Physics Solution Manual for 1100 and 2101

From the figure we see that cos target r 1 r d 10 r

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Unformatted text preview: = vT /(2π), the radius is directly proportional to the speed. Thus, the centripetal force is directly proportional to the speed v of the astronaut. As the astronaut walks from the inner ring to the outer ring, her speed doubles and so does her apparent weight. 14. (d) The skier at A is speeding up, so the direction of the acceleration, and hence the net force, must be parallel to the skier’s velocity. At B the skier is momentarily traveling at a constant speed on a circular path of radius r. The direction of the net force, called the centripetal force, must be toward the center of the path. At C the skier is in free-fall, so the net force, which is the gravitational force, is straight downward. 15. (b) According to Newton’s second law, the net force, FN − mg , must equal the mass m 2 times the centripetal acceleration v /r. Chapter 5 Problems 245 CHAPTER 5 DYNAMICS OF UNIFORM CIRCULAR MOTION PROBLEMS 1. SSM REASONING The speed of the plane is given by Equation 5.1: v = 2 π r / T , where T is the period or the time required for the plane to complete one revolution. SOLUTION Solving Equation 5.1 for T we have T= 2. 2 π r 2 π ( 2850 m ) = = 160 s v 110 m / s 2 REASONING According to ac = v /r (Equation 5.2), the magnitude ac of the centripetal acceleration depends on the speed v of the object and the radius r of its circular path. In Example 2 the object is moving on a path whose radius is infinitely large; in other words, the object is moving along a straight line. SOLUTION Using Equation 5.2, we find the following values for the magnitude of the centripetal acceleration: v 2 (12 m/s ) = = 290 m/s2 r 0.50 m 2 Example 1 ac = Example 2 v 2 ( 35 m/s ) ac = = = 0 m/s2 r ∞ Example 3 v 2 ( 2.3 m/s ) ac = = = 2.9 m/s2 r 1.8 m 2 2 3. REASONING AND SOLUTION Since the speed of the object on and off the circle remains constant at the same value, the object always travels the same distance in equal time intervals, both on and off the circle. Furthermore since the object travels the distance OA in the same time it would have moved from O to P on the circle, we know that the distance OA is equal to the distance along the arc of the circle from O to P. 246 DYNAMICS OF UNIFORM CIRCULAR MOTION The circumference of the circle is 2 π r = 2 π (3.6 m) = 22.6 m . The arc OP subtends an angle of θ = 25° ; therefore, since any circle contains 360°, the arc OP is 25/360 or 6.9 per cent of the circumference of the circle. Thus, b gg b OP = 22.6 m 0.069 = 1.6 m and, from the argument given above, we conclude that the distance OA is 1.6 m . 4. REASONING AND SOLUTION Let s represent the length of the path of the pebble after it is released. From Conceptual Example 2, we know that the pebble will fly off tangentially. Therefore, the path s is perperpendicular to the radius r of the circle. Thus, the distances r, s, and d form a right triangle with hypotenuse d as shown in the figure at the right. From the figure we see that cos α = Target r 1 r = = d 10 r 10 or α = cos –1 s Pebble d α r θ 35° C F1 I = 84° GJ HK 10 Furthermore, from the figure, we see that α + θ + 35° = 180° . Therefore, θ = 145°− α = 145°−84 ° = 61° _____________________________________________________________________________________________ 5. SSM REASONING The magnitude ac of the car’s centripetal acceleration is given by Equation 5.2 as ac = v2 / r , where v is the speed of the car and r is the radius of the track. 3 The radius is r = 2.6 × 10 m. The speed can be obtained from Equation 5.1 as the circumference (2π r) of the track divided by the period T of the motion. The period is the time for the car to go once around the track (T = 360 s). SOLUTION Since ac = v 2 / r and v = ( 2π r ) / T , the magnitude of the car’s centripetal acceleration is 2π r 2 3 2 2 v T = 4π r = 4π ( 2.6 ×10 m ) = 0.79 m/s 2 ac = = 2 r r T2 ( 360 s ) 2 ______________________________________________________________________________ Chapter 5 Problems 6. 247 REASONING Blood traveling through the aortic arch follows a circular path with a diameter of 5.0 cm and, therefore, a radius of r = 2.5 cm = 0.025 m. We know the speed v of v2 the blood flow, so the relation ac = (Equation 5.2) will give the magnitude of the r blood’s centripetal acceleration. SOLUTION With a blood flow speed of v = 0.32 m/s, the magnitude of the centripetal acceleration in the aortic arch is v 2 ( 0.32 m/s ) = = 4.1 m/s 2 r 0.025 m 2 ac = 7. REASONING The relationship between the magnitude ac of the centripetal acceleration and the period T of the tip of a moving clock hand can be obtained by using Equations 5.2 and 5.1: v2 2π r ac = (5.2) v= (5.1) r T The period is the time it takes a clock hand to go once around the circle. In these expressions, v is the speed of the tip of the hand and r is the length of the hand. Substituting Equation 5.1 into Equation 5.2 yields 2π r 2 v 4π 2 r T ac = = =2 r r T 2 (1) SOLUTION The period of the second hand is Tsecond = 60 s. The period of the minute hand is Tminute = 1 h = 3600 s. Using Equation...
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