ps7c19 - Ph 111.01 October 17, 2008 PS 7c (19) Problem 1 =...

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Ph 111.01 October 17, 2008 PS 7c (19) θ r s Problem 1 = Walker P. 10-06 . A spot of paint on a bicycle tire moves in a circular path of radius 0.28 m. When the spot has traveled a linear distance of 1.23 m, through what angle has the tire rotated? Give your answer in radians. rad Solution. The distance moved s is related to the angle θ and the radius r by sr = . So the angle (in radians) is () 1.23 m 0.28 m s r == = 4.393 radians Problem 2 = Walker P. 10-16. After fixing a flat tire on a bicycle you give the wheel a spin. ω 0 ω (a) If its initial angular speed was 7.32 rad/s and it rotated 14.5 revolutions before coming to rest, what was its average angular acceleration? rad/s 2 (b) For what length of time did the wheel rotate? s Solution. The equations for motion with uniform angular acceleration α are very similar to those for uniform linear acceleration a . Here they are: 0 1 00 2 1 0 2 22 0 1 2 3 42 t ωω α θθ ω 2 t tt θω ωα =+ + =++ Δ (a) We have the initial and final angular velocities and the angular distance; so we use the notorious "equation four:" 0 2 ωαθ =
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This note was uploaded on 05/13/2010 for the course PHYSICS 111 taught by Professor Staff during the Fall '08 term at S.F. State.

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ps7c19 - Ph 111.01 October 17, 2008 PS 7c (19) Problem 1 =...

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