure different
382.54 -> velocities in the different frames even
384.58 -> if one of those velocities is zero now
388.42 -> how are the accelerations related well
390.97 -> again we can just take the time
392.56 -> derivative of the velocities to figure
395.53 -> out what the relationship of the
397.12 -> accelerations is so differentiating this
399.43 -> equation I have that D little V prime DT
404.56 -> is equal to D little V DT minus D
411.42 -> capital V DT but here something
415.9 -> interesting happens because remember we
417.46 -> said that capital V is a constant vector
420.34 -> and remember capital V is the velocity
423.49 -> that frame s prime has relative to frame
427 -> s okay so since capital V is a constant
430.78 -> that means that this term goes to zero
434.47 -> and so we see that the acceleration in
438.25 -> frame s prime is equal to the
441.25 -> acceleration in frame a so if I have two
444.94 -> reference frames one moving at a
447.67 -> constant velocity relative to the other
450.1 -> in general I will measure different
453.61 -> positions and different velocities for
456.31 -> an object as measured by the two frames
458.98 -> however the accelerations measured in
462.19 -> both frames will be identical because
467.2 -> the accelerations are identical we'll
469.81 -> see that Newton's laws will look
471.7 -> identical in the two frames and we can
475.27 -> see that in the following way in frame s
479.85 -> we have that the force is equal to the
482.68 -> mass times the acceleration in frame s
486.76 -> I'm the force f prime is equal to the
490.84 -> mass times a prime but a prime is equal
501.61 -> to a as we calculated here so we see
506.32 -> that the forces in the two frames are
508.92 -> identical even though the positions and
511.78 -> velocities in general will be different
514.69 -> as measured in the two frames for the
516.46 -> same object but the accelerations will
519.729 -> be identical and so the forces will be
521.68 -> identical so if one of these reference
525.07 -> frames is an inertial frame one in which
528.73 -> an isolated body moves at constant
530.89 -> velocity then any other frame moving at
534.55 -> constant velocity with respect to the
536.14 -> first frame will also be an inertial
538.81 -> frame what this means is that you are
542.92 -> always free to transform from one
545.98 -> inertial frame to another and what that
551.62 -> means is that you can always transform
553.78 -> to another frame that is moving at
555.85 -> constant velocity with respect to an
558.22 -> original inertial frame
Source: https://www.youtube.com/watch?v=QCA3zOe2xdA