Physics Solution Manual for 1100 and 2101

37 102 ms 2 r 638 106 m 2 b at 300 latitude a person

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Unformatted text preview: (1), we find that the ratio of the centripetal acceleration of the tip of the second hand to that of the minute hand is ac, second ac, minute = 4π 2 r 2 Tsecond 4π 2 r 2 Tminute = 2 Tminute 2 Tsecond ( 3600 s )2 = ( 60 s )2 = 3600 _____________________________________________________________________________________________ 8. 2 REASONING The centripetal acceleration is given by Equation 5.2 as ac = v /r. The value of the radius r is given, so to determine ac we need information about the speed v. But the speed is related to the period T by v = (2π r)/T, according to Equation 5.1. We can substitute this expression for the speed into Equation 5.2 and see that 248 DYNAMICS OF UNIFORM CIRCULAR MOTION b 2π r / T v2 ac = = r r g = 4π 2 T 2 r 2 SOLUTION To use the expression obtained in the reasoning, we need a value for the period T. The period is the time for one revolution. Since the container is turning at 2.0 revolutions per second, the period is T = (1 s)/(2.0 revolutions) = 0.50 s. Thus, we find that the centripetal acceleration is bg bg 2 4 π 2 r 4 π 0.12 m = = 19 m / s 2 ac = 2 2 T 0.50 s _____________________________________________________________________________________________ 9. SSM REASONING AND SOLUTION Since the magnitude of the centripetal 2 acceleration is given by Equation 5.2, a C = v / r , we can solve for r and find that r= v2 ( 98.8 m / s ) 2 = = 332 m a C 3.00(9.80 m / s 2 ) _____________________________________________________________________________________________ 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 = 2 π r / T . Combining these two expressions, we obtain aC = 4π 2 r T2 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 C1 a C2 = 4 π 2 r1 / T 2 4 π r2 / T 2 2 = r1 r2 = 6.7 m = 2.2 3.0 m _____________________________________________________________________________________________ 11. REASONING AND SOLUTION T = 2π r/v. The speed is The sample makes one revolution in time T as given by 2 –2 3 2 v = rac = (5.00 × 10 m)(6.25 × 10 )(9.80 m/s ) so that v = 55.3 m/s Chapter 5 Problems 249 The period is T = 2π (5.00 × 10 –2 m)/(55.3 m/s) = 5.68 × 10 –3 s = 9.47 × 10 –5 min The number of revolutions per minute = 1/T = 10 600 rev/min . _____________________________________________________________________________________________ 12. REASONING AND SOLUTION a. At the equator a person travels in a circle whose radius equals the radius of the earth, 6 r = Re = 6.38 × 10 m, and whose period of rotation is T = 1 day = 86 400 s. We have v = 2πRe/T = 464 m/s The centripetal acceleration is v 2 ( 464 m/s ) ac = = = 3.37 × 10–2 m/s 2 r 6.38 × 106 m 2 b. At 30.0° latitude a person travels in a circle of radius, 6 r = Re cos 30.0° = 5.53 × 10 m Thus, v = 2π r/T = 402 m/s and 2 ac = v /r = 2.92 × 10 –2 m/s 2 _____________________________________________________________________________________________ 13. REASONING The magnitude Fc of the centripetal force that acts on the skater is given by Equation 5.3 as Fc = mv 2 / r , where m and v are the mass and speed of the skater, and r is the distance of the skater from the pivot. Since all of these variables are known, we can find the magnitude of the centripetal force. SOLUTION The magnitude of the centripetal force is mv 2 ( 80.0 kg )( 6.80 m/s ) Fc = = = 606 N r 6.10 m ______________________________________________________________________________ 2 14. REASONING The centripetal acceleration depends on the speed v and the radius r of the curve, according to ac = v2/r (Equation 5.2). The speeds of the cars are the same, and since they are negotiating the same curve, the radius is the same. Therefore, the cars have the same centripetal acceleration. However, the magnitude Fc of the centripetal force depends on the mass m of the car, as well as the speed and the radius of the curve, according to 250 DYNAMICS OF UNIFORM CIRCULAR MOTION 2 Fc = mv /r (Equation 5.3). Since the speed and the radius are the same for each car, the car with the greater mass, which is car B, experiences the greater centripetal acceleration. SOLUTION We find the following values for the magnitudes of the centripetal accelerations and forces: Car A ac = Fc = Car B b 27 m / s v2 = r 120 m mA v 2 r r 6.1 m / s 2 (5.2) 2 120 m b Fc = 2 1100 b kg g27 m / sg = b = 27 m / s v2 ac = = r 120 m mB v 2 g= g= 6700 N (5.3) 2 6.1 m / s 2 1600 b kg g27 m / sg = b = (5.2) 2 120 m 9700 N (5.3) _____________________________________________________________________________________________ 15. SSM REASONING AND SOLUTION The magnitude of the centripetal force on the ball is given by Equation 5.3: FC = mv 2 / r . Solving for v,...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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