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 jitr ? 4 ? 3 += v , where r v is measured in meters when t is measured in seconds. For each of the following, consider only t > 0. The magnitude of the angular momentum of 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 kmimjit t r mrvmrPrL ? 12) ? 3() ? 4 ? 3( d d ?=×+=×=×=×= v vvv v v r Thus the magnitude of the angular momentum of this particle constant/smkg12 2 =?= 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 ?= r when t = 0s. The magnitude of the angular momentum of this particle is （ B ） (A) decreasing (B) constant. (C) increasing. (D) possibly but not necessarily constant. Solution: The angular momentum of this particle about the origin is PrL v v r ×= , so the position vector of the particle when t = 0s is jr ? 2?= v . Assume the mass of the particle is m, the position vector of the particle at any time t is jit m p jivtr ? 2 ?? 2 ? ?=?= v So the angular momentum of this particle is kkpiPjit m p PrL ? /s)mkg20( ? 2 ? ) ? 2 ? ( 2 ?==×?=×= v v r Thus the magnitude of the angular momentum of this particle constant/smkg20 2 =?=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 have 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 angular velocity necessarily has a constant direction. Solution: Using conservation of angular momentum and the definition of the angular momentum , ω r r IL = 4. A particle is located at kjir ? 0 ? 3 ? 0 ++= r , in meter. A constant force kjiF ? 4 ? 0 ? 0 ++= r (in Newton’s) begins to act on the particle. As the particle accelerates 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 ikjikjiFr ? 12) ? 4 ? 0 ? 0() ? 0 ? 3 ? 0( =++×++=×= v vv τ So it is a nonzero constant. II. Filling the Blanks 1. The total angular momentum of the system of particles pictured in Figure 1 about the origin at O is kmv ? )/smkg4.4( 2 0 ? Solution: Using the definition of the angular momentum vmrPrL vv v v v ×=×= , the total angular momentum of the system of particles is )/smkg( ? 4.4 ? 4.2 ? 4 ? 6 ) ? (2) ? 2 ? 2.1() ? (2) ? 2 ? 1() ? (3) ? 2 ? 4( 2 0000 000 ?=+?= ×?+×++?×+?= kmvkmvkmvkmv jvmjiivmjiivmjiL total v 2. 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 angular momentum of the particle about the origin is 62.472 kg·m 2 /s . Solution: Using the definition of the angular momentum, the angular momentum of the particle about the origin is /smkg1025.6380107.131012sin 2233 ?×=××××===×= ??? dmvrmvvmrL θ vv v 3. The rotor of an electric motor has a rotational inertia I m =2.47×10 -3 kg?m 2 about its central axis. The motor is mounted parallel to the axis of a space probe having a rotational inertia I p =12.6kg?m 2 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 is coaxial, the angular momentum is conserved, we have ωω )( pmmm III += y(m) Fig.1 2.00 3m x(m) 2m 2m -4.00 -2.00 1.00 1.20 iv ? 0 iv ? 0 ? jv ? 0 O Integrate both sides of the equation with respect to time, we get rev354 360 )( 2 )( d)(d 2 1 21 00 21 = + == += += ∫∫ m pm pmm pmmm I II n III tIItI o θ π θ θθ ωω θθ 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 momentum of the system about any origin is mvd . Solution: See figure, select any point O and set up coordinate system. So the total angular momentum of the system about the origin is mvdrrmv mvrmvr vmrvmrLLL total =+= += ×+×=+= )sinsin( sinsin 2211 2211 2121 θθ θθ vvvv vvv 5. A particle is located at kjir ? )m85.0( ? )m36.0( ? )m54.0( +?+= r . A constant force of magnitude 2.6 N acts on the particle. When the force acts in the positive x direction, the components of the torque about the origin is kj ? 936.0 ? 21.2 + , and when the force acts in the negative x direction, the components of the torque about the origin is kj ? 936.0 ? 21.2 ?? . Solution: According to the definition of the torque Fr v vv ×=τ , when the force is i ? N6.2 , the torque about the origin is kjikjiFr ? 936.0 ? 21.2) ? 6.2() ? 85.0 ? 36.0 ? 54.0( +=×+?=×= v rv τ when the force is i ? N6.2? , the torque about the origin is kjikjiFr ? 936.0 ? 21.2) ? 6.2() ? 85.0 ? 36.0 ? 54.0( ??=?×+?=×= v rv τ 6. A cylinder having a mass of 1.92kg rotates about its axis of symmetry. Forces are applied as shown in Fig.2: F 1 =5.88N, F 2 =4.13N, and F 3 =2.12N. Also R 1 =4.93cm and R 2 =11.8cm. The magnitude and direction of the angular acceleration of the cylinder are 90.3rad/s 2 . Solution: Using the definition of the torque Fr v vv ×=τ , suppose the clockwise is the positive. The magnitude of the total torque about its axis of symmetry is mN2.101093.412.2108.1113.4108.1188.5 222 132221 ?=××?××?××= ??= ??? RFRFRFτ y x iv ? 0 iv ? 0 ? O d 1 2 r 1 r 2 1 θ 2 θ 1 R 2 R 3 F r 1 F r 2 F r o 30 Fig.2 Using the equation ατ vv I total = , and 222 mkg113.0108.1192.1 2 1 2 1 ?=×××== ? MRI Thus the magnitude of the angular acceleration of the cylinder is 2 rad/s3.90 113.0 2.10 === I τ α The direction of he angular acceleration is counterclockwise, point out of the page. 7. A disk of mass m and radius R is free to turn about a fixed, horizontal axle. The disk has an ideal string wrapped around its periphery from which another mass m (equal to the mass of the disk) is suspended, as indicated in Figure 3. The magnitude of the acceleration of the falling mass is 2g/3 , the magnitude of the angular acceleration of the disk is 2g/3R . Solution: Assume the acceleration of the falling mass is a, and the angular acceleration of the disk isα . We have ? ? ? ? ? ? ? == = ? ? ? ? ? ? ? ? = =? = R g R a g a mR TR maTmg Ra 3 2 3 2 2 2 α α α 8. A uniform beam of length l is in a vertical position with its lower end on a rough surface that prevents this end from slipping. The beam topples. At the instant before impact with the floor, the angular speed of the beam about its fixed end is l g3 . Solution: Use the CWE theorem. Assume the zero PE is at the point of the end of the beam. So l g mglmlmglI mglE IE CM i CMf 3 2 1 3 1 2 1 2 1 2 1 0 2 1 0 2 1 2 222 2 =?=???=? ? ? ? ? ? ? ? += += ωωω ω III. Give the Solutions of the Following Problems 1. A pulley having a rotational inertia of 1.14×10 -3 kg?m 2 and a radius of 9.88cm is acted on by a force, applied tangentially at its rim that varies in time as F=At+Bt 2 , where A=0.496N/s and B=0.305N/s 2 . If the pulley was initially at rest, find its angular speed after 3.60s. Solution: m m R Fig.3 Using the equation ατ v v vv IFr =×= , we have 22 3 2 433.26987.42)305.0496.0( 1014.1 1088.9 tttt I rF z +=+ × × == ? ? α For td dω α = , so 2 433.26987.42 d d tt t +== ω α , If the pulley was initially at rest, thus its angular speed after 3.60s is rad/s6.689811.8494.21d)433.26987.42( 6.3 0 32 6.3 0 2 =+=+= ∫ tttttω 2. Two identical blocks, each of mass M, are connected by a light string over a frictionless pulley of radius R and rotational inertia I (Fig.4). The string does not slip on the pulley, and it is not known whether or not there is friction between the plane and sliding block. When this system is released, it is found that the pulley turns through an angle θ in time t and the acceleration of the blocks is constant. (a) What is the angular acceleration of the pulley? (b) What is the acceleration of the two blocks? (c) What are the tensions in the upper and lower sections of the string? All answer are to be expressed in terms of M, I, R, θ, g, and t. Solution: Sketch the forces diagram of the system shown in figure. Apply the Newton’s second law and the counterpart of Newton’s law for a spinning rigid body, we have ? ? ? ? ? ? ? ? ? = = =? =? 2 21 1 2 1 )( t Ra IRTT MaTMg αθ α α Solving them, we get (a) The angular acceleration of the pulley is 2 2 t θ α = . M M T 2 T 1 I Fig.4 Mg f T 1 T 2 T 1 T 2 N Mg N mg a v (b) The acceleration of the two blocks is 2 2 t R a θ = (c) The tensions in the lower sections of the string is ) 2 ()( 2 1 t R gMagMT θ ?=?= (d) The tensions in the lower sections of the string is 22 12 2 ) 2 ( t I t R gM R I TT θθα ??=?= 3. A disk with moment of inertia I 1 is rotating with initial angular speed ω 0 ; a second disk with moment of inertia I 2 initially is not rotating (see Figure 5). The arrangement is much like a LP record ready to drop onto an unpowered, freely spinning turntable. The second disk drops onto the first and friction between them brings them to a common angular speed. Find the common angular speed ω. Solution: Let two disks be a system, the total torque about the axis is zero, applying the conservation of angular momentum, we have the common angular speed ω. 0 21 1 01210121 )( ωωωωωωω II I IIIIIILL if + =?=+?=+?= rrrr rr 4. Two cylinder having radii R 1 and R 2 and rotational inertias I 1 and I 2 , respectively, are supported by axes perpendicular to the plane of Fig.6. The large cylinder is initially rotating with angular velocity ω 0 . The small cylinder is moved to the fight until it touches the large cylinder and is caused to rotate by the frictional force between the two. Eventually, slipping ceases, and the two cylinders rotate at constant rates in opposite directions. Find the final angular velocity ω 2 of the small cylinder in terms of I 1 , I 2 , R 1 , R 2 , and ω 0 . (Hint: Angular momentum is not conserved. Apply the angular impulse equation to each cylinder.) Solution: Apply the rotational counterpart of Newton’s second law of motion t L total d d v v =τ , we get ? ? ? ? ? ? ? = ?=? ?= ∫ ∫ 2211 01111 222 d 0d RR IItfR ItfR ωω ωω ω Solving the equations, we can get the final angular velocity ω 2 of the small cylinder is 0 ω r I 2 I 1 Direction of spin Fig.5 I 2 I 1 R 2 R 1 ω 0 Fig.6 + + 2 12 2 21 2101 2 RIRI RRI + = ω ω 5. One way to determine the moment of inertia of a disk of radius R and mass M in the laboratory is to suspend a mass m from it to give the disk an angular acceleration. The mass m is attached to a string that is wound around a small hub of radius r, as shown in Figure 7. The mass and moment of the inertia of the hub can be neglected. The mass m is released and takes a time t to fall a distance h to the floor. The total torque on the disk is due to the torque of the tension in the string and the small but unknown frictional torque frictn τ r of the axle on the disk. Once the mass m hits the floor, the disk slowly stops spinning during an additional time t ′ because of the sole influence of the small frictional torque on the disk. Find the moment of inertia of the disk in terms of m, r, g, h, t and t ′. Solution: Assume the acceleration of the falling mass is a, and the angular acceleration of the disk isα . The angular speed of the disk is tαω = while the falling mass reaches the floor. So 2 2 2 2 1 t h aath =?= Using Newton’s second law ? ? ? ? ? = =? =? α ατ ra ITr maTmg fric (1) Apply the rotational analog of Newton’s second law to the disk t LL total total ? ? ≈= rr r dt d τ During the course of the mass hits the floor '' 0 t It I t fric αω τ = ? = (2) Using the equation (1) and (2), we get the moment of inertia of the disk ' 1 )1 2 ( ' 1 1)( 2 2 2 t t h gt mr t t a agmr I + ? = + ? ? = m M R Fig.7 r Hub Wheel