Physics Solution Manual for 1100 and 2101

80 ms 2 400 ms 2 w 475 n 594 n r 650 m 19

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Unformatted text preview: we have v= FC r m = (0.028 N)(0.25 m) = 0.68 m / s 0.015 kg _____________________________________________________________________________________________ 16. REASONING At the maximum speed, the maximum centripetal force acts on the tires, and static friction supplies it. The magnitude of the maximum force of static friction is specified by Equation 4.7 as fsMAX = µs FN , where µs is the coefficient of static friction and FN is the magnitude of the normal force. Our strategy, then, is to find the normal force, substitute it into the expression for the maximum frictional force, and then equate the result to the centripetal force, which is Fc = mv 2 / r , according to Equation 5.3. This will lead us to an expression for the maximum speed that we can apply to each car. SOLUTION Since neither car accelerates in the vertical direction, we can conclude that the car’s weight mg is balanced by the normal force, so FN = mg. From Equations 4.7 and 5.3 it follows that mv 2 fsMAX = µs FN = µs mg = Fc = r Thus, we find that µs mg = mv 2 r or v = µs gr Chapter 5 Problems 251 Applying this result to car A and car B gives vA = µs, A gr vB = µs, B gr and In these two equations, the radius r does not have a subscript, since the radius is the same for either car. Dividing the two equations and noting that the terms g and r are eliminated algebraically, we see that µs, B g r µs, B vB = = vA µs, A g r µs, A or vB = vA µs, B µs, A = ( 25 m/s ) 0.85 = 22 m/s 1.1 17. REASONING AND SOLUTION Initially, the stone executes uniform circular motion in a circle of radius r which is equal to the radius of the tire. At the instant that the stone flies out of the tire, the force of static friction just exceeds its maximum value f sMAX = µ s FN (see Equation 4.7). The force of static friction that acts on the stone from one side of the tread channel is, therefore, fsMAX = 0.90 (1.8 N) = 1.6 N and the magnitude of the total frictional force that acts on the stone just before it flies out is 2 × 1.6 N = 3.2 N . If we assume that only static friction supplies the centripetal force, then, Fc = 3.2 N . Solving Equation 5.3 ( Fc = mv 2 / r ) for the radius r, we have ( ) 6.0 × 10−3 kg (13 m/s)2 mv 2 r= = = 0.31 m Fc 3.2 N _____________________________________________________________________________________________ 18. REASONING AND SOLUTION The force P supplied by the man will be largest when the partner is at the lowest point in the swing. The diagram at the right shows the forces acting on the partner in this situation. The centripetal force necessary to keep the partner swinging along the arc of a circle is provided by the resultant of the force supplied by the man and the weight of the partner. From the figure P − mg = mv 2 r Therefore, P= mv 2 + mg r P mg 252 DYNAMICS OF UNIFORM CIRCULAR MOTION Since the weight of the partner, W, is equal to mg, it follows that m = (W/g) and P= (W/g )v 2 [(475 N)/(9.80 m/s 2 )] (4.00 m/s) 2 +W = + (475 N) = 594 N r (6.50 m) 19. REASONING The centripetal force is the name given to the net force pointing toward the center of the circular path. At the lowest point the net force consists of the tension in the arm pointing upward toward the center and the weight pointing downward or away from the 2 center. In either case the centripetal force is given by Equation 5.3 as Fc = mv /r. SOLUTION (a) The centripetal force is bg b 9 .5 kg 2 .8 m / s mv 2 Fc = = r 0.85 m g= 2 88 N (b) Using T to denote the tension in the arm, at the bottom of the circle we have mv 2 Fc = T − mg = r ( 9.5 kg )( 2.8 m/s ) = 181 N mv 2 T = mg + = ( 9.5kg ) 9.80 m/s 2 + r 0.85 m ( ) 2 20. REASONING When the penny is rotating with the disk (and not sliding relative to it), it is the static frictional force that provides the centripetal force required to keep the penny moving on a circular path. The magnitude fsMAX of the maximum static frictional force is given by f sMAX = µs F N (Equation 4.7), where FN is the magnitude of the normal force and µs is the coefficient of static friction. Solving this relation for µs gives µs = fsMAX FN (1) Since the maximum centripetal force that can act on the penny is the maximum static frictional force, we have F c = f sMAX . Since Fc = mv2/r (Equation 5.3), it follows that f sMAX = mv 2 / r . Substituting this expression into Equation (1) yields Chapter 5 Problems µs = fsMAX FN mv 2 =r FN 253 (2) The speed of the penny can be determined from the period T of the motion and the radius r according to v = 2π r/T (Equation 5.1). Furthermore, since the penny does not accelerate in the vertical direction, the upward normal force must be balanced by the downward-pointing weight, so that FN = mg, where g is the acceleration due to gravity. Substituting these two expressions for v and FN into Equation (2) gives 2 2π r m 2 mv 4π 2 r T µs = = = r FN r ( mg ) gT2 (3) SOLUTION Using Equation (3), we find that the coefficient of static friction required to keep the penny rotating on the disk is µs = 4π 2 ( 0.150 m ) 4π 2 r =...
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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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