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97 Pages

hmwk-3-K11-soln

Course: PHY 317k, Fall 2011
School: University of Texas
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Word Count: 4980

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3 Homework Solutions 1. E F F F F1 F1 F1 x1 x1 1 x x1 x x 3 2 x U = W = F ds = 0 1 F dx. The integral is just the area, so: W for gure 1 is 1 F1x1 , W for gure 2 is F1 x1, and W for gure 3 is 1 F1x1. Then U for gure 1 2 2 is 1 F1 x1, U for gure 2 is F1x1 , and U for gure 3 is 1 F1x1. Hence, from lowest 2 2 to highest, the order is: 2,1,3. 2. E If there is no air resistance, then the mechanical...

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3 Homework Solutions 1. E F F F F1 F1 F1 x1 x1 1 x x1 x x 3 2 x U = W = F ds = 0 1 F dx. The integral is just the area, so: W for gure 1 is 1 F1x1 , W for gure 2 is F1 x1, and W for gure 3 is 1 F1x1. Then U for gure 1 2 2 is 1 F1 x1, U for gure 2 is F1x1 , and U for gure 3 is 1 F1x1. Hence, from lowest 2 2 to highest, the order is: 2,1,3. 2. E If there is no air resistance, then the mechanical energy Emec = K + U of the Earthball system is conserved, meaning it is constant. Figure V shows constant mechanical energy. 3. C The block starts from rest, so its initial kinetic energy is zero. So K = K = U = mg y = (5.0 kg)(9.8 m/s2 )(25 m) = 1225 J. 4. C The change in gravitational potential energy is given with U = mg y , where y is the change in the elevation of the block. y = U/mg = 2000 J/(2.0 kg)(9.8 m/s2 ) = 102 m. Since the starting elevation was 18 m, this corresponds to 120 m above the Earths surface. 5. C Due to the conservation of mechanical energy, K + U = 0. As the block goes up the rise, its kinetic energy decreases and its potential energy increases. At the top, the change (from the bottom) in potential energy is: U = mgh. Thus, the initial kinetic energy must have been at least mgh. Since K = 1 mv 2, the initial speed must satisfy: 2 v 2gh. 1 6. D The initial kinetic energy (at X) is zero, so the kinetic energy at Y is just K = U , where U is the change in potential energy going from X to Y. (Since K + U = 0.) Then K = mg y . Then the speed at Y is: v = 2K/m = 2g y = 2(9.8 m/s2 )(1.85 m) = 6.0 m/s. 7. A The force of 35.0 N compresses the spring by x = F/k = (35.0 N)/(3.50 N/cm) = 10.0 cm. The potential energy stored in the spring is: U = 1 kx2 = 1 (3.50 N/cm)(10.0 cm)2 = 175 N cm = 1.75 J. 2 2 8. D The total mechanical energy is equal to the sum of potential and kinetic energy: Etot = 1 mv 2 + 1 kx2 . When the spring is fully extended the speed of the block is 0, so 2 2 x = 2Etot/k = 2(2.8 J)/(75 N/m) = 0.27 m. 9. D 2 Initially the object has kinetic energy 1 mv0 and the potential energy stored in the 2 2 spring is 0, so the spring-mass system has total mechanical energy Emec = 1 mv0 . When 2 the block starts to compress the spring its kinetic energy will start to transfer into spring 2 potential energy, while keeping the total mechanical energy the same (Emec = 1 mv0 = 2 2 2 K + U ). At the instant K = 3U , it follows that 1 mv0 = 3U + U , so U = 1 mv0 . But 2 8 1 m 2 . U = 1 kx2 , so 1 kx2 = 1 mv0 , meaning: x = v0 2 2 8 2 k 10. B If the object moves around the circular loop, its centripetal acceleration is a = v 2/r. At the top, the normal force n from the loop on the object points vertically downward; the force of gravity also points vertically downward. Since the objects acceleration is also pointing downward at the top of the loop, from Fnet = ma we have: n + mg = mv 2/r. The faster the object moves (i.e., the larger v ), the larger n must be. Likewise, the slower it moves, the less n must be. But n cannot be less than zero. Thus, it must be that mv 2/r mg . If this is not satised, the object will fall down. So, the minimum speed at the top is: vmin = gr = (9.8 m/s2 )(1.5 m) = 3.8 m/s 11. D We see the potential energy U (x) is a parabola, but its minimum is not at the origin. It is at a point to the right of the origin, which we can call x0. Then U (x) = a(x x0)2, where a is a positive constant. Therefore, the force is given by: dU F ( x) = = 2a(x x0) = 2ax + 2ax0 . dx This function F (x) is a straight line which intersects the y -axis at y = 2ax0 (a positive 2 value) and which has a slope 2a (a negative value). This function is represented by graph IV. 12. B Given the potential U (x) = ax + bx2, we can nd the force: d dU = (ax + bx2) = a 2bx. F = dx dx 13. E If M is the mass of the Earth and R its radius, the force on a mass m at the surface Mm has magnitude: F = G 2 ; since F = ma, the acceleration of this object would be R GM/R2 , which give us the standard value of g . If the distance from the center of the Earth is increased to 3R (i.e., one diameter above the surface), then the acceleration will be GM/(3R)2 . Since we know g = GM/R2 = 9.8 m/s2 , it follows that the acceleration at this higher position is (9.8 m/s2 )/9 = 1.1 m/s2 . 14. B The gravitational force on M is due to the four particles of mass m. The conguration is symmetric about the origin, so the contributions to the force in the x-direction will cancel. The gravitational force on M attracts it toward the four particles of mass m, so the net force will be in the downward direction. 15. B For congurations 1, 2, and 3, we can calculate the gravitational force exactly. GMm GMm 5 GMm GMm For 1: F1 = + = = 1.25 . 2 2 2 d (2d) 4d d2 For 2: F2 = 0, since the forces of magnitude GMm/d2 are in opposite directions. GMm GMm 2 GMm GMm For 3: F3 = ( ) + ( 2 )2 = 2 2 = 1.41 . 2 d d d d2 For 4, we dont know the angle between the two masses, but it is less than 90 , so that the net force must be greater than F3. Thus, we can see that the order is: F 2 < F1 < F3 < F4 . 16. D We know g = GM/R2 = 9.8 m/s2 , where M is the mass and R the radius of the Earth. To get the acceleration of a freely falling body near the surface of Venus, we can see the result must be: gv = G Mv (0.0558M ) 0.0558 M 0.0558 =G = G 2= g = 0.384(9.8 m/s2 ) = 3.8 m/s2 . 2 2 2 Rv (0.381R) (0.381) R (0.381)2 3 17. D In this situation, a mass falls from 2R to R, for the Earth and also for some other planet with the same radius. On the other planet the nal speed of the object is four times greater than on Earth; therefore its kinetic energy is 16 times larger. Thus, the change in potential energy is 16 times larger on the other planet. On Earth the change GMm GMm GMm in the potential energy is: U = Uf Ui = + = . R 2R 2R We see the magnitude of U is proportional to M , so the mass of the other planet must be 16 times larger than the mass of Earth. 18. C Fg = GMm/R2 on Earth, if M is the mass of the Earth and R its radius, where m is the mass of the man; we are told Fg = 600 N. To get the gravitational force on the man on another planet, we need to replace the Earth values for M and R with those other = G(0.01M )m/(0.25R)2 . Comparing this to corresponding to the other planet: Fg other = 0.16(600 N) = 96 N. the Earth value, it follows that Fg 19. A One Earth radius above the surface is a distance r = 2R from the center, if R is the radius of the Earth. We use K + U = 0, with the initial value of kinetic energy being Ki = 0, and the initial value of potential energy being Ui = GMm/(2R), where m is the mass of the object. The nal value of potential energy Uf = GMM/R, so U = GMm/(2R). Therefore, the nal value of kinetic energy is: Kf = K = U = GMm/(2R) = 1 mv 2. Then the speed of the object just before it his the 2 surface is v = GM/R . 20. D The particles are at rest both before and after the move takes place, so K = 0. Therefore, W = U ; that is, the work done by the external force equals the change in the potential energy of the (3-particle) system. For a three particle system, with masses m1, m2 and m3 , the potential energy is: U = G m1 m3 m2 m3 m1 m2 G G , r12 r13 r23 where r12 is the distance between particles 1 and 2, etc. Here, we have to particles of mass m a distance d apart, so: m1 = m2 = m and r12 = d. Then m3 = 2m, and initially r13 = and r23 = . Therefore, Ui = Gm2 /d. After the move, r13 = d/2 and r23 = d/2, so Uf = G 2m2 2m2 m2 8m2 m2 G G = G G d (d/2) (d/2) d d Then: W = U = Uf Ui = G 8m2 m2 m2 8m2 G = 8Gm2 /d G = G d d d d 4 21. C The escape speed is the speed for which the kinetic energy equals the absolute value of the potential energy of an object sitting on the surface of a planet; since the potential energy is negative, the sum of these two equals zero. That is, 1 mv 2 = GMm/R, so 2 v = 2GM/R . The question in this problem is, what mass M corresponds to R = 2R and v = 2v ? 2GM RM 1M v R Since =2= = = , we see: M = 8M . v R 2GM RM 2M 22. B The escape speed (vesc = vehicle. 2GM/R ) does not depend on the mass of the escaping 23. B If R is the radius of the Earth, M is its mass, and m is the mass of the projectile, the initial potential energy is Ui = GMm/R. 2 2 The initial kinetic energy is Ki = 1 m 1 vesc = 1 m 1 2GM/R = 1 GMm/R. 2 2 2 2 4 Therefore the initial mechanical energy Ei = Ki + Ui = 3 GMm/R. 4 Mechanical energy will be conserved, so when the projectile stops (Kf = 0), its potential energy will be Uf = 3 GMm/R = GMm/R , where R is the distance from the 4 center of the Earth at which the projectile stops. We see R = 4 R, so the height above 3 the Earth is 1 R. 3 24. C The work W done by the external agent will equal the change in the potential energy U = Uf Ui . The full expression for the potential energy of the ve point masses is long, but since four of the point masses do not move between the initial and nal congurations, many terms do not contribute to U and we can ignore them. The terms which matter are those involving the point mass at the center, which is a distance a/ 2 from each of the other four masses. If this is not clear to you, draw a square (of side a) and gure out the distance from the center to each corner. Then: m2 Ui = 4 G + terms that do not change when the center mass is moved. a/ 2 After the center mass is moved to innity, Uf = 0 + those same terms did not change when the center mass was moved. that So W = U = Uf Ui = 4 2GMm/a. 25. A The object is a distance r = 3R from the center of the Earth. The gravitational force 1 varies as 2 , so the force is 9 times smaller when r = 3R than when r = R. Thus, a r weight of 90 N at the surface has weight 90/9 N = 10 N at this altitude. 5 26. B If a circular orbit has radius R, it has circumference C = 2R. A planet with speed v will complete one orbit in time T = 2R/v , which is the period. Its centripetal acceleration is a = v 2/R, so the magnitude of the gravitational force on the planet must equal ma = mv 2/R; that is, GMm/R2 = mv 2/R, from which it follows: 4 2 R3 2 2 2 . This is consistent with Keplers Third GM/R = (2R/T ) /R. Then: T = GM Law, which says the square of the period is proportional to the cube of the semimajor axis. (Keplers law applies to elliptical orbits; with circular orbits, which are a subset, the semimajor axis equals the radius.) So, the square of the period of a planet which is four times further from the Sun than the Earth will be 43 times larger than the 2 square of the Earths period. That is, T 2 = 64TE , if TE is the Earths period. Then T /TE = 64 = 8. 27. D For a circular orbit, GMm/R2 = mv 2/R; see the previous solution. Then K = 1 mv 2 = 1 GMm/R. 2 2 The potential energy is: U = GMm/R. So K = U/2. 28. D Mechanical energy is conserved, so K + U = 0. Therefore, GMm GMm 1 1 = GMm K = U = R2 R1 R2 R1 = GMm R1 R2 . R1 R2 29. D The planet is acted on by a gravitational force, and from Newtons Second Law, the acceleration of the planet is equal to this force divided by the mass of the planet. Thus, the acceleration is largest where the gravitational force is largest. The gravitational force is proportional to the inverse square of the distance of the planet from the star, so it is largest when the distance is smallest. Therefore, the point at which the planet is closest to the star will have the greatest acceleration. In this case, the magnitude of the acceleration of the planet is greatest at point W. 30. C The center of the mass is: rCM = tions: 1 2 3 m 4 kg 5 kg 6 kg i mi ri . We have 3 particles with masses and posii mi r + 0 0i j + 2m 3m i j 1m + 3m i j 6 Then: rCM = 21m + 28m i j 5 kg(3m + 2m + 6 kg(1m + 3m i j) i j) j i = = 1.4m + 1.9m 4 kg + 5 kg + 6 kg 15 31. E The center of mass of a uniform disk is at the center, by symmetry. 32. D m1y1 + m2y2 At any instant in time, ycm = . Since the vertical positions y1 and y2 of m1 + m2 the two balls are changing versus time, the center-of-mass is moving. To locate it at t = 2 s, we need y1 and y2 at that time: y1 = y01 + v01t 1 gt2 = 30 m 1 (9.8 m/s2 )(2 s)2 = 30 m 19.6 m = 10.4 m. 2 2 y2 = y02 + v02t 1 gt2 = (25 m/s)(2 s) 1 (9.8 m/s2 )(2 s)2 = 50 m 19.6 m = 30.4 m. 2 2 47.3 kg m (0.75 kg)(10.4 m) + (1.3 kg)(30.4 m) = = 23 m. Thus, ycm = (0.75 kg) + (1.3 kg) 2.05 kg 33. C Since there is no friction, the mechanical energy (K + U ) will be conserved for the system consisting of the spring and two masses. Initially, the mechanical energy is entirely in the potential energy of the spring when compressed 10 cm: U = 1 kx2 = 1 (500 N/m)(0.10m)2 = 2.5 J. 2 2 Later, after the masses are moving and the spring returns to its equilibrium length, this energy will have been converted to kinetic energy: 2 2 Kmax = 1 m1 v1 + 1 m2v2 = 2.5 J. 2 2 Linear momentum must also be conserved, so 0 = m1v1 + m2 v2. With two equations in two unknowns, we can solve for the speed of one of the masses: 1 2 2 m v 2 + 1 m2(m1v1/m2 )2 = 1 (m1 + m2/m2 )v1 = 1 m1(1 + m1/m2 )v1 = Kmax 1 2 11 2 2 2 v1 = 2Kmax = m1 (1 + m1 /m2) 2(2.5 J) = 2.0 kg(1 + 2.0 kg/4.0 kg) 2.5 2 2 m /s = 1.3 m/s. 1.5 34. D 2 2 Two bodies A and B have same kinetic energy 1 mA vA = 1 mB vB and the masses are 2 2 vB mB = . Then related by mA = 5mB . Therefore, vA in terms of vB is: vA = vB mA 5 mA vA mA vA 1 pA = = =5 = 5. So, the ratio of pA to pB is 5:1. pB mB pB mB vB 5 35. B The two astronauts are initially not moving, so both have zero momentum. Then Jones pushes Smith, so that Smith has speed 0.4 m/s. Momentum is conserved, so mJ vJ + mSvS = 0. Therefore, vJ = mSvS/mJ = (70 kg)(0.4 m/s)/90 kg = 0.3 m/s. Thus, the speed of Major Jones is 0.3 m/s. 7 36. C No external forces act on the two-cart system, so the center of mass as the velocity after same the collision as before. Before the collision, vA = 4.0 m/s and vB = 0, so mA vA + mB vB 0.25 kg(4.0 m/s) + 0.75 kg(0) vCM = = 1.0 m/s. Hence, the speed of = mA + mB 0.25 kg + 0.75 kg the CM after the collision is 1.0 m/s. 37. B With no external forces acting on the two particles, momentum of the two-particle system will be conserved: Pxi = Pxf . Since all the motion is in along the x-axis, we only need the x-component of momentum. After the collision, the two particles are stuck together, so they behave as a single particle with the combined mass: mA vA + mB vB = (mA + mB )vcombined. Then: mA vA + mB vB 1.5 kg(10 m/s) + 3.5 kg(2.0 m/s) = 4.4 m/s. = vcombined = mA + mB 1.5 kg + 3.5 kg 38. A With no external forces acting on the two particles, momentum of the two-particle system will be conserved: Pxi = Pxf . Since all the motion is in along the x-axis, we only need the x-component of momentum. After the collision, the two particles are stuck together, so they behave as a single particle with the combined mass: mC vC + mD vD = (mC + mD )vcombined. Then: mC vC + mD vD 2.0 kg(4.5 m/s) + 4.0 kg(3.5 m/s) = 0.83 m/s. vcombined = = mC + mD 2.0 kg + 4.0 kg 39. D In the horizontal (x) direction, the nal momentum of the car-bag system equals its initial value. Initally, the sack has zero momentum in the x-direction. Thus, mcar vi = (mcar + msack )vf , so vf = mcar vi /(mcar + msack ) = 2000 kg(3.0 m/s)/2500 kg = 2.4 m/s 40. E The acceleration of the block is a = F/m = 5.0 N/25 kg = 0.2 m/s2 . The speed acquired is: v = at = 0.2 m/s2 (0.2 s) = 0.04 m/s. The distance moved is: d = 1 at2 = 0.5(0.2 m/s2 )(0.2 s)2 = 0.004 m = 0.4 cm. 2 The momentum acquired is: p = mv = 25 kg(0.04 m/s) = 1.0 kg m/s. Equivalently, p = F t = 5.0 N(0.2 s) = 1.0 kg m/s; since the initial momentum was zero, the nal momentum must be 1.0 kg m/s. The kinetic energy acquired is: K = 1 mv 2 = 0.5(25 kg)(0.04 m/s)2 = 0.02 J. 2 8 41. D Initially: pxi = mv sin and pyi = mv cos . y x After bouncing, pxf = mv sin and pyf = mv cos . So px = pxf pxi = 0 and py = pyf pyi = 2mv cos . 42. C The golf ball initially has zero momentum. After being hit, its momentum is p = mv , where m is the balls mass and v its speed. Thus the change in momentum is p = mv . p mv 0.046 kg(50 m/s) Since p = F t, F = = = = 2300 N. t t 1.0 103 s 43. A The impulse J equals the change in momentum: J = p = mvup (mvdown) = 0.4 kg(35 m/s + 20 m/s) = 22 N s. 44. E rad 2 rad 1 min = 3.5 One revolution is 2 rad so 33 1 rpm is: 33.3 3 1 min 60 sec s 45. B rad 55 rev 2 rad = 48 The angular speed is given by = 7.2 sec 1 rev s 46. E The angular displacement is given by: 0 = o t + 1 t2. Solve for . 2 2 2 900 = 2 [( 0) o t] = rad/s2 = 14 rad/s2 [650 rad (25 rad/s)(8.0 s)] = 2 t (8.0 s) 64 47. C 12 rev/s = 24 rad/s. The average angular acceleration is: avg = 48. A 5.0 rev s min 2 rad 1 rev 1 min 60 s = rad/s2 6 9 24 rad/s = = 6 rad/s2 . t 4.0 s 49. C For constant angular acceleration, = 0 +t. So t = 10 rad/s 20 rad/s 0 = = 15 s. 2.0 rad/s2 50. D For constant angular acceleration, = 0 + t, so o 24 rad/s 48 rad/s rad = = = 4.0 2 t 6.0 s s 51. E 12 rev = 24 rad. For constant angular acceleration: 0 = 0 t + 1 t2 . Here 0 = 0, 2 2( 0) 2(24 rad) so t = = = 20 s = 7.9 s. 2.4 rad/s2 52. A If the angular velocity vector points out of the page and the angular acceleration vector points into the page, the body is decelerating. 53. B The circumference of the wheel is C = d = (0.30 m) = 0.94 m. Then, the cord (of length 5.0 m) wraps around the wheel (5.0/0.94) = 5.3 times. So, for the cord to unwind, the wheel must turn through 5.3 revolutions, which is 2 (5.3) rad = 33.3 rad. We need the time for the wheel to rotate 33.3 rad. For constant angular acceleration: 2( 0) 2(33.3 rad) 0 = 0 t + 1 t2 . Here 0 = 0, so t = = = 6.7 s. 2 1.5 rad/s2 54. C At any instant the wheel has an angular velocity . The speed of any point on the wheel will be v = r , where r is the distance of the point from the center of the wheel. Thus, a point on the outer rim will have twice the speed of a point half-way between the outer rim and center. 55. B This is uniform circular motion, and a = v 2/r = 2 r. Here = 4.5 rev/s = 9 rad/s. Then a = (9 rad/s)2(0.12 m) = 96 m/s2 . 56. B N mi ri 2 , where The moment of inertia for a collection of point masses is given by I = i=1 ri is the distance of each mass from the axis of rotation. Hence, I = m(0)2 + m(a)2 + m(a)2 + m(2a)2 = ma2 + ma2 +4ma2 = 6ma2 = 6(4.0 kg)(0.5 m)2 = 6.0 kg m2 10 57. E The density of the material is the same in both objects. Since the volumes are dierent, the masses must be dierent: MA = R2L and MB = (2R)2 (2L) = 8MA IA = 1 MA RA 2 2 and IB = 1 MB RB 2 , 2 1 MB RB 2 IB RB 2 so =1 2 =8 IA RA MA RA 2 2 = 8(22 ) = 32 58. C L 5 2 I = mi ri = M (0)2 + M ( )2 + M (L)2 = ML2 . 2 4 59. D The rotational inertia of a thin cylinder about its central axis is I = MR2 . This is because all the mass is at the same distance R from the axis. 60. E 2 The rotational inertia of a solid uniform sphere is I = MR2 . An axis tangent to the 5 surface of the sphere is distance R from the axis of the sphere. R 2 7 Using the parallel axis theorem: Inew = I + MR2 = MR2 + MR2 = MR2 5 5 61. C L 1 1 1 Apply the parallel axis theorem: Iend = Icenter + M ( )2 = ML2 + ML2 = ML2 . 2 12 4 3 62. D F 1 r 2 r 1 F 2 The torque on an object is given by = r F . We rst need to know the torque of each of the two forces: 1 = r1 F1 sin and 2 = r2F2 sin . Notice that both torques 11 tend to cause a counterclockwise rotation about the pivot, so both torques are positive. So, 1 = (4.0 m)(5.0 N) sin 30 = 10 N m and 2 = (2.0 m)(5.0 N) sin 30 = 5.0 N m. Then, net = 1 + 2 = 15 N m. 63. B We know that = I and = o + t. Since 0 = 0, = t = t/I . The torque on each of the three objects is the same. Therefore, the angular velocity will be larger if the rotational inertia I is smaller. For these shapes, the rotational inertia is: disk: I = 1 MR2 ; hoop: I = MR2 ; sphere : I = 2 MR2 . 2 5 M and R are the same for all three, so the order for largest to smallest I , is: hoop, disk, sphere. Consequently, the order for least to greatest is the same: hoop, disk, sphere. 64. C rF (0.2 m)(5.0N) = = 0.133 rad/s 2 . I 7.5 N m Starting from rest, the angular displacement is given by: = 1 t2 , so we can nd the 2 time for the disk to rotate through rad: 2 2 rad t= = = 6.87 s. Then = t = (0.133 rad/s)(6.87 s) = 0.91 rad/s. 0.133 rad/s 2 The magnitude of the torque is = rF = I, so = 65. C Because the cord connects the wheel and block, the acceleration of the block a = r, where r is the radius of the wheel and its angular acceleration. The cord has tension T , which exerts and upward force on the block and which applies a tangential force at the rim of the wheel. Thus, the magnitude of the (downward) acceleration of the block is given by: mg T = ma, where m is the mass of the block. The angular acceleration of the wheel is given by: = rT = I = Ia/r, where I is the rotational inertia of the wheel. Then r2 T = r2 (mg ma) = Ia, so 16 kg(9.8 m/s2 )(0.2 m)2 mgr2 = = 5.5 m/s2 . a= 2 2 + 16 kg(0.2 m)2 I + mr 0.50 kg m Or, this problem can be solved using energy conservation: K + U = const. That is, the sum of the rotational kinetic energy of the wheel, the kinetic energy of the block, and the potential energy of the block, is constant. Recall v = r . Then: d1212 d 1 v2 1 2 I + mv + mgy = I + mv + mgy dt 2 2 dt 2 r2 2 = dv dy I dv + mv + mg v 2 r dt dt dt dv dy = v , the v s cancel out, and a = , we get: dt dt be rewritten exactly as the earlier expression for a. Since 12 = 0. I + m a = mg, which can r2 66. C The speed of both blocks is the same. The angular velocity of the pulley is = v/r, where v is the speed of the rope (i.e., the speed of the blocks) which is not slipping, and r is the radius of the pulley. Then the total kinetic energy is: 1 1 1 v2 1 I K = m1 v 2 + m2 v 2 + I 2 = m1 + m2 + 2 v 2 2 2 2r 2 r 4.5 103 kg m2 = 0.5 (2.0 kg + 4.0 kg) + (1.6 m/s)2 = 14 J. (0.03 m)2 67. D If the saw blade rotates at constant angular velocity, the net torque must be zero. Therefore, the motor must provide 0.93 N m of torque, which is exactly cancelled by the torque from the wood. Calculate the angle for which the saw rotates in 1 min: 25 rad/s(60 s) = 1500 rad. Then the work done on the blade by the motor is: W = = (0.93 N m)(1500 rad) = 1395 J = 1400 J. 68. D The motor does 1400 J of work in 60 s, so the power is P = 1400 J/60 s = 23 W. 69. E If the wheel rolls without slipping, when the point P is in contact with the road, it cannot be moving. (If it is in contact with the road and moving at the same time, that is called slipping.) 70. C r1 1 v = r1 1 = rr 2 , so 2 = 1 = 1 . r2 2 71. D 121 If h is the height up the incline, conservation of energy tells us: I + Mv 2 = Mgh. 2 2 They are not slipping, so v = r . Then h= 1 v2 1 v2 I (I 2 + Mv 2) = (I 2 + Mv 2 ) = ( + 1). 2Mg 2Mg r 2g Mr2 I is larger. For these shapes, the rotational inertia is: Mr2 disk: I = 1 MR2 ; hoop: I = MR2 ; sphere : I = 2 MR2 . 2 5 I So, increasing order of is: sphere, disk, hoop. Mr2 So, h is larger when 72. E L2 The magnitude of A B is given by |A B | = AB sin = L2 sin 45 = . 2 13 73. A = k , so k = . The rule is, any exchange introduces a factor of 1. You can ij i j verify this from the right hand rule. Then, ( k ) = () = = 0, since the ii i j ij dot product of perpendicular vectors is zero. 74. A = k , so k = , and k = . You can verify this from the right hand rule. ij i j ji j i Then, k ( k ) = k = 0, since the dot product of perpendicular vectors is zero. 75. E Since S and T are not perpendicular, the dot product S T = |S ||T | cos = 0. This is because = /2, so cos = 0. 76. C L = rp sin . Note that = 180 30 = 150 since is the angle between the vector from O to the particle and the momentum vector. Then L = rp sin = r(mv ) sin = (12 m)(6.0 kg)(4.0 m/s) sin 150 = 144 kg m2/s. Note that L is also just p times the moment arm r = (12 m) sin 30 . 77. A For a disk about its central axis, I = 1 MR2 . For a parallel axis a distance h from the 2 central axis, I = 1 MR2 + Mh2 , by the parallel-axis theorem. Then, 2 L = I = ( 1 MR2 + Mh2 ) . 2 78. A If the angular velocity is constant, the torque must be zero. 79. D The angular momentum is given by L = ri pi = mivi ri . Note that both masses have positive angular momentum, since the rotation about the origin is counterclockwise: L = (6 kg)(2 m/s)(1 m) + (3 kg)(3 m/s)(2 m) = (12 + 18) kg m2 /s = 30 kg m2 /s. 80. C The magnitude of the time rate of change of the particles angular momentum about the point O is just the torque. So, we just rank the magnitude of the torque from least to greatest. We know that = rF sin . We can conclude that the maximum torque occurs when the angle is 90 degrees and the minimum torque occurs when the angle is 0, or 180 degrees. Therefore, 1 and 2 tie for the least torque, 3 has the most torque and 4 is somewhere in between. 81. D Since the string does not slip, = a/R, if a is the magnitude of the acceleration of the masses. Let the tension in the string on the m1 side be T1 and the tension on the m2 side be T2. Then the torque acting on the pulley is = R(T1 T2 ) = I. 14 For m1 , T1 m1g = m1a = m1R; the minus sign reects m1 going down. For m2 , T2 m2g = m2a = m2 R. Then T1 T2 = (m1 m2)g (m1 + m2 )R. Using the torque equation, R[(m1 m2 )g (m1 + m2)R] = I. R(m1 m2 )g Solving for , we nd: = . (m1 + m2 )R2 + I 82. D Conservation of angular momentum tells us that the skaters initial angular momentum must be equal to the nal angular momentum. Therefore, Io o = If f . Hence, Io o o = Io =. If = Io f 4o 4 83. E We can apply the principle of conservation of angular momentum where the initial angular momentum is Li and the nal angular momentum is Lf . Initially, the merrygo-round is not moving, but the running child has angular momentum about its center: Li = mvR. The nal angular momentum includes the rotating merry-go-round, with angular velocity f , and the child who is then riding along (a distance R from the center, so his rotational inertial is mR2 ). Then Lf = (I + mR2)f , where I is the rotational inertial of the merry-go-round alone. Li = Lf , so f = mvR/(I + mR2 ). 84. E Again, angular momentum is conserved, when the child walks to the center. Before walking to the center, Li = (I + mR2 ) . Once he has reached the center, his contribution to the rotational inertia vanishes, so Lf = I , where is the new angular velocity. Since Li = Lf , = (I + mR2 )/I . 85. C Since the collapse is due only to internal forces, i.e. there are no external torques present, angular momentum is conserved. That is, it is L before the collapse, so it is also L after. 86. A We can apply the principle of conservation of angular momentum where the initial angular momentum is Li and the nal angular momentum is Lf : Li = I + 2I 2 and Lf = (I + 2I )f , so f = I + 4I 5 I + 2I 2 = = . I + 2I I + 2I 3 87. A Because the force acts along the radius there is no torque applied to the block. Then the angular momentum of M must be constant. 15 88. D As the string is pulled, the radius decreases. Since L = MRv is constant, v must increase. Thus, the kinetic energy increases, and W = K . 2 Initially kinetic energy is: Ki = 1 Mvi = 0.5(0.6 kg)(2.0 m/s)2 = 1.2 J. 2 Since the angular momentum L = MRv is constant, as R is reduced from 40 cm to 20 cm, the speed v must be doubled to 4.0 m/s. 2 The nal kinetic energy is: Kf = 1 Mvf = 0.5(0.6 kg)(4.0 m/s)2 = 4.8 J. 2 Thus K = 4.8 J 1.2 J = 3.6 J, and that is the work done. 16
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Newton's 2nd Law The acceleration of an object is directly proportional to the force acting on it and inversely proportional to its mass.FormulaFNET = ma F = force(vector) m = mass(scalar) a = acceleration(vector)Small forceBig massSmall Acceleratio
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Impulse and MomentumAP Physics B10/26/11 Aim: What is momentum?L.O.'s Calculate momentum and impulse Derive units for p and JAgenda 1. DNQ 2. Notes-p and JThe linear momentum p of an object of mass m with avelocity of v is p = mvIt is a vector and
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Aarhus Universitet, Handels- og IngeniørHøjskolen - PHYSICS - 20339841
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Aarhus Universitet, Handels- og IngeniørHøjskolen - PHYSICS - 20339841
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Aarhus Universitet, Handels- og IngeniørHøjskolen - PHYSICS - 20339841
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Aarhus Universitet, Handels- og IngeniørHøjskolen - PHYSICS - 20339841
LO'sDefine volume flow rate Apply Bernoulli's equationAim: What is Bernoulli's Equation?Bernoulli's Principle &amp; Fluid ContinuityFluid FlowUp till now, we have pretty much focused on fluids at rest. Now let's look at fluids in motion It is important t