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# Recit%206%20ch%205 - PHY 2053 Spring 2011 Recitation 6...

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PHY 2053 Spring 2011 Recitation # 6 Chapter # 4 Feb 21 – Feb 25, 2011 10. The large blade of a helicopter is rotating in a horizontal circle. The length of the blade is 6.7 m, measured from its tip to the center of the circle. Find the ratio of the centripetal acceleration at the end of the blade to that which exists at a point located 3.0 m from the center of the circle. 10. REASONING The magnitude of the centripetal acceleration of any point on the helicopter blade is given by Equation 5.2, a C = v 2 / r , where r is the radius of the circle on which that point moves. From Equation 5.1: v r T = 2 π / . Combining these two expressions, we obtain a r T C = 4 2 2 All points on the blade move with the same period T . SOLUTION The ratio of the centripetal acceleration at the end of the blade (point 1) to that which exists at a point located 3.0 m from the center of the circle (point 2) is a a r T r T r r C1 C2 6.7 m 3.0 m 2.2 = = = = 4 4 2 1 2 2 2 2 1 2 / / 15. A 0.015-kg ball is shot from the plunger of a pinball machine. Because of a centripetal force of 0.028 N, the ball follows a circular arc with a radius of 0.25 m. What is the speed of the ball? 15.

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## This note was uploaded on 09/21/2011 for the course PHY 2053 taught by Professor Vellaris during the Spring '08 term at University of Central Florida.

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Recit%206%20ch%205 - PHY 2053 Spring 2011 Recitation 6...

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