11 Angular Momentum CHAPTER OUTLINE 11.1 The Vector Product and Torque 11.2 Analysis Model: Nonisolated System (Angular Momentum) 11.3 Angular Momentum of a Rotating Rigid Object 11.4 Analysis Model: Isolated System (Angular Momentum) 11.5 The Motion of Gyroscopes and Tops * An asterisk indicates a question or problem new to this edition ANSWERS TO OBJECTIVE QUESTIONS OQ11.1 Answer (b) Her angular momentum stays constant as I is cut in half and ω doubles Then Iω doubles OQ11.2 The angular momentum of the mouse-turntable system is initially zero, with both at rest The frictionless axle isolates the mouseturntable system from outside torques, so its angular momentum must stay constant with the value of zero (i) Answer (a) The mouse makes some progress north, or counterclockwise (ii) Answer (b) The turntable will rotate clockwise The turntable rotates in the direction opposite to the motion of the mouse, for the angular momentum of the system to remain zero (iii) No Mechanical energy changes as the mouse converts some chemical into mechanical energy, positive for the motions of both the mouse and the turntable (iv) No Linear momentum is not conserved The turntable has zero momentum while the mouse has a bit of northward momentum Initially, momentum is zero; later, when the mouse moves 584 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 585 north, the fixed axle prevents the turntable from moving south (v) Yes Angular momentum is constant, with the value of zero OQ11.3 (i) Answer (a), (ii) Answer (e), (3 m, down) × (2 N, toward you) = N · m, left OQ11.4 Answer c = e > b = d > a = The unit vectors have magnitude 1, so the magnitude of each cross product is |1 · · sin θ| where θ is the angle between the factors Thus for (a) the magnitude of the cross product is sin 0° = For (b), |sin 135°| = 0.707, (c) sin 90° = 1, (d) sin 45° = 0.707, (e) sin 90° = OQ11.5 (a) No (b) No An axis of rotation must be defined to calculate the torque acting on an object The moment arm of each force is measured from the axis, so the value of the torque depends on the location of the axis OQ11.6 (i) ( ) Answer (e) Down–cross–left is away from you: − ˆj × − ˆi = − kˆ , as in the first picture (ii) ( ) Answer (d) Left–cross–down is toward you: − ˆi × − ˆj = kˆ , as in the second picture ANS FIG OQ11.6 OQ11.7 (i) Answer (a) The angular momentum is constant The moment of inertia decreases, so the angular speed must increase (ii) No Mechanical energy increases The ponies must work to push themselves inward (iii) Yes Momentum stays constant, with the value of zero (iv) Yes Angular momentum is constant with a nonzero value No outside torque can influence rotation about the vertical axle OQ11.8 Answer (d) As long as no net external force, or torque, acts on the system, the linear and angular momentum of the system are constant © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 586 Angular Momentum ANSWERS TO CONCEPTUAL QUESTIONS CQ11.1 The star is isolated from any outside torques, so its angular momentum is conserved as it changes size As the radius of the star decreases, its moment of inertia decreases, resulting in its angular speed increasing CQ11.2 The suitcase might contain a spinning gyroscope If the gyroscope is spinning about an axis that is oriented horizontally passing through the bellhop, the force he applies to turn the corner results in a torque that could make the suitcase swing away If the bellhop turns quickly enough, anything at all could be in the suitcase and need not be rotating Since the suitcase is massive, it will tend to follow an inertial path This could be perceived as the suitcase swinging away by the bellhop CQ11.3 The long pole has a large moment of inertia about an axis along the rope An unbalanced torque will then produce only a small angular acceleration of the performer-pole system, to extend the time available for getting back in balance To keep the center of mass above the rope, the performer can shift the pole left or right, instead of having to bend his body around The pole sags down at the ends to lower the system’s center of gravity CQ11.4 (a) Frictional torque arises from kinetic friction between the inside of the roll and the child’s fingers As with all friction, the magnitude of the friction depends on the normal force between the surfaces in contact As the roll unravels, the weight of the roll decreases, leading to a decrease in the frictional force, and, therefore, a decrease in the torque (b) As the radius R of the paper roll shrinks, the roll’s angular v speed ω = must increase because the speed v is constant R (c) If we think of the roll as a uniform disk, then its moment of inertia is I = MR But the roll’s mass is proportional to its base area π R ; therefore, the moment of inertia is proportional to R4 The moment of inertia decreases as the roll shrinks When the roll is given a sudden jerk, its angular acceleration may not be great enough to set the roll moving in step with the paper, so the paper breaks The roll is most likely to break when its radius is large, when its moment of inertia is large, than when its radius is small, when its moment of inertia is small © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 CQ11.5 587 Work done by a torque results in a change in rotational kinetic energy about an axis Work done by a force results in a change in translational kinetic energy Work by either has the same units: W = FΔx = [ N ][ m ] = N ⋅ m = J W = τ Δθ = [ N ⋅ m ][ rad ] = N ⋅ m = J CQ11.6 Suppose we look at the motorcycle moving to the right Its drive wheel is turning clockwise The wheel speeds up when it leaves the ground No outside torque about its center of mass acts on the airborne cycle, so its angular momentum is conserved As the drive wheel’s clockwise angular momentum increases, the frame of the cycle acquires counterclockwise angular momentum The cycle’s front end moves up and its back end moves down CQ11.7 Its angular momentum about that axis is constant in time You cannot conclude anything about the magnitude of the angular momentum CQ11.8 No The angular momentum about any axis that does not lie along the instantaneous line of motion of the ball is nonzero CQ11.9 The Earth is an isolated system, so its angular momentum is conserved when the distribution of its mass changes When its mass moves away from the axis of rotation, its moment of inertia increases, its angular speed decreases, so its period increases Most of the mass of Earth would not move, so the effect would be small: we would not have more hours in a day, but more nanoseconds CQ11.10 As the cat falls, angular momentum must be conserved Thus, if the upper half of the body twists in one direction, something must get an equal angular momentum in the opposite direction Rotating the lower half of the body in the opposite direction satisfies the law of conservation of angular momentum CQ11.11 Energy bar charts are useful representations for keeping track of the various types of energy storage in a system: translational and rotational kinetic energy, various types of potential energy, and internal energy However, there is only one type of angular momentum Therefore, there is no need for bar charts when analyzing a physical situation in terms of angular momentum © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 588 Angular Momentum SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 11.1 The Vector Product and Torque P11.1 ˆi ˆj kˆ   ˆ M × N = −3 = ˆi(6 − 5) − ˆj(−4 − 4) + k(10 + 12) = ˆi + 8.00ˆj + 22.0kˆ −2 P11.2 (a)   area = A × B = ABsin θ = ( 42.0 cm ) ( 23.0 cm ) sin ( 65.0° − 15.0° ) = 740 cm (b) The longer diagonal is equal to the sum of the two vectors   A + B = [(42.0 cm)cos15.0° + (23.0 cm)cos65.0°]ˆi +[(42.0 cm)]sin 15.0° + (23.0 cm)sin 65.0°]ˆj P11.3   A + B = ( 50.3 cm ) ˆi + ( 31.7 cm ) ˆj   2 length = A + B = ( 50.3 cm ) + ( 31.7 cm ) = 59.5 cm   We take the cross product of each term of A with each term of B, using the cross-product multiplication table for unit vectors Then we use the identification of the magnitude of the cross product as AB sin θ to find θ We assume the data are known to three significant digits (a) We use the definition of the cross product and note that ˆi × ˆi = ˆj × ˆj = 0:   A × B = 1ˆi + ˆj × ˆi + 3ˆj   A × B = ˆi × ˆi + 3ˆi × ˆj − 4ˆj × ˆi + 6ˆj × ˆj ( (b) ) ( ) = + 3kˆ − ( −kˆ ) + = 7.00kˆ   Since A × B = ABsin θ , we have   ⎛ A×B ⎞ ⎞ −1 ⎛ θ = sin ⎜ = 60.3° ⎟ = sin ⎜ ⎝ + 2 2 + 32 ⎟⎠ ⎝ AB ⎠ −1 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 P11.4 589 ˆi × ˆi = ⋅ ⋅ sin 0° = ˆj × ˆj and kˆ × kˆ are zero similarly since the vectors being multiplied are parallel ˆi × ˆj = ⋅ ⋅ sin 90° = ANS FIG P11.4 P11.5 We first resolve all of the forces shown in Figure P11.5 into components parallel to and perpendicular to the beam as shown in ANS FIG P11.5 (a) ANS FIG P11.5 The torque about an axis through point O is given by τ O = + [( 25 N ) cos 30°⎤⎦ ( 2.0 m ) − [( 10 N ) sin 20°]( 4.0 m ) = +30 N ⋅m or (b) τ = 30 N ⋅ m counterclockwise The torque about an axis through point C is given by τ C = + ⎡⎣( 30 N ) sin 45° ⎤⎦ (2.0 m) − ⎡⎣( 10 N ) sin 20° ⎤⎦ (2.0 m) = + 36 N ⋅ m or P11.6 τ C = 36 N ⋅ m counterclockwise   A ⋅ B = −3.00 ( 6.00 ) + 7.00 ( −10.0 ) + ( −4.00 ) ( 9.00 ) = −124 AB = (a) ( −3.00)2 + (7.00)2 + ( −4.00)2 ⋅ (6.00)2 + ( −10.0)2 + ( 9.00)2 = 127   ⎛ ⎞ −1 A ⋅ B cos ⎜ = cos −1 ( −0.979 ) = 168° ⎟ ⎝ AB ⎠ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 590 Angular Momentum (b) ˆi   A × B = −3.00 ˆj kˆ 7.00 −4.00 = 23.0ˆi + 3.00ˆj − 12.0kˆ 6.00 −10.0 9.00   2 A × B = ( 23.0 ) + ( 3.00 ) + ( −12.0 ) = 26.1   ⎛ A×B ⎞ −1 sin −1 ⎜ ⎟ = sin ( 0.206 ) = 11.9° or 168° ⎜⎝ AB ⎟⎠ (c) P11.7 Only the first method gives the angle between the vectors unambiguously because sin(180° – θ ) = sin θ but cos (180° – θ ) = – cos θ ; in other words, the vectors can only be at most 180° apart and using the second method cannot distinguish θ from 180° – θ     We are given the condition A × B = A ⋅ B This says that ABsin θ = ABcosθ so tan θ = θ = 45.0° satisfies this condition P11.8 (a) The torque acting on the particle about the origin is ˆi ˆj kˆ    τ = r × F = = ˆi ( − ) − ˆj( − ) + kˆ ( − 18 ) = ( −10.0 N ⋅ m ) kˆ (b) Yes The point or axis must be on the other side of the line of action of the force, and half as far from this line along which the force acts Then the lever arm of the force about this new axis will be half as large and the force will produce counterclockwise instead of clockwise torque (c) Yes There are infinitely many such points, along a line that passes through the point described in (b) and parallel the line of action of the force (d) Yes, at the intersection of the line described in (c) and the y axis © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 (e) (f) 591 No, because there is only one point of intersection of the line described in (d) with the y axis Let (0, y) represent the coordinates of the special axis of rotation located on the y axis of Cartesian coordinates Then the displacement from this point to the particle feeling the force is  rnew = 4ˆi + (6 − y)ˆj in meters The torque of the force about this new axis is ˆi ˆj kˆ    τ new = rnew × F = − y = ˆi ( − ) − ˆj( − ) + kˆ ( − 18 + 3y ) = ( +5 N ⋅ m ) kˆ Then, − 18 + 3y = → 3y = 15 → y=5 The position vector of the new axis is 5.00ˆj m P11.9 (a) The lever arms of the forces about O are all the same, equal to length OD, L     If F3 has a magnitude F3 = F1 + F2 , the net torque is zero: ∑ τ = F1L + F2 L − F3 L = F1L + F2 L − ( F1 + F2 ) L = (b)  The torque produced by F3 depends on the perpendicular  distance OD, therefore translating the point of application of F3 to any other point along BC will not change the net torque ANS FIG P11.9 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 592 P11.10 Angular Momentum (a) (b) No The cross-product vector must be perpendicular to both of the factors, so its dot product with either factor must be zero To check: (2ˆi − 3ˆj + 4kˆ ) ⋅ ( 4ˆi + 3ˆj − kˆ ) = (ˆi ⋅ ˆi ) + −9(ˆj ⋅ ˆj) − 4(kˆ ⋅ kˆ ) = − − = −5 The answer is not zero No The cross product could not work out that way Section 11.2 P11.11 Analysis Model: Nonisolated System (Angular Momentum) Taking the geometric center of the compound object to be the pivot, the angular speed and the moment of inertia are ω = v/r = (5.00 m/s)/0.500 m = 10.0 rad/s and I = ∑ mr = ( 4.00 kg )( 0.500 m ) + ( 3.00 kg )( 0.500 m ) 2 = 1.75 kg · m By the right-hand rule, we find that the angular velocity is directed out of the plane So the object’s angular momentum, with magnitude ) L = I ω = ( 1.75 kg ⋅ m (10.0 rad/s) is the vector  L = ( 17.5 kg ⋅ m /s ) kˆ ANS FIG P11.11 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 11 P11.12    We use L = r × p:  L = 1.50ˆi + 2.20ˆj m × ( 1.50 kg ) 4.20ˆi − 3.60ˆj m/s  L = −8.10kˆ − 13.9kˆ kg ⋅ m /s = ( −22.0 kg ⋅ m /s ) kˆ ( ( P11.13 593 )    We use L = r × p:  L= ˆi x mvx ( ) ) kˆ = ˆi ( − ) − ˆj ( − ) + kˆ mxvy − myvx ˆj y mvy ( )  L = m xvy − yvx kˆ ( P11.14 ) Whether we think of the Earth’s surface as curved or flat, we interpret the problem to mean that the plane’s line of flight extended is precisely tangent to the mountain at its peak, and nearly parallel to the wheat field Let the positive x direction be eastward, positive y be northward, and positive z be vertically upward  (a) r = ( 4.30 km ) kˆ = ( 4.30 × 103 m ) kˆ ( )   p = mv = ( 12 000 kg ) −175ˆi m/s = −2.10 × 106 ˆi kg ⋅ m/s    L = r × p = 4.30 × 103 kˆ m × −2.10 × 106 ˆi kg ⋅ m/s ( = (b) No ( −9.03 × 10 ) ( kg ⋅ m /s ) ˆj )   L = r p sin θ = mv ( r sin θ ) , and r sin θ is the altitude of the plane Therefore, L = constant as the plane moves in level flight with constant velocity P11.15 (c) Zero The position vector from Pike’s Peak to the plane is anti- (a) parallel to the velocity of the plane That is, it is directed along the same line and opposite in direction Thus, L = mvr sin 180° =     Zero because L = r × p and r = (b) At the highest point of the trajectory, vi2 sin 2θ and x= R= 2g y = hmax vi sin θ ) ( = 2g © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part