Sm chapter6

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Sm chapter6

  1. 1. 6 Circular Motion and Other Applications of Newton’s Laws CHAPTER OUTLINE 6.1 Newton’s Second Law Applied to Uniform Circular Motion 6.2 Nonuniform Circular Motion 6.3 Motion in Accelerated Frames 6.4 Motion in the Presence of Resistive Forces ANSWERS TO QUESTIONS *Q6.1 (i) nonzero. Its direction of motion is changing. (ii) zero. Its speed is not changing. (iii) zero: when v = 0, v2 րr = 0. (iv) nonzero: its velocity is changing from, say 0.1 mրs north to 0.1 mրs south. Q6.2 (a) The object will move in a circle at a constant speed. (b) The object will move in a straight line at a changing speed. Q6.3 The speed changes. The tangential force component causes tangential acceleration. *Q6.4 (a) A > C = D > B = E. At constant speed, centripetal acceleration is largest when radius is smallest. A straight path has infinite radius of curvature. (b) Velocity is north at A, west at B, and south at C. (c) Acceleration is west at A, nonexistent at B, and east at C, to be radially inward. *Q6.5 (a) yes, point C. Total acceleration here is centripetal acceleration, straight up. (b) yes, point A. Total acceleration here is tangential acceleration, to the right and downward perpendicular to the cord. (c) No. (d) yes, point B. Total acceleration here is to the right and upward. Q6.6 I would not accept that statement for two reasons. First, to be “beyond the pull of gravity,” one would have to be infinitely far away from all other matter. Second, astronauts in orbit are moving in a circular path. It is the gravitational pull of Earth on the astronauts that keeps them in orbit. In the space shuttle, just above the atmosphere, gravity is only slightly weaker than at the Earth’s surface. Gravity does its job most clearly on an orbiting spacecraft, because the craft feels no other forces and is in free fall. Q6.7 This is the same principle as the centrifuge. All the material inside the cylinder tends to move along a straight-line path, but the walls of the cylinder exert an inward force to keep everything moving around in a circular path. Q6.8 The water has inertia. The water tends to move along a straight line, but the bucket pulls it in and around in a circle. Q6.9 Blood pressure cannot supply the force necessary both to balance the gravitational force and to provide the centripetal acceleration, to keep blood flowing up to the pilot’s brain. *Q6.10 (a) The keys shift backward relative to the student’s hand. The cord then pulls the keys upward and forward, to make them gain speed horizontally forward along with the airplane. (b) The angle stays constant while the plane has constant acceleration. This experiment is described in the book Science from Your Airplane Window by Elizabeth Wood. 127 13794_06_ch06_p127-152.indd 12713794_06_ch06_p127-152.indd 127 12/2/06 1:58:33 PM12/2/06 1:58:33 PM
  2. 2. 128 Chapter 6 Q6.11 The person in the elevator is in an accelerating reference frame. The apparent acceleration due to gravity, “g,” is changed inside the elevator. “g” = ±g a Q6.12 From the proportionality of the drag force to the speed squared and from Newton’s second law, we derive the equation that describes the motion of the skydiver: m d dt mg D Ay y v v= − ρ 2 2 where D is the coefficient of drag of the parachutist, and A is the projected area of the parachu- tist’s body. At terminal speed, a d dt y y = = v 0 and vT mg D A = ⎛ ⎝⎜ ⎞ ⎠⎟ 2 1 2 ρ When the parachute opens, the coefficient of drag D and the effective area A both increase, thus reducing the speed of the skydiver. Modern parachutes also add a third term, lift, to change the equation to m d dt mg D A L Ay y x v v v= − − ρ ρ 2 2 2 2 where vy is the vertical velocity, and vx is the horizontal velocity. The effect of lift is clearly seen in the “paraplane,” an ultralight airplane made from a fan, a chair, and a parachute. Q6.13 (a) Static friction exerted by the roadway where it meets the rubber tires accelerates the car forward and then maintains its speed by counterbalancing resistance forces. (b) The air around the propeller pushes forward on its blades. Evidence is that the propeller blade pushes the air toward the back of the plane. (c) The water pushes the blade of the oar toward the bow. Evi- dence is that the blade of the oar pushes the water toward the stern. Q6.14 The larger drop has higher terminal speed. In the case of spheres, the text demonstrates that terminal speed is proportional to the square root of radius. When moving with terminal speed, an object is in equilibrium and has zero acceleration. *Q6.15 (a) Speed increases, before she reaches terminal speed. (b) The magnitude of acceleration decreases, as the air resistance force increases to counterbalance more and more of the gravita- tional force. Q6.16 The thesis is false. The moment of decay of a radioactive atomic nucleus (for example) cannot be predicted. Quantum mechanics implies that the future is indeterminate. On the other hand, our sense of free will, of being able to make choices for ourselves that can appear to be random, may be an illusion. It may have nothing to do with the subatomic randomness described by quantum mechanics. 13794_06_ch06_p127-152.indd 12813794_06_ch06_p127-152.indd 128 12/2/06 1:58:34 PM12/2/06 1:58:34 PM
  3. 3. Circular Motion and Other Applications of Newton’s Laws 129 SOLUTIONS TO PROBLEMS Section 6.1 Newton’s Second Law Applied to Uniform Circular Motion P6.1 m = 3 00. kg, r = 0 800. m. The string will break if the tension exceeds the weight corresponding to 25.0 kg, so T Mgmax . .= = ( ) =25 0 9 80 245 N When the 3.00 kg mass rotates in a horizontal circle, the tension causes the centripetal acceleration, so T m r = = ( )v v2 2 3 00 0 800 . . Then v2 0 800 3 00 0 800 3 00 0 800 24 = = ( ) ≤ ( ) = rT m T T. . . . .max 55 3 00 65 3 ( ) = . . m s2 2 and 0 65 3≤ ≤v . or 0 8 08≤ ≤v . m s P6.2 In F m r ∑ = v2 , both m and r are unknown but remain constant. Therefore, F∑ is proportional to v2 and increases by a factor of 18 0 14 0 2 . . ⎛ ⎝ ⎞ ⎠ as v increases from 14.0 mրs to 18.0 mրs. The total force at the higher speed is then Ffast N N∑ = ⎛ ⎝ ⎞ ⎠ ( ) = 18 0 14 0 130 215 2 . . Symbolically, write F m r slow m s∑ = ⎛ ⎝ ⎞ ⎠ ( )14 0 2 . and F m r fast m s∑ = ⎛ ⎝ ⎞ ⎠ ( )18 0 2 . . Dividing gives F F fast slow ∑ ∑ = ⎛ ⎝ ⎞ ⎠ 18 0 14 0 2 . . , or F Ffast slow∑ ∑= ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ 18 0 14 0 18 0 14 0 130 2 2 . . . . N N( ) = 215 This force must be horizontally inward to produce the driver’s centripetal acceleration. P6.3 (a) F m r = = ×( ) ×( ) × − v2 31 6 2 9 11 10 2 20 10 0 530 1 . . . kg m s 00 8 32 1010 8 − − = × m N inward. (b) a r = = ×( ) × = ×− v2 6 2 10 22 2 20 10 0 530 10 9 13 10 . . . m s m m s inward2 P6.4 (a) F may y∑ = , mg m r moon down down= v2 v = = ( ) × + ×( ) =g rmoon 2 m s m m1 52 1 7 10 100 10 16 3 . . .665 103 × m s (b) v = 2πr T , T = ×( ) × = × = 2 1 8 10 1 65 10 6 84 10 1 90 6 3 3 π . . . . m m s s hh FIG. P6.1 13794_06_ch06_p127-152.indd 12913794_06_ch06_p127-152.indd 129 12/2/06 1:58:34 PM12/2/06 1:58:34 PM
  4. 4. 130 Chapter 6 FIG. P6.9 P6.5 (a) static friction (b) ma f n mgˆ ˆ ˆ ˆi i j j= + + −( ) F n mgy∑ = = −0 thus n mg= and F m r f n mgr∑ = = = = v2 µ µ . Then µ = = ( ) ( )( ) = v2 2 50 0 30 0 980 0 085 rg . . . cm s cm cm s2 00 . P6.6 Neglecting relativistic effects. F ma m r c= = v2 F = × ×( ) ×( ) ( − 2 1 661 10 2 998 10 0 480 27 7 2 . . . kg m s m)) = × − 6 22 10 12 . N P6.7 Standing on the inner surface of the rim, and moving with it, each person will feel a normal force exerted by the rim. This inward force causes the 3 00. m s2 centripetal acceleration: a rc = v2 / so v = = ( )( ) =a rc 3 00 60 0 13 4. . .m s m m s2 The period of rotation comes from v = 2πr T : T r = = ( ) = 2 2 60 0 13 4 28 1 π π v . . . m m s s so the frequency of rotation is f T = = = ⎛ ⎝ ⎞ ⎠ = 1 1 28 1 1 28 1 60 2 14 . . . s s s 1 min rev miin . P6.8 T mgcos . . .5 00 80 0 9 80° = = ( )( )kg m s2 (a) T = 787 N: T i j= ( ) + ( )68 6 784. ˆ ˆN N (b) T macsin .5 00° = : ac = 0 857. m s2 toward the center of the circle. The length of the wire is unnecessary information. We could, on the other hand, use it to find the radius of the circle, the speed of the bob, and the period of the motion. P6.9 n mg= since ay = 0 The force causing the centripetal acceleration is the frictional force f. From Newton’s second law f ma m r c= = v2 . But the friction condition is f ns≤ µ i.e., m r mgs v2 ≤ µ v ≤ = ( )( )µsrg 0 600 35 0 9 80. . .m m s2 v ≤ 14 3. m s FIG. P6.8 13794_06_ch06_p127-152.indd 13013794_06_ch06_p127-152.indd 130 12/2/06 1:58:35 PM12/2/06 1:58:35 PM
  5. 5. Circular Motion and Other Applications of Newton’s Laws 131 P6.10 (a) v = = 235 6 53 m 36.0 s m s. The radius is given by 1 4 2 235πr = m so r = 150 m (b) ar r = ⎛ ⎝⎜ ⎞ ⎠⎟ = ( ) v2 2 6 53 150 toward center m s m . at 35.0 north of west m s2 ° °= ( )0 285 35 0. cos . −−( )+( ) = − + ˆ sin . ˆ . ˆ . ˆ i j i 35 0 0 233 0 163 ° m s m s2 2 jj (c) a v v j i a g f i t v = −( ) = −( )6 53 6 53 36 0 . ˆ . ˆ . m s m s ss m s m s2 2 = − +0 181 0 181. ˆ . ˆi j P6.11 F mgg = = ( )( )=4 9 8 39 2kg m s N2 . . sin . . θ θ = = 1 5 48 6 m 2 m ° r = ( ) =2 48 6 1 32m mcos . .° F ma m r T T x x a b ∑ = = + = ( ) v2 48 6 48 6 4 6 cos . cos .° ° kg m ss m N N ( ) + = = 2 1 32 109 48 6 165 . cos . T Ta b ° F ma T T T T y y a b a b ∑ = + − − = − sin . sin . .48 6 48 6 39 2 0° ° N == = 39 2 48 6 52 3 . sin . . N N ° (a) To solve simultaneously, we add the equations in Ta and Tb: T T T Ta b a b+ + − = +165 52 3N N. Ta = = 217 108 N 2 N (b) T Tb a= − = − =165 165 108 56 2N N N N. FIG. P6.11 θ 39.2 N Ta Tb forces motion v ac 13794_06_ch06_p127-152.indd 13113794_06_ch06_p127-152.indd 131 12/2/06 1:58:36 PM12/2/06 1:58:36 PM
  6. 6. 132 Chapter 6 Section 6.2 Nonuniform Circular Motion P6.12 (a) a r c = = ( ) = v2 2 4 00 1 33 . . m s 12.0 m m s2 (b) a a ac t= +2 2 a = ( ) + ( ) =1 33 1 20 1 79 2 2 . . . m s2 at an angleθ = ⎛ ⎝⎜ ⎞ ⎠⎟ =− tan .1 48 0 a a c t ° inward P6.13 M = 40 0. kg, R = 3 00. m, T = 350 N (a) F T Mg M R ∑ = − =2 2 v v2 2= −( )⎛ ⎝ ⎞ ⎠ T Mg R M v2 700 40 0 9 80 3 00 40 0 23 1= − ( )( )[ ]⎛ ⎝ ⎞ ⎠ =. . . . . m s2 2 (( ) v = 4 81. m s (b) n Mg F M R − = = v2 n Mg M R = + = +⎛ ⎝ ⎞ ⎠ = v2 40 0 9 80 23 1 3 00 700. . . . N P6.14 (a) v = 20 0. m s, n = force of track on roller coaster, and R = 10 0. m. F M R n Mg∑ = = − v2 From this we find n Mg M R = + = ( )( )+ ( )v2 500 9 80 500 20 0 kg m s kg m2 . . ss m N N N 2 ( ) = + = × 10 0 4 900 20 000 2 49 104 . .n (b) At B, n Mg M R − = − v2 The maximum speed at B corresponds to n Mg M R Rg = − = − ⇒ = = ( ) = 0 15 0 9 80 12 1 2 v vmax max . . . m s FIG. P6.12 FIG. P6.13(a) T T Mg child + seat FIG. P6.13(b) Mg child alone n FIG. P6.14 10 m 15 m B A C 13794_06_ch06_p127-152.indd 13213794_06_ch06_p127-152.indd 132 12/2/06 1:58:37 PM12/2/06 1:58:37 PM
  7. 7. Circular Motion and Other Applications of Newton’s Laws 133 continued on next page FIG. P6.15 P6.15 Let the tension at the lowest point be T. F ma T mg ma m r T m g r T c∑ = − = = = + ⎛ ⎝⎜ ⎞ ⎠⎟ = ( : . v v 2 2 85 0 kg)) + ( )⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = >9 80 8 00 10 0 1 38 1 2 . . . .m s m s m kN2 0000 N He doesn’t make it across the river because the vine breaks. *P6.16 (a) Consider radial forces on the object, taking inward as positive. ΣF ma T r r r= − = = : ( . )( . )cos .0 5 9 8 20 0 5kg °m s2 2 mv kkg m N N N ( ) . . . 8 2 4 60 16 0 20 6 2 m s T = + = (b) We already found the radial component of acceleration, ( ) . .8 2 32 0m s m s2 2 m inward= Consider tangential forces on the object. ΣF ma a a t t t t = = = : ( . )( . )sin .0 5 9 8 20 0 5kg m s kg2 33 35. tanm s2 downward gent to the circle (c) a = +[ . ]32 3 352 2 1 2 2 m s inward and below the cord at angle tan ( . )−1 3 35 32 = 32 2. m s2 inward and below the cord at 5.98° (d) No change. If the object is swinging down it is gaining speed. If the object is swinging up it is losing speed but its acceleration is the same size and its direction can be described in the same terms. P6.17 F m r mg ny∑ = = + v2 But n = 0 at this minimum speed condition, so m r mg gr v v 2 9 80 1 00 3 13= ⇒ = = ( )( ) =. . .m s m m s2 P6.18 (a) a r c = v2 r ac = = ( ) ( ) = v2 2 13 0 2 9 80 8 62 . . . m s m s m2 (b) Let n be the force exerted by the rail. Newton’s second law gives Mg n M r + = v2 n M r g M g g Mg= − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −( )= v2 2 , downward FIG. P6.17 FIG. P6.18 13794_06_ch06_p127-152.indd 13313794_06_ch06_p127-152.indd 133 12/2/06 1:58:37 PM12/2/06 1:58:37 PM
  8. 8. 134 Chapter 6 (c) a r c = v2 ac = ( ) = 13 0 20 0 8 45 2 . . . m s m m s2 If the force exerted by the rail is n1 then n Mg M r Mac1 2 + = = v n M a gc1 = −( ) which is < 0 since ac = 8 45. m s2 Thus, the normal force would have to point away from the center of the curve. Unless they have belts, the riders will fall from the cars. To be safe we must require n1 to be positive. Then a gc > . We need v2 r g> or v > = ( )( )rg 20 0 9 80. .m m s2 , v > 14 0. m s Section 6.3 Motion in Accelerated Frames P6.19 (a) F Max∑ = , a T M = = = 18 0 3 60 . . N 5.00 kg m s2 to the right. (b) If v = const, a = 0, so T = 0 (This is also an equilibrium situation.) (c) Someone in the car (noninertial observer) claims that the forces on the mass along x are T and a fictitious force (–Ma). Someone at rest outside the car (inertial observer) claims that T is the only force on M in the x-direction. P6.20 The water moves at speed v = = ( ) = 2 2 0 12 7 25 0 104 π πr T . . . m s m s. The top layer of water feels a downward force of gravity mg and an outward fictitious force in the turntable frame of reference, m r m m v2 2 20 104 0 12 9 01 10= ( ) = × −. . . m s m m s2 It behaves as if it were stationary in a gravity field pointing downward and outward at tan . . .− =1 0 090 1 9 8 0 527 m s m s 2 2 ° Its surface slopes upward toward the outside, making this angle with the horizontal. FIG. P6.19 5.00 kg 13794_06_ch06_p127-152.indd 13413794_06_ch06_p127-152.indd 134 12/2/06 1:58:38 PM12/2/06 1:58:38 PM
  9. 9. Circular Motion and Other Applications of Newton’s Laws 135 P6.21 The only forces acting on the suspended object are the force of gravity mg and the force of tension T forward and upward at angle θ with the vertical, as shown in the free-body diagram. Applying Newton’s second law in the x and y directions, F T max∑ = =sinθ (1) F T mgy∑ = − =cosθ 0 or T mgcosθ = (2) (a) Dividing equation (1) by (2) gives tan . . .θ = = = a g 3 00 9 80 0 306 m s m s 2 2 Solving for θ, θ = 17 0. ° (b) From Equation (1), T ma = = ( )( ) ( ) = sin . . sin . . θ 0 500 3 00 17 0 5 1 kg m s2 ° 22 N P6.22 Consider forces on the backpack as it slides in the Earth frame of reference. F ma n mg ma n m g a f m g a F y y k k x ∑ ∑ = + − = = +( ) = +( ) = : , , µ mma m g a max k x: − +( ) =µ The motion across the floor is described by L t a t t g a tx k= + = − +( )v v 1 2 1 2 2 2 µ . We solve for µk : vt L g a tk− = +( ) 1 2 2 µ , 2 2 vt L g a t k −( ) +( ) = µ . P6.23 F F magmax = + = 591 N F F magmin = − = 391 N (a) Adding, 2 982Fg = N, Fg = 491 N (b) Since F mgg = , m = = 491 50 1 N 9.80 m s kg2 . (c) Subtracting the above equations, 2 200ma = N ∴ =a 2 00. m s2 FIG. P6.21 T cos T sin mg θ θ 13794_06_ch06_p127-152.indd 13513794_06_ch06_p127-152.indd 135 12/2/06 1:58:39 PM12/2/06 1:58:39 PM
  10. 10. 136 Chapter 6 P6.24 In an inertial reference frame, the girl is accelerating horizontally inward at v2 2 5 70 2 40 13 5 r = ( ) = . . . m s m m s2 In her own non-inertial frame, her head feels a horizontally outward fictitious force equal to its mass times this acceleration. Together this force and the weight of her head add to have a magni- tude equal to the mass of her head times an acceleration of g r 2 2 2 2 2 9 80 13 5 16 7+ ⎛ ⎝⎜ ⎞ ⎠⎟ = ( ) + ( ) = v . . .m s m s2 2 This is larger than g by a factor of 16 7 9 80 1 71 . . .= . Thus, the force required to lift her head is larger by this factor, or the required force is F = ( ) =1 71 55 0 93 8. . .N N . P6.25 a R T r e = ⎛ ⎝⎜ ⎞ ⎠⎟ = 4 35 0 0 027 6 2 2 π cos . .° m s2 We take the y axis along the local vertical. a a a y r y x net 2 net m s m ( ) = − ( ) = ( ) = 9 80 9 77 0 015 8 . . . ss2 θ = =arctan . a a x y 0 092 8° Section 6.4 Motion in the Presence of Resistive Forces P6.26 m = 80 0. kg, vT = 50 0. m s, mg D A D A mgT T = ∴ = = ρ ρv v 2 2 2 2 0 314. kg m (a) At v = 30 0. m s a g D A m = − = − ( )( ) = ρ v2 2 2 9 80 0 314 30 0 80 0 6 27 / . . . . . m ss downward2 (b) At v = 50 0. m s, terminal velocity has been reached. F mg R R mg y∑ = = − ⇒ = = ( )( )= 0 80 0 9 80 784. .kg m s N d2 iirected up (c) At v = 30 0. m s D Aρ v2 2 2 0 314 30 0 283= ( )( ) =. . N upward FIG. P6.25 N ar g0 anet Equator 35.0° θ 35.0° (exaggerated size) 13794_06_ch06_p127-152.indd 13613794_06_ch06_p127-152.indd 136 12/2/06 1:58:40 PM12/2/06 1:58:40 PM
  11. 11. Circular Motion and Other Applications of Newton’s Laws 137 P6.27 (a) a g b= − v When v v= T , a = 0 and g b T= v b g T = v The Styrofoam falls 1.50 m at constant speed vT in 5.00 s. Thus, vT y t = = = 1 50 0 300 . . m 5.00 s m s Then b = = −9 80 0 300 32 7 1. . . m s m s s 2 (b) At t = 0, v = 0 and a g= = 9 80. m s2 down (c) When v = 0 150. m s, a g b= − = − ( )( ) =− v 9 80 32 7 0 150 4 901 . . . .m s s m s m2 ss2 down P6.28 (a) ρ = m V , A = 0 020 1. m2 , R AD mgT= 1 = 2 2 ρair v m V= = ( )⎡ ⎣⎢ ⎤ ⎦⎥ =ρ πbead 3 g cm cm0 830 4 3 8 00 1 78 3 . . . kg Assuming a drag coefficient of D = 0 500. for this spherical object, and taking the density of air at 20°C from the endpapers, we have vT = ( )( ) ( ) 2 1 78 9 80 0 500 1 20 0 02 . . . . . kg m s kg m 2 3 00 1 53 8 m m s2 ( ) = . (b) v vf i gh gh2 2 2 0 2= + = + : h g f = = ( ) ( ) = v2 2 2 53 8 2 9 80 148 . . m s m s m2 P6.29 Since the upward velocity is constant, the resultant force on the ball is zero. Thus, the upward applied force equals the sum of the gravitational and drag forces (both downward): F mg b= + v. The mass of the copper ball is m r = = ⎛ ⎝ ⎞ ⎠ ×( ) ×( −4 3 4 3 8 92 10 2 00 10 3 3 2πρ π . .kg m m3 )) = 3 0 299. kg The applied force is then F mg b= + = ( )( )+ ( ) ×( )=− v 0 299 9 80 0 950 9 00 10 32 . . . . .001 N P6.30 The resistive force is R D A= = ( )( )( )1 2 1 2 0 250 1 20 2 20 27 82 ρ v . . . .kg m m3 2 m s N N 1200 kg m s ( ) = = − = − = − 2 255 255 0 212 R a R m . 22 13794_06_ch06_p127-152.indd 13713794_06_ch06_p127-152.indd 137 12/2/06 1:58:40 PM12/2/06 1:58:40 PM
  12. 12. 138 Chapter 6 P6.31 (a) At terminal velocity, R b mgT= =v ∴ = = ×( )( ) × − − b mg Tv 3 00 10 9 80 2 00 10 3 2 . . . kg m s m 2 ss N s m= ⋅1 47. (b) In the equation describing the time variation of the velocity, we have v v= −( )− T bt m e1 v v= 0 632. T when e bt m− = 0 368. or at time t m b = − ⎛ ⎝ ⎞ ⎠ ( ) = × − ln . .0 368 2 04 10 3 s (c) At terminal velocity, R b mgT= = = × − v 2 94 10 2 . N *P6.32 (a) Since the window is vertical, the normal force is horizontal. n = 4 N f = n =k kµ 0.9(4 N) = 3.6 N upward, to oppose downward motion ΣF may y= : + − + =3 6 0 16 9 8 0. ( . )( . )N kg m s2 Py Py = − =2 03 2 03. .N N down (b) ΣF may y= : + − − =3 6 0 16 9 8 1 25 2 03 0 16. ( . )( . ) . ( . ) .N kg N kgm s2 ay ay = − = − =0 508 0 16 3 18 3 18. . . .N kg downm s m s2 2 (c) At terminal velocity, ΣF may y= : + ⋅ − − =( ) ( . )( . ) . ( . )20 0 16 9 8 1 25 2 03 0N s m kg NvT m s2 vT = ⋅ =4 11 20 0 205. ( ) .N N s m downm s P6.33 v = ⎛ ⎝ ⎞ ⎠ − −⎛ ⎝ ⎞ ⎠ ⎡ ⎣⎢ ⎤ ⎦⎥ mg b bt m 1 exp where exp x ex ( ) = is the exponential function. At t →∞ v v→ =T mg b At t = 5 54. s 0 500 1 5 54 9 00 . exp . . v vT T b = − − ( )⎛ ⎝⎜ ⎞ ⎠⎟ ⎡ ⎣ ⎢ ⎤ ⎦ s kg ⎥⎥ exp . . . ; . − ( )⎛ ⎝⎜ ⎞ ⎠⎟ = − ( ) b b 5 54 9 00 0 500 5 54 s kg s 99 00 0 500 0 693 9 00 0 693 5 . ln . . ; . . kg kg = = − = ( )( ) b .. . 54 1 13 s kg s= (a) vT mg b = vT = ( )( ) = 9 00 9 80 1 13 78 3 . . . . kg m s kg s m s 2 (b) 0 750 1 1 13 9 00 . exp . . v vT T t = − −⎛ ⎝ ⎞ ⎠ ⎡ ⎣⎢ ⎤ ⎦⎥s exp . . . −⎛ ⎝ ⎞ ⎠ = 1 13 9 00 0 250 t s t = ( ) − = 9 00 0 250 1 13 11 1 . ln . . .s s continued on next page 13794_06_ch06_p127-152.indd 13813794_06_ch06_p127-152.indd 138 12/2/06 1:58:41 PM12/2/06 1:58:41 PM
  13. 13. Circular Motion and Other Applications of Newton’s Laws 139 (c) dx dt mg b bt m = ⎛ ⎝ ⎞ ⎠ − −⎛ ⎝ ⎞ ⎠ ⎡ ⎣⎢ ⎤ ⎦⎥1 exp ; dx mg b bt m dt x x t 0 1 0 ∫ ∫= ⎛ ⎝ ⎞ ⎠ − −⎛ ⎝ ⎞ ⎠ ⎡ ⎣⎢ ⎤ ⎦⎥exp x x mgt b m g b bt m mgt b m g t − = + ⎛ ⎝⎜ ⎞ ⎠⎟ −⎛ ⎝ ⎞ ⎠ = +0 2 2 0 2 exp bb bt m2 1 ⎛ ⎝⎜ ⎞ ⎠⎟ −⎛ ⎝ ⎞ ⎠ − ⎡ ⎣⎢ ⎤ ⎦⎥exp At t = 5 54. s, x = ( ) + ( ) 9 00 5 54 9 00 . . . kg 9.80 m s s 1.13 kg s kg2 22 2 9 80 1 13 0 693 1 . . exp . m s kg s 2 ( ) ( ) ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ −( )−[ ]] x = + −( ) =434 626 0 500 121m m m. P6.34 F ma∑ = − = − = − = − −( ∫ ∫ − km m d dt kdt d k dt d k t t v v v v v v v v 2 2 0 2 0 0)) = − = − + = + = + = + − v v v v v v v v v v v1 0 0 0 0 0 1 1 1 1 1 1 1 0 kt kt vv0kt P6.35 (a) From Problem 34, v v v v v v = = + = + =∫ ∫ dx dt kt dx dt kt k kdt x t 0 0 0 0 00 0 1 1 1 1++ = +( ) − = +( ) ∫ v v v 00 0 0 0 0 1 1 0 1 1 kt x k kt x k kt t x t ln ln −−⎡⎣ ⎤⎦ = +( ) ln ln 1 1 1 0x k ktv (b) We have ln 1 0+( ) =v kt kx 1 0+ =v kt ekx so v v v v v v= + = = =−0 0 0 0 1 kt e ekx kx 13794_06_ch06_p127-152.indd 13913794_06_ch06_p127-152.indd 139 12/2/06 1:58:42 PM12/2/06 1:58:42 PM
  14. 14. 140 Chapter 6 P6.36 We write − = −km D Av v2 21 2 ρ so k D A m = = ( ) ×( )− ρ 2 0 305 1 20 4 2 10 2 0 145 3 . . . . kg m m3 2 kg m m s ( ) = × = = ( ) − − − × − 5 3 10 40 2 3 0 5 3 10 . . . v v e ekx 33 18 3 36 5 m m m s ( )( ) = . . P6.37 (a) v vt ei ct ( ) = − v v20 0 5 00 20 0 . . . s( ) = = − i c e , vi = 10 0. m s So 5 00 10 0 20 0 . . . = − e c and − = ⎛ ⎝ ⎞ ⎠ 20 0 1 2 . lnc c = − ( ) = × − −ln . . 1 2 2 1 20 0 3 47 10 s (b) At t = 40 0. s v = ( ) = ( )( ) =− 10 0 10 0 0 250 2 5040 0 . . . .. m s m s m se c (c) v v= − i ct e a d dt c e ci ct = = − = −−v v v P6.38 In R D A= 1 2 2 ρ v , we estimate that D = 1 00. , ρ = 1 20. kg m3 , A = ( )( ) = × − 0 100 0 160 1 60 10 2 . . .m m m2 and v = 27 0. m s. The resistance force is then R = ( )( ) ×( )(−1 2 1 00 1 20 1 60 10 27 02 . . . .kg m m m s3 2 )) = 2 7 00. N or R ~ 101 N Additional Problems *P6.39 Let v0 represent the speed of the object at time 0. We have d b m dt b m t tv v v v v v v 0 00 0∫ ∫= − = −ln tt b m t bt m e ln ln / / v v v v v v − = − −( ) ( ) = − = 0 0 0 0 ln −− − =bt m bt m e/ / v v0 From its original value, the speed decreases rapidly at first and then more and more slowly, asymptotically approaching zero. In this model the object keeps losing speed forever. It travels a finite distance in stopping. The distance it travels is given by dr e dt r m b e b m dt r bt m t bt m t 0 0 0 0 0 ∫ ∫ ∫ = = − − ⎛ ⎝ ⎞ − − v v / / ⎠⎠ = − = − −( )= −− − −m b e m b e m b ebt m t bt m bt v v v 0 0 0 0 1 1/ / //m t ( ) As goes to infinity, the distance appproaches m b m b v v0 01 0−( ) = / FIG. P6.39 t 0 0 13794_06_ch06_p127-152.indd 14013794_06_ch06_p127-152.indd 140 12/2/06 1:58:43 PM12/2/06 1:58:43 PM
  15. 15. Circular Motion and Other Applications of Newton’s Laws 141 P6.40 At the top of the vertical circle, T m R mg= − v2 or T = ( ) ( ) − ( )( ) =0 400 4 00 0 500 0 400 9 80 8 88 2 . . . . . . N P6.41 (a) The speed of the bag is 2 7 46 38 1 23 π . . m s m s ( ) = . The total force on it must add to mac = ( )( ) = 3 1 23 7 46 6 12 2 0 kg m s m N . . . F ma f n F ma f x x s y y s ∑ ∑ = − = = : cos sin . : sin 20 20 6 12 2 N 00 20 30 9 8 0 20 6 12 + − ( )( )= = − n n fs cos . cos . kg m s2 NN sin20 Substitute: f fs ssin cos sin . cos sin 20 20 20 6 12 20 20 29 2 + − ( ) =N 44 2 92 294 16 8 106 N N N N f f s s . .( ) = + = (b) v = ( ) = 2 7 94 34 1 47 π . . m s m s mac = ( )( ) = 30 1 47 7 94 8 13 2 kg m s m N . . . f n f n n s s cos sin . sin cos 20 20 8 13 20 20 294 − = + = = N N ff f f s s s cos . sin sin cos sin 20 8 13 20 20 20 20 8 2 − + − N .. cos sin . . 13 20 20 294 2 92 294 22 4 N N N ( ) = ( ) = +fs N N N N N f n s = = ( ) − = 108 108 20 8 13 20 273 cos . sin µµs sf n = = = 108 273 0 396 N N . FIG. P6.41 n mg fs ac x y 13794_06_ch06_p127-152.indd 14113794_06_ch06_p127-152.indd 141 12/2/06 1:58:44 PM12/2/06 1:58:44 PM
  16. 16. 142 Chapter 6 P6.42 When the cloth is at a lower angle θ, the radial component of F ma∑ = reads n mg m r + =sinθ v2 At θ = 68 0. °, the normal force drops to zero and g r sin68 2 ° = v . v = = ( )( ) =rgsin . . sin .68 0 33 9 8 68 1 73° °m m s m s2 The rate of revolution is angular speed = ( )⎛ ⎝ ⎞ ⎠ ( ) ⎛ ⎝⎜ ⎞ ⎠⎟ =1 73 1 2 2 2 0 33 0. . m s rev mπ π πr r .. .835 50 1rev s rev min= P6.43 (a) v = ( ) ⎛ ⎝⎜ ⎞ ⎠⎟ ⎛ ⎝ ⎞ ⎠ =30 1 1 000 8 3km h h 3 600 s m 1 km . 33 m s F may y∑ = : + − = −n mg m r v2 n m g r = − ⎛ ⎝⎜ ⎞ ⎠⎟ = − ( )v2 2 1800 9 8 8 33 20 kg m s m s2 . . .44 1 15 104 m N up ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ = ×. (b) Take n = 0. Then mg m r = v2 . v = = ( )( ) = =gr 9 8 20 4 14 1 50 9. . . .m s m m s km h2 P6.44 (a) F ma m R y y∑ = = v2 mg n m R − = v2 n mg m R = − v2 (b) When n = 0 mg m R = v2 Then, v = gR A more gently curved bump, with larger radius, allows the car to have a higher speed with- out leaving the road. This speed is proportional to the square root of the radius. P6.45 (a) slope = − = 0 160 0 9 9 0 016 2 . . . N m s kg m2 2 (b) slope = = = R D A D A v v v2 1 2 2 2 1 2 ρ ρ FIG. P6.43 n mg FIG. P6.42 R 68° p mg p mg cos 68° mg sin 68° continued on next page 13794_06_ch06_p127-152.indd 14213794_06_ch06_p127-152.indd 142 12/2/06 1:58:44 PM12/2/06 1:58:44 PM
  17. 17. Circular Motion and Other Applications of Newton’s Laws 143 (c) 1 2 0 016 2D Aρ = . kg m D = ( ) ( ) ( ) = 2 0 016 2 1 20 0 105 0 7782 . . . . kg m kg m m3 π (d) From the table, the eighth point is at force mg = ×( )( )=− 8 1 64 10 9 8 0 1293 . . .kg m s N2 and horizontal coordinate 2 80 2 . m s( ) . The vertical coordinate of the line is here 0 016 2 2 8 0 127 2 . . .kg m m s N( )( ) = .The scatter percentage is 0 129 0 127 1 5 . . . % N N 0.127 N − = . (e) The interpretation of the graph can be stated thus: For stacked coffee filters falling at terminal speed, a graph of air resistance force as a function of squared speed demonstrates that the force is proportional to the speed squared within the experimental uncertainty estimated as 2%. This proportionality agrees with that described by the theoretical equation R D A= 1 2 2 ρ v . The value of the constant slope of the graph implies that the drag coefficient for coffee filters is D = ±0 78 2. %. *P6.46 (a) The forces acting on the ice cube are the Earth’s gravitational force, straight down, and the basin’s normal force, upward and inward at 35° with the vertical. We choose the x and y axes to be horizontal and vertical, so that the acceleration is purely in the x direction. Then ∑Fx = max : n sin 35° = mv2 րR ∑Fy = may : n cos 35° − mg = 0 Dividing eliminates the normal force: n sin 35°ր n cos 35° = mv2 րRmg tan 35° = v2 րRg v = = ( )Rg Rtan . .35 0 6 86° m s2 (b) The massis unnecessary. (c) The answer to (a) indicates that the speed is proportional to the square root of the radius, so doubling the radius will make the required speed increase by 2 times . (d) The period of revolution is given by T R R Rg R= = = ( )2 2 35 0 2 40 π π v tan . . ° s m When the radius doubles, the period increases by 2 times . (e) On the larger circle the ice cube moves 2 times faster but also takes longer to get around, because the distance it must travel is 2 times larger. Its period is also proportional to the square root of the radius. P6.47 Take x-axis up the hill F ma T mg ma a T m g F m x x y ∑ ∑ = + − = = − = : sin sin sin sin θ φ θ φ aa T mg T mg a g y: cos cos cos cos cos sin c + − = = = θ φ φ θ φ θ 0 oos sin cos tan sin θ φ φ θ φ − = −( ) g a g 13794_06_ch06_p127-152.indd 14313794_06_ch06_p127-152.indd 143 12/2/06 1:58:45 PM12/2/06 1:58:45 PM
  18. 18. 144 Chapter 6 P6.48 (a) v = ( ) ⎛ ⎝⎜ ⎞ ⎠⎟ =300 88 0 60 0 440mi h ft s mi h ft s . . At the lowest point, his seat exerts an upward force; therefore, his weight seems to increase. His apparent weight is ′ = + = + ⎛ ⎝ ⎞ ⎠ ( ) =F mg m r g v2 2 160 160 32 0 440 1 200 967 . lbb (b) At the highest point, the force of the seat on the pilot is directed down and ′ = − = −F mg m r g v2 647 lb Since the plane is upside down, the seat exerts this downward force as a normal force. (c) When ′ =Fg 0, then mg m R = v2 . If we vary the aircraft’s R and v such that this equation is satisfied, then the pilot feels weightless. P6.49 (a) Since the centripetal acceleration of a person is downward (toward the axis of the earth), it is equivalent to the effect of a falling elevator. Therefore, ′ = −F F m r g g v2 or F Fg g> ′ (b) At the poles v = 0 and ′ = = = ( ) =F F mgg g 75 0 9 80 735. . N down. At the equator, ′ = − = − ( ) =F F mag g c 735 75 0 0 033 7 732N N N. . down. *P6.50 (a) Since the object of mass m2 is in equilibrium, F T m gy∑ = − =2 0 or T m g= 2 (b) The tension in the string provides the required centripetal acceleration of the puck. Thus, F T m gc = = 2 (c) From F m R c = 1 2 v we have v = = ⎛ ⎝⎜ ⎞ ⎠⎟ RF m m m gRc 1 2 1 (d) The puck will spiral inward,gainingspeedasit doessso. It gains speed because the extra- large string tension produces forward tangential acceleration as well as inward radial acceleration of the puck, pulling at an angle of less than 90° to the direction of the inward-spiraling velocity. (e) The puck will spiral outward,slowingdown asit doeesso. FIG. P6.49 13794_06_ch06_p127-152.indd 14413794_06_ch06_p127-152.indd 144 12/2/06 1:58:46 PM12/2/06 1:58:46 PM
  19. 19. Circular Motion and Other Applications of Newton’s Laws 145 *P6.51 (a) The only horizontal force on the car is the force of friction, with a maximum value determined by the surface roughness (described by the coefficient of static friction) and the normal force (here equal to the gravitational force on the car). (b) ∑Fx = max −f = ma a = −fրm = (v2 − v0 2 )ր2(x − x0 ) x − x0 = (v2 − v0 2 )mր2f = (02 − [20 mրs]2 )1200 kgր2(−7000 N) = 34 3. m (c) Now f = mv2 րr r = mv2 րf = 1200 kg [20 mրs]2 ր7000 N = 68 6. m A top view shows that you can avoid running into the wall by turning through a quarter-circle, if you start at least this far away from the wall. (d) Braking is better. You should not turn the wheel. If you used any of the available fric- tion force to change the direction of the car, it would be unavailable to slow the car, and the stopping distance would be longer. (e) The conclusion is true in general. The radius of the curve you can barely make is twice your minimum stopping distance. P6.52 v = = ( ) ( ) = 2 2 9 00 15 0 3 77 π πr T . . . m s m s (a) a r r = = v2 1 58. m s2 (b) F m g arlow N= +( ) = 455 (c) F m g arhigh N= −( ) = 328 (d) F m g armid N upward and= + =2 2 397 at θ = = =− − tan tan . . .1 1 1 58 9 8 9 15 a g r °inward . P6.53 (a) The mass at the end of the chain is in vertical equilibrium. Thus T mgcosθ = . Horizontally T ma m r rsinθ = = v2 r r = +( ) = +( ) = 2 50 4 00 2 50 28 0 4 00 . sin . . sin . . θ m m° 55 17. m Then ar = v2 5 17. m . By division tan . θ = = a g g r v2 5 17 v v 2 5 17 5 17 9 80 28 0 5 = = ( )( )( ) = . tan . . tan .g θ ° m s2 2 ..19 m s (b) T mgcosθ = T mg = = ( )( ) ° = cos . . cos .θ 50 0 9 80 28 0 555 kg m s N 2 FIG. P6.53 T R = 4.00 m θ l = 2.50 m r mg 13794_06_ch06_p127-152.indd 14513794_06_ch06_p127-152.indd 145 12/2/06 1:58:47 PM12/2/06 1:58:47 PM
  20. 20. 146 Chapter 6 P6.54 (a) The putty, when dislodged, rises and returns to the original level in time t. To find t, we use v vf i at= + : i.e., − = + −v v gt or t g = 2v where v is the speed of a point on the rim of the wheel. If R is the radius of the wheel, v = 2πR t , so t g R = = 2 2v v π . Thus, v2 = πRg and v = πRg . (b) The putty is dislodged when F, the force holding it to the wheel, is F m R m g= = v2 π *P6.55 (a) n m R = v2 f mg− = 0 f ns= µ v = 2πR T T R g s = 4 2 π µ (b) T = 2 54. s # . rev min rev 2.54 s s min rev min = ⎛ ⎝ ⎞ ⎠ = 1 60 23 6 (c) The gravitational and frictional forces remain constant. The normal force increases. The person remains in motion with the wall. (d) The gravitational force remains constant. The normal and frictional forces decrease. The person slides relative to the wall and downward into the pit. P6.56 Let the x-axis point eastward, the y-axis upward, and the z-axis point southward. (a) The range is Z g i i = v2 2sin θ The initial speed of the ball is therefore vi i gZ = = ( )( ) ° = sin . sin . . 2 9 80 285 96 0 53 0 θ m s The time the ball is in the air is found from ∆y t a tiy y= +v 1 2 2 as 0 53 0 48 0 4 90 2 = ( )( ) − ( ). sin . .m s m s2 ° t t giving t = 8 04. s . continued on next page FIG. P6.55 f mg nn 13794_06_ch06_p127-152.indd 14613794_06_ch06_p127-152.indd 146 12/2/06 1:58:47 PM12/2/06 1:58:47 PM
  21. 21. Circular Motion and Other Applications of Newton’s Laws 147 (b) vix e iR = = ×( ) °2 86 400 2 6 37 10 35 06 π φ πcos . cos . s m 886 400 379 s m s= (c) 360° of latitude corresponds to a distance of 2πRe, so 285 m is a change in latitude of ∆φ π π = ⎛ ⎝⎜ ⎞ ⎠⎟ ( ) = ×( ) ⎛ ⎝ ⎜ S Re2 360 285 6 37 106 ° m 2 m. ⎞⎞ ⎠ ⎟ ( ) = × − 360 2 56 10 3 ° . degrees The final latitude is then φ φ φf i= − = − =∆ 35 0 0 002 56 34 997 4. . .° ° °. The cup is moving eastward at a speed vfx e fR = 2 86 400 π φcos s , which is larger than the eastward velocity of the tee by ∆v v vx fx fi e f i eR R = − = −⎡⎣ ⎤⎦ = 2 86 400 2π φ φ π s cos cos 886 400 2 86 400 s s cos cos cos φ φ φ π i i eR −( )−⎡⎣ ⎤⎦ = ∆ φφ φ φ φ φi i icos sin sin cos∆ ∆+ −[ ] Since ∆φ is such a small angle, cos ∆φ ≈ 1and ∆ ∆vx e i R ≈ 2 86 400 π φ φ s sin sin . ∆vx ≈ ×( )2 6 37 10 86 400 35 0 0 002 56 6 π . sin . sin . m s ° °° = × − 1 19 10 2 . m s (d) ∆ ∆x tx= ( ) = ×( )( ) = =− v 1 19 10 8 04 0 095 5 92 . . .m s s m ..55 cm P6.57 (a) If the car is about to slip down the incline, f is directed up the incline. F n f mgy∑ = + − =cos sinθ θ 0 where f ns= µ gives n mg s = +( )cos tanθ µ θ1 and f mgs s = +( ) µ θ µ θcos tan1 Then, F n f m R x∑ = − =sin cos min θ θ v2 yields vmin tan tan = −( ) + Rg s s θ µ µ θ1 When the car is about to slip up the incline, f is directed down the incline. Then, F n f mgy∑ = − − =cos sinθ θ 0 with f ns= µ yields n mg s = −( )cos tanθ µ θ1 and f mgs s = −( ) µ θ µ θcos tan1 In this case, F n f m R x∑ = + =sin cos max θ θ v2 , which gives vmax tan tan = +( ) − Rg s s θ µ µ θ1 (b) If vmin tan tan = −( ) + = Rg s s θ µ µ θ1 0, then µ θs = tan . continued on next page 13794_06_ch06_p127-152.indd 14713794_06_ch06_p127-152.indd 147 12/2/06 1:58:48 PM12/2/06 1:58:48 PM
  22. 22. 148 Chapter 6 (c) vmin . tan . . . = ( )( ) −( ) + 100 9 80 10 0 0 100 1 0 1 m m s2 ° 000 10 0 8 57 ( ) = tan . . ° m s vmax . tan . . . = ( )( ) +( ) − 100 9 80 10 0 0 100 1 0 1 m m s2 ° 000 10 0 16 6 ( ) = tan . . ° m s *P6.58 (a) We let R represent the radius of the hoop and T represent the period of its rotation. The bead moves in a circle with radius v = Rsinθ at a speed of v = = 2 2π π θr T R T sin The normal force has an inward radial component of nsinθ and an upward component of n cosθ F ma n mgy y∑ = − =: cosθ 0 or n mg = cosθ Then F n m r x∑ = =sinθ v2 becomes mg m R R Tcos sin sin sin θ θ θ π θ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ 2 2 which reduces to g R T sin cos sinθ θ π θ = 4 2 2 This has two solutions: sinθ θ= ⇒ = °0 0 (1) and cosθ π = gT R 2 2 4 (2) If R = 15 0. cm and T = 0 450. s, the second solution yields cos . . . .θ π = ( )( ) ( ) = 9 80 0 450 4 0 150 0 335 2 2 m s s m 2 and θ = 70 4. ° Thus, in this case, the bead can ride at two positions θ = 70 4. ° and θ = 0° . (b) At this slower rotation, solution (2) above becomes cos . . . .θ π = ( )( ) ( ) = 9 80 0 850 4 0 150 1 20 2 2 m s s m 2 , which is impossible. In this case, the bead can ride only at the bottom of the loop, θ = 0° . (c) The equation that the angle must satisfy has two solutions whenever 4π2 R > gT2 but only the solution 0° otherwise. The loop’s rotation must be faster than a certain threshold value in order for the bead to move away from the lowest position. Zero is always a solution for the angle. There are never more than two solutions. FIG. P6.58(a) FIG. P6.57 13794_06_ch06_p127-152.indd 14813794_06_ch06_p127-152.indd 148 12/2/06 1:58:49 PM12/2/06 1:58:49 PM
  23. 23. Circular Motion and Other Applications of Newton’s Laws 149 P6.59 At terminal velocity, the accelerating force of gravity is balanced by frictional drag: mg ar br= +v v2 2 (a) mg = ×( ) + ×( )− − 3 10 10 0 870 109 10 2 . .v v For water, m = = ( )⎡ ⎣⎢ ⎤ ⎦⎥ − ρ πV 1 000 4 3 10 5 3 kg m m3 4 11 10 3 10 10 0 870 1011 9 10 2 . . .× = ×( ) + ×( )− − − v v Assuming v is small, ignore the second term on the right hand side: v = 0 013 2. m s . (b) mg = ×( ) + ×( )− − 3 10 10 0 870 108 8 2 . .v v Here we cannot ignore the second term because the coefficients are of nearly equal magnitude. 4 11 10 3 10 10 0 870 10 3 1 8 8 8 2 . . . . × = ×( ) + ×( ) = − − − − v v v 00 3 10 4 0 870 4 11 2 0 870 1 03 2 ± ( ) + ( )( ) ( ) = . . . . . m s (c) mg = ×( ) + ×( )− − 3 10 10 0 870 107 6 2 . .v v Assuming v > 1 m s, and ignoring the first term: 4 11 10 0 870 105 6 2 . .× = ×( )− − v v = 6 87. m s P6.60 (a) t ds m . . . . . . . . . ( ) ( ) 1 00 2 00 3 00 4 00 5 00 6 00 7 00 8 00 9 000 10 0 11 0 12 0 13 0 14 0 15 0 16 0 17 0 18 0 19 0 2 . . . . . . . . . . 00 0 4 88 18 9 42 1 73 8 112 154 199 246 296 347 399 45 . . . . . 22 505 558 611 664 717 770 823 876 (b) (c) A straight line fits the points from t = 11 0. s to 20.0 s quite precisely. Its slope is the termi- nal speed. vT = = − − =slope m m 20.0 s s m s 876 399 11 0 53 0 . . (s) (m) 900 800 700 600 500 400 300 200 100 0 0 2 4 6 8 10 12 14 16 18 20 t d 13794_06_ch06_p127-152.indd 14913794_06_ch06_p127-152.indd 149 12/2/06 1:58:50 PM12/2/06 1:58:50 PM
  24. 24. 150 Chapter 6 P6.61 F L T mg L T may y y y∑ = − − = − − =cos . sin . .20 0 20 0 7 35° ° N == = + = + ° = = ∑ 0 20 0 20 0 0 2 2 F L T L T m r m r x x x sin . cos .° v v .. . . cos . .750 35 0 60 0 20 0 16 3 2 kg m s m N We ( ) ( ) = ° have the simultaneous equations L sin .20 0° +TT L T cos . . cos . sin . . 20 0 16 3 20 0 20 0 7 35 ° ° ° = − = N N LL T L T + = − cos . sin . . sin . sin 20 0 20 0 16 3 20 0 20 ° ° ° N .. cos . . cot . tan . 0 20 0 7 35 20 0 20 ° ° ° ° = + N cos20.0 T 00 16 3 20 0 7 35 20 0 3 11 ° ° ° ( ) = − ( ) = . sin . . cos . . N N T 339 8 12 8 . . N NT = *P6.62 (a) v v= +i kx implies the acceleration is a d dt k dx dt k= = + = + v v0 Then the total force is F ma m k∑ = = +( )v As a vector, the force is parallel or antiparallel to the velocity: F v∑ = km . (b) For k positive, some feedback mechanism could be used to impose such a force on an object for a while. The object’s speed rises exponentially. Riding on such an object would be more scary than riding on a skyrocket. It would be a good opportunity for learning about exponential growth in population or in energy use. (c) For k negative, think of a duck landing on a lake, where the water exerts a resistive force on the duck proportional to its speed. ANSWERS TO EVEN PROBLEMS P6.2 215 N horizontally inward P6.4 (a) 1 65. km s (b) 6 84 103 . × s P6.6 6 22 10 12 . × − N P6.8 (a) 68.6 N toward the center of the circle and 784 N up (b) 0.857 mրs2 P6.10 (a) − +( )0 233 0 163. ˆ . ˆ m s2 i j (b) 6 53. m s (c) − +( )0 181 0 181. ˆ . ˆ m s2 i j P6.12 (a)1 33. m s2 (b)1 79. m s2 forward and 48.0° inward P6.14 (a) 2.49 × 104 N up (b) 12.1 mրs FIG. P6.61 13794_06_ch06_p127-152.indd 15013794_06_ch06_p127-152.indd 150 12/2/06 1:58:51 PM12/2/06 1:58:51 PM
  25. 25. Circular Motion and Other Applications of Newton’s Laws 151 P6.16 (a) 20.6 N (b) 3.35 mրs2 downward tangent to the circle; 32.0 mրs2 radially inward (c) 32.2 mրs2 at 5.98° to the cord, pointing toward a location below the center of the circle. (d) No change. If the object is swinging down it is gaining speed. If it is swinging up it is losing speed but its acceleration is the same size and its direction can be described in the same terms. P6.18 (a) 8.62 m (b) Mg downward (c) 8 45. m s2 Unless they are belted in, the riders will fall from the cars. P6.20 0.527° P6.22 µk t L g a t = −( ) +( ) 2 2 v P6.24 93.8 N P6.26 (a) 6 27. m s downward2 (b) 784 N up (c) 283 N up P6.28 (a) 53 8. m s (b) 148 m P6.30 −0 212. m s2 P6.32 (a) 2.03 N down (b) 3.18 mրs2 down (c) 0.205 mրs down P6.34 see the solution P6.36 36 5. m s P6.38 ~101 N P6.40 8.88 N P6.42 0 835. rev s P6.44 (a) mg m R − v2 upward (b) v = gR P6.46 (a) v = = ( )Rg Rtan . .35 0 6 86° m/s2 (b) The mass is unnecessary. (c) Increase by 2 times (d) Increase by 2 times (e) On the larger circle the ice cube moves 2 times faster but also takes longer to get around, because the distance it must travel is 2 times larger. Its period is described byT R R Rg R= = = ( )2 2 35 0 2 40 π v π tan . . ° s m/ . P6.48 (a) The seat exerts 967 lb up on the pilot. (b) The seat exerts 647 lb down on the pilot. (c) If the plane goes over the top of a section of a circle with v2 = Rg, the pilot will feel weightless. P6.50 (a) m g2 (b) m g2 (c) m m gR2 1 ⎛ ⎝⎜ ⎞ ⎠⎟ (d) The puck will move inward along a spiral, gaining speed as it does so. (e) The puck will move outward along a spiral as it slows down. P6.52 (a) 1.58 m/s2 (b) 455 N (c) 329 N (d) 397 N upward and 9.15° inward P6.54 (a) v = πRg (b) m gπ P6.56 (a) 8.04 s (b) 379 m s (c) 1 19. cm s (d) 9.55 cm 13794_06_ch06_p127-152.indd 15113794_06_ch06_p127-152.indd 151 12/2/06 1:58:51 PM12/2/06 1:58:51 PM
  26. 26. 152 Chapter 6 P6.58 (a) either 70.4° or 0° (b) 0° (c) The equation that the angle must satisfy has two solutions whenever 4π2 R > gT2 but only the solution 0° otherwise. (Here R and T are the radius and period of the hoop.) Zero is always a solution for the angle. There are never more than two solutions. P6.60 (a) and (b) see the solution (c) 53 0. m s P6.62 (a) ΣF v= mk (b) For k positive, some feedback mechanism could be used to impose such a force on an object for a while. The object’s speed rises exponentially. Riding on such an object would be more scary than riding on a skyrocket. It would be a good opportunity for learning about exponential growth in population or in energy use. (c) For k negative, think of a duck landing on a lake, where the water exerts a resistive force on the duck proportional to its speed. 13794_06_ch06_p127-152.indd 15213794_06_ch06_p127-152.indd 152 12/2/06 1:58:52 PM12/2/06 1:58:52 PM

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