Solutions 4 - 3.30 IDENTIFY Each blade tip moves in a...

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3.30. IDENTIFY: Each blade tip moves in a circle of radius 3.40 m R = and therefore has radial acceleration 2 rad / av R = . SET UP: 550 rev/min 9.17 rev/s = , corresponding to a period of 1 0.109 s 9.17 rev/s T == . EXECUTE: (a) 2 196 m/s R v T π . (b) 2 42 3 rad 1.13 10 m/s 1.15 10 v ag R == × = × . EVALUATE: 2 rad 2 4 R a T = gives the same results for rad a as in part (b). 3.33. IDENTIFY: Uniform circular motion. SET UP: Since the magnitude of v r is constant. tan 0 d v dt = = v r and the resultant acceleration is equal to the radial component. At each point in the motion the radial component of the acceleration is directed in toward the center of the circular path and its magnitude is given by 2 /. vR EXECUTE: (a) 22 2 rad (7.00 m/s) 3.50 m/s , 14.0 m v a R = upward. (b) The radial acceleration has the same magnitude as in part (a), but now the direction toward the center of the circle is downward. The acceleration at this point in the motion is 2 3.50 m/s , downward. (c) SET UP: The time to make one rotation is the period T , and the speed v is the distance for one revolution divided by T .
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This note was uploaded on 04/03/2008 for the course PHYSICS 140 taught by Professor Evrard during the Fall '07 term at University of Michigan.

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Solutions 4 - 3.30 IDENTIFY Each blade tip moves in a...

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