Physics Solution Manual for 1100 and 2101

10 kg m 2 0037 rads 0067 rads 050 kg 040

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Unformatted text preview: 7.0 m/s 55. SSM REASONING AND SOLUTION The only force that does work on the cylinders as they move up the incline is the conservative force of gravity; hence, the total mechanical energy is conserved as the cylinders ascend the incline. We will let h = 0 m on the horizontal plane at the bottom of the incline. Applying the principle of conservation of mechanical energy to the solid cylinder, we have 2 2 mghs = 1 mv0 + 1 Isω0 2 2 (1) where, from Table 9.1, Is = 1 mr 2 . In this expression, v0 and ω0 are the initial translational 2 and rotational speeds, and hs is the final height attained by the solid cylinder. Since the cylinder rolls without slipping, the rotational speed ω0 and the translational speed v0 are related according to Equation 8.12, ω0 = v0 / r . Then, solving Equation (1) for hs , we obtain hs = 2 3v0 4g Repeating the above for the hollow cylinder and using I h = mr 2 we have 476 ROTATIONAL DYNAMICS hh = 2 v0 g The height h attained by each cylinder is related to the distance s traveled along the incline and the angle θ of the incline by h h ss = s and sh = h sin θ sin θ Dividing these gives ss = 3/4 sh 56. REASONING a. The kinetic energy of the rolling wheel is the sum of its translational rotational ( 12 mv2 ) and ( 12 Iω 2 ) kinetic energies. In these expressions m and I are, respectively, the mass and moment of inertia of the wheel, and v and ω are, respectively, its linear and angular speeds. The sliding wheel only has translational kinetic energy, since it does not rotate. b. As the wheels move up the incline plane, the total mechanical energy is conserved, since only the conservative force of gravity does work on each wheel. Thus, the initial kinetic energy at the bottom of the incline is converted entirely into potential energy when the wheels come to a momentary halt. The potential energy PE is given by PE = mgh (Equation 6.5), where h is the height of the wheel above an arbitrary zero level. SOLUTION a. Since the rolling wheel is a disk, its moment of inertia is I = 1 mR 2 2 (see Table 9.1), where R is the radius of the disk. Furthermore, its angular speed ω is related to the linear speed v of its center of mass by Equation 8.12 as ω = v/R. Thus, the total kinetic energy of the rolling wheel is Rolling Wheel KE = 1 2 2 1 2 mv + = 3 mv 2 = 4 3 4 I ω = mv + 2 1 2 2 ( 11 22 ( 2.0 kg )( 6.0 m/s )2 = mR 2 54 J The kinetic energy of the sliding wheel is Sliding Wheel KE = 1 mv 2 2 = 1 2 ( 2.0 kg )( 6.0 m/s )2 = ) 36 J v R 2 Chapter 9 Problems 477 b. As each wheel rolls up the incline, its total mechanical energy is conserved. The initial kinetic energy KE at the bottom of the incline is converted entirely into potential energy PE when the wheels come to a momentary halt. Thus, the potential energies of the wheels have the values calculated in part a for the total kinetic energies. The potential energy of a wheel is given by Equation 6.5 as PE = mgh, where g is the acceleration due to gravity and h is the height relative to an arbitrary zero level. Therefore, the height reached by each wheel is as follows: Rolling Wheel h= PE 54 J = = 2.8 m mg ( 2.0 kg ) 9.80 m/s2 Sliding Wheel h= PE 36 J = = 1.8 m mg 2.0 kg ) 9.80 m/s 2 ( ( ) ( ) 57. REASONING AND SOLUTION The conservation of energy gives 2 2 2 mgh + (1/2) mv + (1/2) Iω = (1/2) mv0 + (1/2) Iω0 2 If the ball rolls without slipping, ω = v/R and ω0 = v0/R. We also know that I = (2/5) mR2. Substitution of the last two equations into the first and rearrangement gives 2 v = v0 − 10 gh = 7 ( 3.50 m/s )2 − 10 ( 9.80 m/s2 ) ( 0.760 m ) = 7 1.3 m/s 58. REASONING We first find the speed v0 of the ball when it becomes airborne using the conservation of mechanical energy. Once v0 is known, we can use the equations of kinematics to find its range x. SOLUTION When the tennis ball starts from rest, its total mechanical energy is in the form of gravitational potential energy. The gravitational potential energy is equal to mgh if we take h = 0 m at the height where the ball becomes airborne. Just before the ball becomes airborne, its mechanical energy is in the form of rotational kinetic energy and translational 2 kinetic energy. At this instant its total energy is 1 mv0 + 1 I ω 2 . If we treat the tennis ball as 2 2 a thin-walled spherical shell of mass m and radius r, and take into account that the ball rolls down the hill without slipping, its total kinetic energy can be written as 1 2 mv0 2 + Iω = 1 2 2 1 2 mv0 2 + v 12 ( mr 2 ) 0 23 r 2 5 2 = 6 mv0 478 ROTATIONAL DYNAMICS Therefore, from conservation of mechanical energy, he have 5 6 2 mgh = mv0 or v0 = 6 gh 5 The range of the tennis ball is given by x = v0 xt = v0 (cos θ ) t , where t is the flight time of the ball. From Equation 3.3b, we find that the flight time t is given by t= v − v0 y ay = (−v0 y ) − v0 y ay =– 2v0 sin θ ay Therefore, the range of the tennis ball is 2v sin θ x = v0 xt = v0 (cos θ ) – 0 ay If we take upward as the positive direction, then using the fact that a y = – g and the expression for v0 given above, we find 2 2cos θ sin θ 2 2cos θ sin θ 6 gh 12 x= = h cos θ sin θ v0 = g g 5 5 12 (1.8 m) (cos 35°)(sin 35°)= 2.0 m 5 = 59. SSM WWW REASONI...
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