University Physics Al No 8 Spin and Orbital Motion Class Number ame L. Choose the Correct Answer A particle moves with position given by P= 3ti +4j, where F is measured in meters when t is measured in seconds. For each of the following, consider only 1>0. The magnitude of the angular momentum of this particle about the origin is (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 L=F×P=Fxm=Pmx,=(3n+4)×m(31)=-12mk Thus the magnitude of the angular momentum of this particle L=12mkg m /s=constant 2. A particle moves with constant momentum p=(10kg. m/s)i. The particle has an angul momentum about the origin of L=(20kg. m/s)k when [=Os. The magnitude of the angul momentum of this particle is (A)decreasing (B) not necessarily constant Solution: The angular momentum of this particle about the origin is L=rx P, so the position vector of the particle when (=Os is r=-2j Assume the mass of the particle is m, the position vector of the particle at any time t is F=vti-2j=Pni-2 So the angular momentum of this particle L=fxP=(Pti-2j) Pi=2pk=20kg. m/s)k Thus the magnitude of the angular momentum of this particle L=20kg. m"/s=constant 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
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 r ti j = 3 ˆ + 4 ˆ 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 ti j m i mk t r L r P r mv r m ˆ (3 ˆ 4 ˆ) (3ˆ) 12 d d = × = × = × = + × = − v v v v v v r Thus the magnitude of the angular momentum of this particle 12 kg m /s constant 2 L = m ⋅ = . 2. A particle moves with constant momentum p i ˆ = (10kg ⋅m/s) r . The particle has an angular momentum about the origin of L k ˆ (20kg m /s) 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 L r P v v r = × , so the position vector of the particle when t = 0s is r j = −2 ˆ v . Assume the mass of the particle is m, the position vector of the particle at any time t is ti j m p r vti j = ˆ − 2 ˆ = ˆ − 2 ˆ v So the angular momentum of this particle is ti j Pi pk k m p L r P ˆ (20kg m /s) ˆ ( ˆ 2 ˆ) ˆ 2 2 = × = − × = = ⋅ v v r Thus the magnitude of the angular momentum of this particle 20kg m /s constant 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 L=I 4. A particle is located at r=Oi+3j+Ok, in meter. A constant force F=Oi+0j+4k (in Newton's)begins to act on the particle. As the particle accelerates under the action of this force, the torque as measured ab out the origin (A)increases.(B)decrease (C)is zero. (D)is a nonzero constant Solution: The torque as measured about the origin is z=F×F=(0+3j+0k)×(0i+0j+4k)=12i So it is a nonzero constant I Filling the blank 1. The total angular momentum of the system of i at o --202 (4.4mvo kg m /s)k -4.00 Solution: Using the definition of the angular -200 Vo momentum=F×P=F×m, the total angular momentum of the system of particles is Loa=(-4+2j)×3m(-v0i)+(li+2))×2m(v01)+(1.21-2j×2m(vj) 6mvok-4mvok+2.4my k= 4.4mvok(kg. m2/s) 2. A particle of mass 137g 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 / s Solution: Using the definition of the angular momentum, the angular momentum of the particle about the L=F×m=msin6=lm=12×10-3×137×10-3×380=625×10-2kg·m2/s 3. The rotor of an electric motor has a rotational inertia Im=2.47x10. m about its central axis The motor is mounted parallel to the axis of a space probe having a rotational inertia Ip=12.6kg. m about its axis. The number of revolutions of the motor required to turn the probe through 25.00 about its axis is 3 54rev Solution: assume the two axes is coaxial. the angular momentum is conserved we have LOm=(m+l)o
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 L = I 4. A particle is located at r i j k ˆ = 0ˆ + 3 ˆ + 0 r , in meter. A constant force F i j k ˆ = 0ˆ + 0 ˆ + 4 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 r F i j k i j k i 12ˆ )ˆ ) (0ˆ 0 ˆ 4 ˆ = × = (0ˆ + 3 ˆ + 0 × + + = v v v 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 mv k ˆ (4.4 kg m /s) 2 0 ⋅ Solution: Using the definition of the angular momentum L r P r mv v v v v v = × = × , the total angular momentum of the system of particles is (kg m /s) ˆ 4.4 ˆ 2.4 ˆ 4 ˆ 6 )ˆ ) 2 ( 2ˆ 2 ˆ ) (1. ˆ ) 2 ( 1ˆ 2 ˆ ) ( ˆ ) 3 ( 4ˆ 2 ˆ ( 2 0 0 0 0 0 0 0 = − + = ⋅ = − + × − + + × + − × mv k mv k mv k mv k L i j m v i i j m v i i j m v j 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·m2 /s . Solution: Using the definition of the angular momentum, the angular momentum of the particle about the origin is sin 12 10 13.7 10 380 6.25 10 kg m /s 3 3 2 2 = × = = = × × × × = × ⋅ − − − L r mv rmv θ dmv v v v 3. The rotor of an electric motor has a rotational inertia Im=2.47×10-3kg⋅m 2 about its central axis. The motor is mounted parallel to the axis of a space probe having a rotational inertia Ip=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 Imω m = (Im + I p )ω y(m) Fig.1 2.00 3m x(m) 2m 2m -4.00 -2.00 1.00 1.20 v i ˆ 0 v i ˆ − 0 v j ˆ 0 O
Integrate both sides of the equation with respect to time, we get LO dt=l(I +I ad m=(m+ln)02 e(Im+1,e2 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 md Solution See figure, select any point O and set up coordinate stem. So the total angular momentum of the system about the origin is tol =l+l,=F xmv+F, xmv = rmvsin 8,+r,mosin 82 =mv(r sin 8,+r, sin 82)=md 5. A particle is located at r=(0.54m)i+(-036m)j+(0.85m)k. 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 2.21j+0.936k, and when the force acts in the negative x direction, the components of the torque about the origin is -2. 21j-0.936k Solution: According to the definition of the torque t=Px F when the force is 2.6Ni, the torque about the origin is z=F×F=(0.54-0.36j+0.85k)×(26i)=2.21j+0.936 when the force is -2.6Ni, the torque about the origin is z=F×F=(0.54-0.36j+0.85k)×(-261)=-2.21j-0.936k 6. A cylinder having a mass of l%2kg rotates about its axis of symmetry. F. Forces are applied as shown in Fig. 2: F1 5.88N, F2=4. 13N, and F3=2. 12N. Also R =4.93cm and R2=11.8cm. The magnitude and direction R of the angular acceleration of the cylinder are_ 90.brad/s R↓F30 Solution: Using the definition of the torque t rx F, suppose the clockwise is the positive. The magnitude of the total torque about its axis [=FR-FR2-F,R 588×118×10-2-4.13×118×10-2-212×493×10-2=102N.m
Integrate both sides of the equation with respect to time, we get 354rev 360 ( ) 2 ( ) d ( ) d 1 2 1 2 0 0 1 2 = + = = = + = + ∫ ∫ m m p m m p m m m p I I I n I I I I t I I t o θ π θ θ θ ω ω θ θ 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 mv r r mvd r mv r mv L L L r mv r mv total = + = = + = + = × + × ( sin sin ) sin sin 1 1 2 2 1 1 2 2 1 2 1 2 θ θ θ θ v v v v v v v 5. A particle is located at r i j k ˆ (0.85m) ˆ ( 0.36m) ˆ = (0.54m) + − + 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 j k ˆ 2.21ˆ + 0.936 , and when the force acts in the negative x direction, the components of the torque about the origin is j k ˆ − 2.21ˆ − 0.936 . Solution: According to the definition of the torque r F v v v τ = × , when the force is i 6Nˆ 2. , the torque about the origin is r F i j k i j k ˆ ) (2.6ˆ) 2.21ˆ 0.936 ˆ = × = (0.54ˆ − 0.36 ˆ + 0.85 × = + r v v τ when the force is i 6Nˆ − 2. , the torque about the origin is r F i j k i j k ˆ ) ( 2.6ˆ) 2.21ˆ 0.936 ˆ = × = (0.54ˆ − 0.36 ˆ + 0.85 × − = − − r v v τ 6. A cylinder having a mass of 1.92kg rotates about its axis of symmetry. Forces are applied as shown in Fig.2: F1=5.88N, F2=4.13N, and F3=2.12N. Also R1=4.93cm and R2=11.8cm. The magnitude and direction of the angular acceleration of the cylinder are 90.3rad/s2 . Solution: Using the definition of the torque r F v v v τ = × , suppose the clockwise is the positive. The magnitude of the total torque about its axis of symmetry is 5.88 11.8 10 4.13 11.8 10 2.12 4.93 10 10.2 N m 2 2 2 1 2 2 2 3 1 = × × − × × − × × = ⋅ = − − − − − τ F R F R F R y x v i ˆ 0 v i ˆ − 0 O d 1 2 r1 r2 θ1 θ 2 R1 R2 F3 r F1 r F2 r o 30 Fig.2
Using the equation T=la,andl=MR2=×1.92×118×10-2=0.113kgm2 Thus the magnitude of the angular acceleration of the cylinder is 102 90.3rad/s2 0.113 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 is a We have 3 a= Ra TP≈mR2a R 3R 8. A uniform beam of length / 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 Solution: Use the Cwe theorem. Assume the zero pe is at the point of the end of the beam. So E l,2+0 2 2 23m12D=mgl→0=y1 E==mgl+0 III. Give the Solutions of the following problems 1. A pulley having a rotational inertia of 1. 14x10 kg m" and a radius of 9.88cm is acted on by force, applied tangentially at its rim that varies in time as F=At+ BI where A=0.496N/s and B=0.305N/s". If the pulley was initially at rest, find its angular speed after 3. 60s Solution
Using the equation τ α v v I total = , and 2 2 2 1.92 11.8 10 0.113 kg m 2 1 2 1 = = × × × = ⋅ − I MR Thus the magnitude of the angular acceleration of the cylinder is 2 90.3 rad/s 0.113 10.2 = = = 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 mg T ma a R 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 3g . Solution: Use the CWE theorem. Assume the zero PE is at the point of the end of the beam. So l g I mgl ml mgl E mgl E I CM i f CM 3 2 1 3 1 2 1 2 1 2 1 0 2 1 0 2 1 2 2 2 2 2 ⇒ = ⇒ ⋅ ⋅ = ⇒ = ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ = + = + ω ω ω ω III. Give the Solutions of the Following Problems 1. A pulley having a rotational inertia of 1.14×10-3kg⋅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/s2 . If the pulley was initially at rest, find its angular speed after 3.60s. Solution: m m R Fig.3
Using the equation t=FXF=la, we have rF9.88×10 11.14x10~3(04961+030512)=42.9871+20643y3 42.987t+26.433t If the pulley was initially at rest, thus its angular speed after 3. 60s is O=(42.987+264332)d=21492+88r1=6896rads 2. Two identical blocks, each of mass M, are connected by a 72 ight 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 8 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 wo 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,,R, e, g, and t Solution: Sketch the forces diagram of the system shown in figure N Mg m8↓T1 Apply the Newton's second law and the counterpart of Newtons law for a spinning rigid body, we Mg -l= Ma (l-TR=la a=aR Solving them, we get a)The angular acceleration of the pulley is a
Using the equation τ α v v v v = r × F = I , we have 2 2 3 2 (0.496 0.305 ) 42.987 26.433 1.14 10 9.88 10 t t t t I rF z + = + × × = = − − α For dt dω α = , so 2 42.987 26.433 d d t t t = = + ω α , If the pulley was initially at rest, thus its angular speed after 3.60s is (42.987 26.433 )d 21.494 8.811 689.6 rad/s 3.6 0 2 3 3.6 0 2 = + = + = ∫ ω t t t t t 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 1 2 1 2 1 ( ) t a R T T R I Mg T Ma θ α α α Solving them, we get (a) The angular acceleration of the pulley is 2 2 t θ α = . M M T2 T1 I Fig.4 Mg f T1 T2 T1 T2 N Mg N mg a v
(b) The acceleration of the two blocks is a= < op 2OR (c) The tensions in the lower sections of the string is T=M(g-a)=M(g-) 2R 201 (d)The tensions in the lower sections of the string is T,=T M(g--2)--2 R 3. A disk with moment of inertia I, is rotating with initial angular speed oo; a second disk with moment of inertia I2 initially is not rotating(see Figure 5). The arrangement is much like a LP record ready to drop Direction of onto an unpowered, freely spinning turntable. The h second disk drops onto the first and friction between them brings them to a common angular speed. Find the common angular speed o 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 o L 1o+120=1o0→(l1+12)b=l 1+12 4. Two cylinder having radii R, and R2 and rotational inertias 1, and 12, respectively, are supported by axes perpendicular to the plane of E如时四h( l1 linder 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 h of the small cylinder in terms of 11, 12, R1, R2, and @o(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 dt ∫R,r=102-0 we get 「R,t=1a-1mo O,R=O2R2 Solving the equations, we can get the final angular velocity a of the small cylinder is
(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 ( ) ( 1 2 t R T M g a M g θ = − = − (d) The tensions in the lower sections of the string is 2 1 2 2 2 ) 2 ( t I t R M g R I T T α θ θ = − = − − 3. A disk with moment of inertia I1 is rotating with initial angular speed ω0; a second disk with moment of inertia I2 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 1 2 1 1 2 1 0 1 2 1 0 ω ω ω ( )ω ω ω ω I I I L L I I I I I I f i + = ⇒ + = ⇒ + = ⇒ = r r r r r r 4. Two cylinder having radii R1 and R2 and rotational inertias I1 and I2, respectively, are supported by axes perpendicular to the plane of Fig.6. The large cylinder is initially rotating with angular velocity ω0. 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 ω2 of the small cylinder in terms of I1, I2, R1, R2, and ω0. (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 ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = − = − = − ∫ ∫ 1 1 2 2 1 1 1 1 0 2 2 2 d d 0 R R R f t I I R f t I ω ω ω ω ω Solving the equations, we can get the final angular velocity ω2 of the small cylinder is ω0 r I2 I1 Direction of spin Fig.5 I2 I1 R2 R1 ω0 Fig.6 + +
10R1R R2+12R2 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 Fig. 7 distance h to the floor. The total torque on the disk is due to the frictional torque t frictm 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 is a. The angular speed of the disk is @=at while the falling mass reaches the floor. So 2h h Using newton nd law T=ma Tr-t.=la Apply the rotational analog of Newton's second law to the disk △t During the course of the mass hits the floor Using the equation(1)and (2), we get the moment of inertia of the disk 1=m(8-a)1
2 2 1 2 1 2 1 0 1 2 2 I R I R I R R + = 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 ω = αt while the falling mass reaches the floor. So 2 2 2 2 1 t h h = at ⇒ a = Using Newton’s second law ⎪ ⎩ ⎪ ⎨ ⎧ = − = − = α τ α a r Tr I mg T ma fric (1) Apply the rotational analog of Newton’s second law to the disk t Ltotal L total ∆ ∆ = ≈ r r r dt d τ During the course of the mass hits the floor ' ' 0 t It I t fric ω α 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 a t mr g a I + − = + ⋅ − = m M R Fig.7 r Wheel Hub