Physics Solution Manual for 1100 and 2101

500 s2 0189 n b the centripetal force varies as the

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Unformatted text preview: s provided by the normal force FN and the radial component of the weight. From the drawing, the radial component of the weight is given by FN φ mg θ mg cosφ = mg cos (90 ° – θ ) = mg sinθ Therefore, with inward taken as the positive direction, Equation 5.3 ( Fc = mv 2 / r ) gives FN + mg sinθ = mv 2 r At the instant that a piece of clothing loses contact with the surface of the drum, FN = 0 N, and the above expression becomes mv 2 mg sinθ = r According to Equation 5.1, v = 2π r / T , and with this substitution we obtain g sinθ = ( 2π r / T ) 2 4π 2 r = r T2 This expression can be solved for the period T. Since the period is the required time for one revolution, the number of revolutions per second can be found by calculating 1/T. SOLUTION Solving for the period, we obtain T= 4π 2 r = 2π g sinθ r = 2π g sinθ 0.32 m 9.80 sin c m / s h 70.0° = 1.17 s 2 Therefore, the number of revolutions per second that the cylinder should make is 1 1 = = 0.85 rev / s T 1.17 s Chapter 5 Problems 273 49. SSM REASONING In Example 3, it was shown that the magnitudes of the centripetal acceleration for the two cases are Radius = 33 m a C = 35 m / s 2 Radius = 24 m a C = 48 m / s 2 According to Newton's second law, the centripetal force is FC = ma C (see Equation 5.3). SOLUTION a. Therefore, when the sled undergoes the turn of radius 33 m, FC = ma C = ( 350 kg)(35 m / s 2 ) = 1.2 × 10 4 N b. Similarly, when the radius of the turn is 24 m, FC = ma C = ( 350 kg)(48 m / s 2 ) = 1.7 × 10 4 N 50. REASONING Two pieces of information are provided. One is the fact that the magnitude of the centripetal acceleration ac is 9.80 m/s2. The other is that the space station should not rotate faster than two revolutions per minute. This rate of twice per minute corresponds to thirty seconds per revolution, which is the minimum value for the period T of the motion. With these data in mind, we will base our solution on Equation 5.2, which gives the centripetal acceleration as ac = v 2 / r , and on Equation 5.1, which specifies that the speed v on a circular path of radius r is v = 2π r / T . SOLUTION From Equation 5.2, we have ac = v2 r or r= v2 ac Substituting v = 2π r / T into this result and solving for the radius gives v 2 ( 2π r / T ) r= = ac ac 2 or r= acT 2 4π 2 (9.80 m/s2 ) ( 30.0 s )2 = 223 m = 4π 2 51. REASONING Since the tip of the blade moves on a circular path, it experiences a centripetal acceleration whose magnitude ac is given by Equation 5.2 as, ac = v 2 / r , where v is the speed of blade tip and r is the radius of the circular path. The radius is known, and the speed can be obtained by dividing the distance that the tip travels by the time t of travel. 274 DYNAMICS OF UNIFORM CIRCULAR MOTION Since an angle of 90° corresponds to one fourth of the circumference of a circle, the distance is 1 ( 2π r ) . 4 SOLUTION Since ac = v 2 / r and v = acceleration of the blade tip is 1 4 ( 2π r ) / t = π r / ( 2t ) , the magnitude of the centripetal 2 πr 2 π 2 r π ( 0.45 m ) v 2 2t ac = = = 2= = 6.9 m/s 2 2 r r 4t 4 ( 0.40 s ) 52. REASONING AND SOLUTION a. In terms of the period of the motion, the centripetal force is written as Fc = 4π2mr/T2 = 4π2 (0.0120 kg)(0.100 m)/(0.500 s)2 = 0.189 N b. The centripetal force varies as the square of the speed. Thus, doubling the speed would increase the centripetal force by a factor of 2 2 = 4 . 53. REASONING The astronaut in the chamber is subjected to a centripetal acceleration ac that is given by ac = v 2 / r (Equation 5.2). In this expression v is the speed at which the astronaut in the chamber moves on the circular path of radius r. We can solve this relation for the speed. SOLUTION Using Equation 5.2, we have ac = 54. REASONING v2 r or ( ) v = ac r = 7.5 9.80 m/s 2 (15 m ) = 33 m/s The person feels the centripetal force acting on his back. This force is 2 Fc = mv /r, according to Equation 5.3. This expression can be solved directly to determine the radius r of the chamber. SOLUTION Solving Equation 5.3 for the radius r gives bg b 83 kg 3.2 m / s mv 2 r= = FC 560 N g= 2 1.5 m _____________________________________________________________________________________________ Chapter 5 Problems 275 55. SSM REASONING As the motorcycle passes over the top of the hill, it will experience a centripetal force, the magnitude of which is given by Equation 5.3: FC = mv 2 / r . The centripetal force is provided by the net force on the cycle + driver system. At that instant, the net force on the system is composed of the normal force, which points upward, and the weight, which points downward. Taking the direction toward the center of the circle (downward) as the positive direction, we have FC = mg − FN . This expression can be solved for FN , the normal force. SOLUTION a. The magnitude of the centripetal force is FC = mv 2 (342 kg)(25.0 m / s) 2 = = 1.70 × 10 3 N r 126 m b. The magnitude of the normal force is FN = mg − FC = (342 kg)(9.80 m/s 2 ) − 1.70 × 103 N = 1.66 × 103 N _____________________________________________...
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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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