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1、University Physics AI No. 8 Spin and Orbital Motion Class Number Name I. Choose the Correct Answer 1. A particle moves with position given by ji tr43+=v, where rvis measured in meters when t is measured in seconds. For each of the following, consider only t 0. The magnitude of the angular momentum o
2、f this particle about the origin is （ B ） (A) increasing in time. (B) constant in time. (C) decreasing in time. (D) undefined Solution: Assume the mass of the particle is m, so the angular momentum of this particle about the origin is kmimji ttrmrvmrPrL12)3()43(dd=+=vvvvvvrThus the magnitude of the
3、angular momentum of this particle constant/smkg122=mL. 2. A particle moves with constant momentum ip)m/skg10(=r. The particle has an angular momentum about the origin of kL)/smkg20(2=rwhen t = 0s. The magnitude of the angular momentum of this particle is （ B ） (A) decreasing (B) constant. (C) increa
4、sing. (D) possibly but not necessarily constant. Solution: The angular momentum of this particle about the origin is PrLvvr=, so the position vector of the particle when t = 0s is jr2=v. Assume the mass of the particle is m, the position vector of the particle at any time t is ji tmpjivtr22=vSo the
5、angular momentum of this particle is kkpiPji tmpPrL/s)mkg20(2)2(2=vvrThus the magnitude of the angular momentum of this particle constant/smkg202=L. 3. A solid object is rotating freely without experiencing any external torques. In this case （ A ） (A) Both the angular momentum and angular velocity h
6、ave constant direction. (B) The direction of angular momentum is constant but the direction of the angular velocity might not be constant. (C) The direction of angular velocity is constant but the direction of the angular momentum might not be constant. (D) Neither the angular momentum nor the angul
7、ar velocity necessarily has a constant direction. Solution: Using conservation of angular momentum and the definition of the angular momentum , rrIL = 4. A particle is located at kjir030+=r, in meter. A constant force kjiF400+=r(in Newtons) begins to act on the particle. As the particle accelerates
8、under the action of this force，the torque as measured about the origin is （ D ） (A) increases. (B) decreases. (C) is zero. (D) is a nonzero constant. Solution: The torque as measured about the origin is ikjikjiFr12)400()030(=+=vvv So it is a nonzero constant. II. Filling the Blanks 1. The total angu
9、lar momentum of the system of particles pictured in Figure 1 about the origin at O is kmv)/smkg4 . 4(2 0 Solution: According to the definition of the angular momentum vmrPrLvvvvv=, the total angular momentum of the system of particles is )/smkg(4 . 44 . 246)(2)22 . 1 ()(2)21 ()(3)24(2 0000000=+=+=km
10、vkmvkmvkmvjvmjiivmjiivmjiLtotalv2. A particle of mass 13.7g is moving with a constant velocity of magnitude 380m/s. The particle, moving in a straight line, passes with the distance 12cm to the origin. The magnitude of the angular momentum of the particle about the origin is 62.472 kgm2/s . Solution
11、: By the definition of the angular momentum, the angular momentum of the particle about the origin is /smkg1025. 6380107 .131012sin2233=dmvrmvvmrLvvv3. The rotor of an electric motor has a rotational inertia Im=2.4710-3kgm2 about its central axis. The motor is mounted parallel to the axis of a space
12、 probe having a rotational inertia Ip=12.6kgm2 about its axis. The number of revolutions of the motor required to turn the probe through 25.0 about its axis is 354rev . Solution: Assume the two axes are coaxial, the angular momentum is conserved, we have )(pmmmIII+= Integrate both sides of the equat
13、ion with respect to time, we get y(m) Fig.1 2.00 3m x(m) 2m 2m -4.00 -2.00 1.00 1.20 iv0iv0jv0Orev354360)( 2)(d)(d21210021=+=+=+=mpmpmmpmmmIIInIIItIItIo 4. Two particles each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. The total angular moment
14、um of the system about any origin is mvd . Solution: Select a point O as an origin and set up coordinate system as shown in figure. The total angular momentum of the system about the origin is 5. A particle is located at kjir)m85. 0()m36. 0()m54. 0(+=r. A constant force of magnitude 2.6 N acts on th
15、e particle. When the force acts in the positive x direction, the components of the torque about the origin is kj936. 021. 2+, and when the force acts in the negative x direction, the components of the torque about the origin is kj936. 021. 2. Solution: According to the definition of the torque Frvvv=, If the force is iN6 . 2, the torque about the origin is kjikjiFr936. 021. 2)6 . 2()85. 036. 054. 0(+=+=vrv If the for
