Physics Solution Manual for 1100 and 2101

80 ms2 450 m 210 ms chapter 5 problems 271 47

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Unformatted text preview: g (Equation 4.5), so we can divide the expression for the centripetal force by the expression for the weight and obtain that Fc = FN − W W = mv 2 mg r FN or W −1 = v2 gr Solving for the ratio FN /W, we find that ( 230 m/s ) v2 = 1+ = 1+ = 8.8 W gr 9.80 m/s 2 ( 690 m ) FN 2 ( ) 45. REASONING The magnitude Fc of the centripetal force is Fc = mv 2 / r (Equation 5.3). Since the speed v, the mass m, and the radius r are fixed, the magnitude of the centripetal force is the same at each point on the circle. When the ball is at the three o’clock position, the force of gravity, acting downward, is perpendicular to the stick and cannot contribute to the centripetal force. (See Figure 5.21 in the text, point 2, for a similar situation.) At this point, only the tension of T = 16 N contributes to the centripetal force. Considering that the centripetal force is the same everywhere, we can conclude that it has a magnitude of 16 N everywhere. At the twelve o’clock position the tension T and the force of gravity mg both act downward (the negative direction) toward the center of the circle, with the result that the centripetal force at this point is –T – mg. (See Figure 5.21, point 3.) At the six o’clock position the tension points upward toward the center of the circle, while the force of gravity points downward, with the result that the centripetal force at this point is T – mg. (See Figure 5.21, point 1.) SOLUTION Assuming that upward is the positive direction, we find at the twelve and six o’clock positions that 270 DYNAMICS OF UNIFORM CIRCULAR MOTION − T − mg = −16 N 1 24 43 Twelve o' clock Centripetal force b g c h g c h T = 16 N − 0.20 kg 9 .80 m / s 2 = 14 N T − mg = 16 N 123 Six o' clock Centripetal force b T = 16 N + 0.20 kg 9 .80 m / s 2 = 18 N 46. REASONING Because the crest of the hill is a circular arc, the motorcycle’s speed v is related to the centripetal force Fc acting FN on the motorcycle: Fc = mv 2 r (Equation 5.3), where m is the mass of the motorcycle and r is the radius of the circular crest. Solving Equation 5.3 for the speed, we obtain v 2 = Fc r m or mg v = Fc r m . The free-body diagram shows that two vertical Free-body diagram of forces act on the motorcycle. One is the weight mg of the the motorcycle motorcycle, which points downward. The other is the normal force FN exerted by the road. The normal force points directly opposite the motorcycle’s weight. Note that the motorcycle’s weight must be greater than the normal force. The reason for this is that the centripetal force is the net force produced by mg and FN and must point toward the center of the circle, which lies below the motorcycle. Only if the magnitude mg of the weight exceeds the magnitude FN of the normal force will the centripetal force point downward. Therefore, we can express the magnitude of the centripetal force as Fc = mg − FN. With this identity, the relation v = Fc r m becomes v= ( mg − FN ) r m (1) SOLUTION When the motorcycle rides over the crest sufficiently fast, it loses contact with the road. At that point, the normal force FN is zero. In that case, Equation (1) yields the motorcycle’s maximum speed: v= ( mg − 0 ) r m = m gr = gr = m (9.80 m/s2 ) ( 45.0 m) = 21.0 m/s Chapter 5 Problems 271 47. REASONING When the stone is whirled in a horizontal circle, the centripetal force is provided by the tension Th in the string and is given by Equation 5.3 as mv 2 Th = { r (1) Centripetal force where m and v are the mass and speed of the stone, and r is the radius of the circle. When the stone is whirled in a vertical circle, the maximum tension occurs when the stone is at the lowest point in its path. The free-body diagram shows the forces that act on the stone in this situation: the tension Tv in the string and the weight mg of the stone. The centripetal force is the net force that points toward the center of the circle. Setting the centripetal force equal to mv 2 / r , as per Equation 5.3, we have mv 2 +Tv − mg = 14 3 24 r Tv Stone mg (2) Centripetal force Here, we have assumed upward to be the positive direction. We are given that the maximum tension in the string in the case of vertical motion is 15.0% larger than that in the case of horizontal motion. We can use this fact, along with Equations 1 and 2, to find the speed of the stone. Solution Since the maximum tension in the string in the case of vertical motion is 15.0% larger than that in the horizontal motion, Tv = (1.000 + 0.150) Th . Substituting the values of Th and Tv from Equations (1) and (2) into this relation gives Tv = (1.000 + 0.150 ) Th mv 2 mv 2 + mg = (1.000 + 0.150 ) r r Solving this equation for the speed v of the stone yields v= gr (9.80 m/s 2 ) (1.10 m) = = 8.48 m/s 0.150 0.150 272 DYNAMICS OF UNIFORM CIRCULAR MOTION Clothes 48. REASONING The drawing at the right shows the two forces that act on a piece of clothing just before it loses contact with the wall of the cylinder. At that instant the centripetal force i...
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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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