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lecture 9

# lecture 9 - Lecture 9 Example of 2D motion Circular motion...

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1 Lecture 9 Example of 2D motion: Circular motion Conditions for circular motion Examples

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2 θ is the angular position. Angular displacement: i f θ - = Note: angles measured CW are negative and angles measured CCW are positive. θ is measured in radians. 2 π radians = 360 ° =1 revolution x y θ i θ f ∆θ Circular Motion
3 x y θ i θ f ∆θ r arclength = s = r r s = ∆θ is a ratio of two lengths, it is a dimensionless ratio!

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4 x y θ r (x,y) y x Consider an object in motion around a circular path of radius r. A point can be described by its location: ( 29 y x r ˆ sin ˆ cos θ r r t + =
5 The velocity of the object is ( 29 y x y x v ˆ cos ˆ sin ˆ cos ˆ sin θ ϖ r r dt d r dt d r t + - = + - = The magnitude of the (tangential) velocity is r v = Assume the radius of the path is fixed. ϖ is the angular velocity of the particle.

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6 The average and instantaneous angular velocities are dt d t t t θ ϖ = = = 0 av lim and ϖ is measured in rads/sec. ϖ really is a vector, but we will ignore this for now. Take ϖ > 0 as motion CCW and ϖ < 0 as motion CW.
7 x y v v v v The direction of the velocity is The velocity is tangent to the path!

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8 The acceleration of the object is ( 29 y x y x a ˆ sin ˆ cos ˆ sin ˆ cos 2 2 θ ϖ r r dt d r dt d r t - - = - - = The magnitude of the (radial) acceleration is v r v r a r = = = 2 2 Assume that ϖ is constant
9 The direction of the acceleration is The acceleration is radial.

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lecture 9 - Lecture 9 Example of 2D motion Circular motion...

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