ch04-p095

# ch04-p095 - 95 We write our magnitude-angle results in the...

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(b) At t = 7.5 s, the particle has traveled a fraction of 7.5/20 = 3/8 of a revolution around the circle (starting at the origin). Relative to the circle-center, the particle is therefore at φ = 3/8 (360°) = 135° measured from vertical in the manner discussed above. Referring to Fig. 4-56, we compute that this position corresponds to x = (3.00 m)sin 135° = 2.1 m y = (3.0 m) – (3.0 m)cos 135° = 5.1 m relative to the coordinate origin. In our magnitude-angle notation, this is expressed as ( R θ ) = (5.5 68°). 95. We write our magnitude-angle results in the form R θ bg with SI units for the magnitude understood (m for distances, m/s for speeds, m/s 2 for accelerations). All angles are measured counterclockwise from + x , but we will occasionally refer to angles which are measured counterclockwise from the vertical line between the circle-center and the coordinate origin and the line drawn from the circle-center to the particle location (see r in the figure). We note that the speed of the particle is v = 2 π r / T where r = 3.00 m and T = 20.0 s; thus, v = 0.942 m/s. The particle is moving counterclockwise in Fig. 4-56.

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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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ch04-p095 - 95 We write our magnitude-angle results in the...

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