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Chapter 8: Relative MotionPowerPoint Presentation

Chapter 8: Relative Motion

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Chapter 8: Relative Motion

Chapter 8 Goals:

- To remind ourselves of the meaning of reference frames, and in particular inertial ones
- To discuss the subtleties of Galilean relativity, whereby one can understand kinematics in different reference frames
- To briefly contrast this state of affairs with Einstein’s special relativity, which is necessary when the relative velocity of one frame referred to the other, and/or the object’s velocity in a frame, approaches the speed of light c = 2.99 x 108 m/s

Book’s notation for kinematical quantities

- an inertial frame of reference K is a cartesian system (x,y,z) in which motion occurs and in which N1 applies
- One can think of an infinite 3d lattice of meter sticks along each axis, with a clock at each vertex, and a little gnome sitting there taking notes

- inertial frame of reference K’ is moving at constant velocity with respect to K, with its axes parallel to those of K (also N1)
- let the origin O’ be at position R(t) as measured from O, in K
- Thus dR/dt = v = constant

- a point P is moving; its position in K is r(t) and its position in K’ is r’(t)
- r(t) = r’(t) + R(t) is the fundamental relationship
- taking a time derivative, we get u(t) = u’(t) + v
- in English: the velocity of a point with respect to frame K is the velocity of the point with respect to frame K’ plus the velocity of K’ with respect to K
- in practice, there will be a ‘fixed’ frame and a ‘moving’ frame and a moving point
- taking one more time derivative yields a(t) = a’(t)
- therefore, N2 is the same in either inertial frame!!
- YAY!!

An alternative way of writing the same stuff

- u = uP/K u’ = uP/K’ v = vK’/K
- P/K P as measured in K
- note how the algebra works: P/K = (P/K’)(K’/K)
- addition of vectors multiply subscripts
- thus, uP/K’ = uP/K –vK’/K or P/K’ = (P/K)/(K’/K)
- subtraction of vectors divide subscripts
- the challenge is to identify the frames and the object

Simple 1d example: An empty water bottle rolls toward the back of a bus at 2.5 m/s. The bus is traveling at 55 mph ≈ 25 m/s. What is the velocity of the bottle as seen from the ground?

Ground is frame K; bus is frame K’; bottle is object P

uP/K = uP/K’ +v K’/K uP/K = – 2.5 + 25 = 22.5 m/s

Trickier 2d example: An empty water bottle rolls sideways across the back of a bus at 5 m/s. The bus is traveling at 55 mph ≈ 25 m/s. What is the velocity of the bottle as seen from the ground? What is the speed of the bottle as seen from the ground? In what direction is the bottle moving as seen from the ground?

Ground is frame K; bus is frame K’; bottle is object P

Take x direction to be along bus’s motion as seen from ground

Take y direction to be perpendicular to that: across the road

uP/K = uP/K’ +v K’/K uP/K = 5 m/s j+ 25 m/s i

Speed is |uP/K | = √(52 + 252) = 25.5 m/s

Direction is Arctan(5/25) = 11.3° away from the x direction

y

y

x

x

uP/K’

uP/K

As seen from K’

As seen from K

Trickier 2d example: You are canoeing across a river that is 75 m wide, and in which the current moves at 2 m/s. You can paddle at 2.5 m/s. How should you paddle to go directly across the river, and how much time will it take [hint: if you aim straight across, you will reach the other side in time t = 75 m/2.5 m/s = 30 s but will have drifted downstream a distance d = 2 m/s(30 s) = 60 m!!!]

- Ground is frame K; river is frame K’; canoe is object P
- Take x direction to be along river’s motion as seen from ground
- Take y direction to be perpendicular to that: across the river
- uP/K = uP/K’ +v K’/K uP/K = uP/K j = uP/K’ + 2 m/s i
- uP/K’ = uP/K j – 2 m/s i
- But we know | uP/K’ | = 2.5 m/s so we can solve:

uP/K’ = 1.5 m/sj – 2 m/s i and uP/K = 1.5 m/sj

Speed is 1.5 m/s with respect to the shore

Canoe is aimed at Arctan(1.5/2) = 36.9° away from UPSTREAM!

It will require a time t = 75 m/1.5 m/s = 50 s to get across, but at least you will be where you intended to be!!

y

y

x

x

uP/K

uP/K’

As seen from K

As seen from K’

Extremely tricky 2d example: High-altitude winds are blowing at 450 km/hr at an angle of 35° S of E. A plane with airspeed 1100 km/hr desires to fly from New York to Los Angeles, a distance of 5000 km at a bearing of 25°S of W [some numbers different from class]. How should the pilot aim the plane? What will the plane’s speed be with respect to the ground?

Ground is frame K; air is frame K’; plane is object P

Take x direction to be E and y direction to be N

Solve for t and substitute back in… quadratic equation!!

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