Physics Solution Manual for 1100 and 2101

Substituting i p0 mr02 and i pf mrf2 equation 96 for

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Unformatted text preview: NG Let the two disks constitute the system. Since there are no external torques acting on the system, the principle of conservation of angular momentum applies. Therefore we have Linitial = L final , or I A ω A + I Bω B = ( I A + I B ) ω final This expression can be solved for the moment of inertia of disk B. SOLUTION Solving the above expression for I B , we obtain IB = IA F ω G ω H final B –ωA – ω final I = ( 3.4 kg ⋅ m ) L –2.4 rad / s – 7.2 rad / s O 4.4 kg ⋅ m = J M rad / s – (–2.4 rad / s) P –9.8 K N Q 2 2 Chapter 9 Problems 479 60. REASONING Before any sand strikes the disk, only the disk is rotating. After the sand has landed on the disk, both the sand and the disk are rotating. If the sand and disk are taken to be the system of objects under consideration, we note that there are no external torques acting on the system. As the sand strikes the disk, each exerts a torque on the other. However, these torques are exerted by members of the system, and, as such, are internal torques. The conservation of angular momentum states that the total angular momentum of a system remains constant (is conserved) if the net average external torque acting on the system is zero. We will use this principle to find the final angular velocity of the system. SOLUTION The angular momentum of the system (sand plus disk) is given by Equation 9.10 as the product of the system’s moment of inertia I and angular velocity ω, or L = Iω. The conservation of angular momentum can be written as I3 1ω 2 Final angular momentum = I 0ω0 13 2 Initial angular momentum where ω and ω0 are the final and initial angular velocities, respectively, and I and I0 are final and initial moments of inertia. The initial moment of inertia is given, while the final moment of inertia is the sum of the values for the rotating sand and disk, I = Isand + I0. We note that the sand forms a thin ring, so its moment of inertia is given by (see Table 9.1) 2 Isand = M sand Rsand , where Msand is the mass of the sand and Rsand is the radius of the ring. Thus, the final angular velocity of the system is, then, I0 I0 I0 = ω0 = ω0 2 I M + I0 Rsand + I 0 I sand sand ω = ω0 0.10 kg ⋅ m 2 = 0.037 rad/s = ( 0.067 rad/s ) ( 0.50 kg )( 0.40 m )2 + 0.10 kg ⋅ m 2 61. REASONING We consider a system consisting of the person and the carousel. Since the carousel rotates on frictionless bearings, no net external torque acts on this system. Therefore, its angular momentum is conserved, and we can set the total angular momentum of the system before the person hops on equal to the total angular momentum afterwards. SOLUTION The initial angular momentum of the carousel is I carousel ω0 (Equation 9.10) , where Icarousel is the moment of inertia of the carousel and ω0 is its initial angular velocity. After the person climbs aboard, the total angular momentum of the carousel and person is I carouselωf + I personωf , where ωf is the final angular velocity. According to Equation 9.6, the 480 ROTATIONAL DYNAMICS person’s moment of inertial is I person = MR 2 , since he is at the outer edge of the carousel, which has a radius R. Applying the conservation of angular momentum, we have I carouselωf + I personωf 144424443 = I carouselω 1 24 0 43 Initial total angular momentum Final total angular momentum ωf = = I carouselω0 I carouselω0 = I carousel + I person I carousel + MR 2 (125 kg ⋅ m2 ) ( 3.14 rad/s) 125 kg ⋅ m2 + ( 40.0 kg ) (1.50 m ) 2 = 1.83 rad/s 62. REASONING Once the motorcycle is in the air, it is subject to no net external torque, since gravity and air resistance are being ignored. Therefore, its total angular momentum is conserved: Lf = L0 . The total angular momentum of the motorcycle is the sum of the angular momentum LE = I EωE (Equation 9.10) of the engine and the angular momentum LM = I MωM of the rest of the motorcycle (including the rider): I EωEf + I ω = I EωE0 + I ω 144 M Mf 144 M M0 244 3 244 3 Lf (1) L0 We will use Equation (1) to find the ratio IE/IM of the moments of inertia of the engine and the rest of the motorcycle. We note that, so long as all angular velocities are expressed in rev/min, there is no need to convert to SI units (rad/s). SOLUTION Initially, only the engine is rotating, so the rest of the motorcycle has no angular velocity: ωM0 = 0 rad/s. Solving Equation (1) for the ratio IE/IM, we obtain I EωEf + I MωMf = I EωE0 + I M ( 0) or I E (ωEf − ωE0 ) = − I MωMf or −ωMf IE = I M ωEf − ωE0 As usual, clockwise rotation is negative, and counterclockwise is positive. The ratio of the moments of inertia is, then, IE IM = −ωMf ωEf − ωE0 = − ( −3.8 rev/min ) ( +12 500 rev/min ) − ( +7700 rev/min ) = 7.9 ×10−4 Chapter 9 Problems 481 63. REASONING The rod and bug are taken to be the system of objects under consideration, and we note that there are no external torques acting on the system. As the bug crawls out to the end of the rod, each exerts a torque on the other. However, these torques are internal torques. The conservation of angular momentum states that the total angular momentum of a system re...
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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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