How do figure skaters manage to spin so elegantly? The answer lies in a simple physical principle We’re all familiar with the magic of the Olympic games. Once every four years the entire family sits down together to watch a selection of sports that many of us are not routinely exposed to. As a child I was in awe of the spectacular abilities of the athletes, and especially the figure skaters at the Winter Olympics. I just couldn’t understand how they could change the pace of their spin so quickly and elegantly. Today I know: it’s all about angular momentum conservation. Angular momentum is a conserved physical quantity, similar to the way that energy is a conserved quantity. Roughly, it is a measure of the rotational momentum of a rotating object or body. The Law of Conservation of Angular Momentum is what allows the figure skater to control the pace of her spin, just as it prevents us from falling every time we ride a bicycle. In order to understand angular momentum, let’s take a point around which an object rotates and calculate the radius, or the distance of the object from that middle point, and the angular velocity of the object, which is based on the number of revolutions per second. A “regular” or linear momentum, which describes motion along a straight line, is the product of the mass times the velocity of an object, characterizing the magnitude of the motion. If a car and a truck are traveling at the same speed, the truck has a larger momentum since its mass is much larger. This is also why much more energy is required in order to stop it. The angular momentum is a product of the mass times the angular velocity times the squared radius, implying that the energy of an object rotating further from the central point is larger than that of an identical object rotating at the same velocity in a smaller circle. Another physical quantity is torque, also referred to as the rotational force, which, in its most basic form, is the product of the force times the length of the axis exerting that force. If a swing is hanging on long chains, we will need to apply more force to swing a child to a certain height, than we would on a swing hanging on shorter chains. The moment of inertia of an object describes the ratio between the rotational force and the angular acceleration of an object along a certain rotational axis. The angular momentum is equal to the moment of inertia times the angular velocity. An object with a higher moment of inertia will spin slower than would the same object with a lower moment of inertia, when a similar force is applied. Spinning While SkatingLet’s get back to the spinning figure skater. Given that no outside force is applied, the angular momentum is conserved. When the skater extends her arms or legs, she effectively increases her radius, and thus changes her moment of inertia. Since the angular momentum remains constant, what changes is the angular velocity of the spin. For example, when the skater extends her arms outwards, increasing twofold the moment of inertia, the velocity of her spin also decreases twofold. While tucking her arms in, she decreases the moment of inertia significantly and thus gains high rotational velocity. One doesn’t have to be a skilled dancer to experience this phenomenon – you can try it out (carefully!) on a swivel chair or a rotating stool, in the following way. Now that we understand the meaning of the conservation of angular momentum, we can enjoy watching figure skating competitions even more.
By the end of this section, you will be able to:
So far, we have looked at the angular momentum of systems consisting of point particles and rigid bodies. We have also analyzed the torques involved, using the expression that relates the external net torque to the change in angular momentum, (Figure). Examples of systems that obey this equation include a freely spinning bicycle tire that slows over time due to torque arising from friction, or the slowing of Earth’s rotation over millions of years due to frictional forces exerted on tidal deformations. However, suppose there is no net external torque on the system, [latex] \sum \overset{\to }{\tau }=0. [/latex] In this case, (Figure) becomes the law of conservation of angular momentum.
The angular momentum of a system of particles around a point in a fixed inertial reference frame is conserved if there is no net external torque around that point: [latex] \frac{d\overset{\to }{L}}{dt}=0 [/latex] or [latex] \overset{\to }{L}={\overset{\to }{l}}_{1}+{\overset{\to }{l}}_{2}\,\text{+}\,\text{⋯}\,\text{+}\,{\overset{\to }{l}}_{N}=\text{constant}\text{.} [/latex] Note that the total angular momentum [latex] \overset{\to }{L} [/latex] is conserved. Any of the individual angular momenta can change as long as their sum remains constant. This law is analogous to linear momentum being conserved when the external force on a system is zero. As an example of conservation of angular momentum, (Figure) shows an ice skater executing a spin. The net torque on her is very close to zero because there is relatively little friction between her skates and the ice. Also, the friction is exerted very close to the pivot point. Both [latex] |\overset{\to }{F}|\,\text{and}\,|\overset{\to }{r}| [/latex] are small, so [latex] |\overset{\to }{\tau }| [/latex] is negligible. Consequently, she can spin for quite some time. She can also increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that [latex] {L}^{\prime }=L [/latex] or [latex] {I}^{\prime }{\omega }^{\prime }=I\omega , [/latex] where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because [latex] {I}^{\prime } [/latex] is smaller, the angular velocity [latex] {\omega }^{\prime } [/latex] must increase to keep the angular momentum constant.
 Figure 11.14 (a) An ice skater is spinning on the tip of her skate with her arms extended. Her angular momentum is conserved because the net torque on her is negligibly small. (b) Her rate of spin increases greatly when she pulls in her arms, decreasing her moment of inertia. The work she does to pull in her arms results in an increase in rotational kinetic energy. It is interesting to see how the rotational kinetic energy of the skater changes when she pulls her arms in. Her initial rotational energy is [latex] {K}_{\text{Rot}}=\frac{1}{2}I{\omega }^{2}, [/latex] whereas her final rotational energy is [latex] {{K}^{\prime }}_{\text{Rot}}=\frac{1}{2}{I}^{\prime }{({\omega }^{\prime })}^{2}. [/latex] Since [latex] {I}^{\prime }{\omega }^{\prime }=I\omega , [/latex] we can substitute for [latex] {\omega }^{\prime } [/latex] and find [latex] {{K}^{\prime }}_{\text{Rot}}=\frac{1}{2}{I}^{\prime }{({\omega }^{\prime })}^{2}=\frac{1}{2}{I}^{\prime }{(\frac{I}{{I}^{\prime }}\omega )}^{2}=\frac{1}{2}I{\omega }^{2}(\frac{I}{{I}^{\prime }})={K}_{\text{Rot}}(\frac{I}{{I}^{\prime }})\text{}. [/latex] Because her moment of inertia has decreased, [latex] {I}^{\prime }<I, [/latex] her final rotational kinetic energy has increased. The source of this additional rotational kinetic energy is the work required to pull her arms inward. Note that the skater’s arms do not move in a perfect circle—they spiral inward. This work causes an increase in the rotational kinetic energy, while her angular momentum remains constant. Since she is in a frictionless environment, no energy escapes the system. Thus, if she were to extend her arms to their original positions, she would rotate at her original angular velocity and her kinetic energy would return to its original value. The solar system is another example of how conservation of angular momentum works in our universe. Our solar system was born from a huge cloud of gas and dust that initially had rotational energy. Gravitational forces caused the cloud to contract, and the rotation rate increased as a result of conservation of angular momentum ((Figure)).
Figure 11.15 The solar system coalesced from a cloud of gas and dust that was originally rotating. The orbital motions and spins of the planets are in the same direction as the original spin and conserve the angular momentum of the parent cloud. (credit: modification of work by NASA) We continue our discussion with an example that has applications to engineering.
A flywheel rotates without friction at an angular velocity [latex] {\omega }_{0}=600\,\text{rev}\text{/}\text{min} [/latex] on a frictionless, vertical shaft of negligible rotational inertia. A second flywheel, which is at rest and has a moment of inertia three times that of the rotating flywheel, is dropped onto it ((Figure)). Because friction exists between the surfaces, the flywheels very quickly reach the same rotational velocity, after which they spin together. (a) Use the law of conservation of angular momentum to determine the angular velocity [latex] \omega [/latex] of the combination. (b) What fraction of the initial kinetic energy is lost in the coupling of the flywheels?
Figure 11.16 Two flywheels are coupled and rotate together. StrategyPart (a) is straightforward to solve for the angular velocity of the coupled system. We use the result of (a) to compare the initial and final kinetic energies of the system in part (b). Solutiona. No external torques act on the system. The force due to friction produces an internal torque, which does not affect the angular momentum of the system. Therefore conservation of angular momentum gives [latex] \begin{array}{c}{I}_{0}{\omega }_{0}=({I}_{0}+3{I}_{0})\omega ,\hfill \\ \omega =\frac{1}{4}{\omega }_{0}=150\,\text{rev}\text{/}\text{min}=15.7\,\text{rad}\text{/}\text{s}.\hfill \end{array} [/latex] b. Before contact, only one flywheel is rotating. The rotational kinetic energy of this flywheel is the initial rotational kinetic energy of the system, [latex] \frac{1}{2}{I}_{0}{\omega }_{0}^{2} [/latex]. The final kinetic energy is[latex] \frac{1}{2}(4{I}_{0}){\omega }^{2}=\frac{1}{2}(4{I}_{0}){(\frac{{\omega }_{0}}{4})}^{2}=\frac{1}{8}{I}_{0}{\omega }_{0}^{2}. [/latex] Therefore, the ratio of the final kinetic energy to the initial kinetic energy is [latex] \frac{\frac{1}{8}{I}_{0}{\omega }_{0}^{2}}{\frac{1}{2}{I}_{0}{\omega }_{0}^{2}}=\frac{1}{4}. [/latex] Thus, 3/4 of the initial kinetic energy is lost to the coupling of the two flywheels. SignificanceSince the rotational inertia of the system increased, the angular velocity decreased, as expected from the law of conservation of angular momentum. In this example, we see that the final kinetic energy of the system has decreased, as energy is lost to the coupling of the flywheels. Compare this to the example of the skater in (Figure) doing work to bring her arms inward and adding rotational kinetic energy.
A merry-go-round at a playground is rotating at 4.0 rev/min. Three children jump on and increase the moment of inertia of the merry-go-round/children rotating system by [latex] 25% [/latex]. What is the new rotation rate?
An 80.0-kg gymnast dismounts from a high bar. He starts the dismount at full extension, then tucks to complete a number of revolutions before landing. His moment of inertia when fully extended can be approximated as a rod of length 1.8 m and when in the tuck a rod of half that length. If his rotation rate at full extension is 1.0 rev/s and he enters the tuck when his center of mass is at 3.0 m height moving horizontally to the floor, how many revolutions can he execute if he comes out of the tuck at 1.8 m height? See (Figure).
Figure 11.17 A gymnast dismounts from a high bar and executes a number of revolutions in the tucked position before landing upright. StrategyUsing conservation of angular momentum, we can find his rotation rate when in the tuck. Using the equations of kinematics, we can find the time interval from a height of 3.0 m to 1.8 m. Since he is moving horizontally with respect to the ground, the equations of free fall simplify. This will allow the number of revolutions that can be executed to be calculated. Since we are using a ratio, we can keep the units as rev/s and don’t need to convert to radians/s. SolutionThe moment of inertia at full extension is [latex] {I}_{0}=\frac{1}{12}m{L}^{2}=\frac{1}{12}80.0\,\text{kg}(1{.8\,\text{m})}^{2}=21.6\,\text{kg}·{\text{m}}^{2} [/latex]. The moment of inertia in the tuck is [latex] {I}_{\text{f}}=\frac{1}{12}m{L}_{\text{f}}^{\text{2}}=\frac{1}{12}80.0\,\text{kg}(0{.9\,\text{m})}^{2}=5.4\,\text{kg}·{\text{m}}^{2} [/latex]. Conservation of angular momentum: [latex] {I}_{\text{f}}{\omega }_{\text{f}}={I}_{0}{\omega }_{0}⇒{\omega }_{\text{f}}=\frac{{I}_{0}{\omega }_{0}}{{I}_{\text{f}}}=\frac{21.6\,\text{kg}·{\text{m}}^{2}(1.0\,\text{rev}\text{/}\text{s})}{5.4\,\text{kg}·{\text{m}}^{2}}=4.0\,\text{rev}\text{/}\text{s} [/latex]. Time interval in the tuck: [latex] t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(3.0-1.8)\text{m}}{9.8\,\text{m}\text{/}\text{s}}}=0.5\,\text{s} [/latex]. In 0.5 s, he will be able to execute two revolutions at 4.0 rev/s. SignificanceNote that the number of revolutions he can complete will depend on how long he is in the air. In the problem, he is exiting the high bar horizontally to the ground. He could also exit at an angle with respect to the ground, giving him more or less time in the air depending on the angle, positive or negative, with respect to the ground. Gymnasts must take this into account when they are executing their dismounts.
A bullet of mass [latex] m=2.0\,\text{g} [/latex] is moving horizontally with a speed of [latex] 500.0\,\text{m}\text{/}\text{s}. [/latex] The bullet strikes and becomes embedded in the edge of a solid disk of mass [latex] M=3.2\,\text{kg} [/latex] and radius [latex] R=0.5\,\text{m}\text{.} [/latex] The cylinder is free to rotate around its axis and is initially at rest ((Figure)). What is the angular velocity of the disk immediately after the bullet is embedded?
Figure 11.18 A bullet is fired horizontally and becomes embedded in the edge of a disk that is free to rotate about its vertical axis. StrategyFor the system of the bullet and the cylinder, no external torque acts along the vertical axis through the center of the disk. Thus, the angular momentum along this axis is conserved. The initial angular momentum of the bullet is [latex] mvR [/latex], which is taken about the rotational axis of the disk the moment before the collision. The initial angular momentum of the cylinder is zero. Thus, the net angular momentum of the system is [latex] mvR [/latex]. Since angular momentum is conserved, the initial angular momentum of the system is equal to the angular momentum of the bullet embedded in the disk immediately after impact. SolutionThe initial angular momentum of the system is [latex] {L}_{i}=mvR. [/latex] The moment of inertia of the system with the bullet embedded in the disk is [latex] I=m{R}^{2}+\frac{1}{2}M{R}^{2}=(m+\frac{M}{2}){R}^{2}. [/latex] The final angular momentum of the system is [latex] {L}_{f}=I{\omega }_{f}. [/latex] Thus, by conservation of angular momentum, [latex] {L}_{i}={L}_{f} [/latex] and [latex] mvR=(m+\frac{M}{2}){R}^{2}{\omega }_{f}. [/latex] Solving for [latex] {\omega }_{f}, [/latex] [latex] {\omega }_{f}=\frac{mvR}{(m+M\text{/}2){R}^{2}}=\frac{(2.0\,×\,{10}^{-3}\,\text{kg})(500.0\,\text{m}\text{/}\text{s})}{(2.0\,×\,{10}^{-3}\,\text{kg}\,+\,1.6\,\text{kg})(0.50\,\text{m})}=1.2\,\text{rad}\text{/}\text{s}. [/latex] The system is composed of both a point particle and a rigid body. Care must be taken when formulating the angular momentum before and after the collision. Just before impact the angular momentum of the bullet is taken about the rotational axis of the disk. Summary
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As the rope of a tethered ball winds around a pole, what happens to the angular velocity of the ball?
Suppose the polar ice sheets broke free and floated toward Earth’s equator without melting. What would happen to Earth’s angular velocity?
Explain why stars spin faster when they collapse.
Competitive divers pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down (see below). Explain the effect of both actions on their angular velocities. Also explain the effect on their angular momentum.
A disk of mass 2.0 kg and radius 60 cm with a small mass of 0.05 kg attached at the edge is rotating at 2.0 rev/s. The small mass suddenly separates from the disk. What is the disk’s final rotation rate?
The Sun’s mass is [latex] 2.0\,×\,{10}^{30}\text{kg,} [/latex] its radius is [latex] 7.0\,\text{×}\,{10}^{5}\,\text{km,} [/latex] and it has a rotational period of approximately 28 days. If the Sun should collapse into a white dwarf of radius [latex] 3.5\,\text{×}\,{10}^{3}\,\text{km,} [/latex] what would its period be if no mass were ejected and a sphere of uniform density can model the Sun both before and after?
A cylinder with rotational inertia [latex] {I}_{1}=2.0\,\text{kg}·{\text{m}}^{2} [/latex] rotates clockwise about a vertical axis through its center with angular speed [latex] {\omega }_{1}=5.0\,\text{rad}\text{/}\text{s}. [/latex] A second cylinder with rotational inertia [latex] {I}_{2}=1.0\,\text{kg}·{\text{m}}^{2} [/latex] rotates counterclockwise about the same axis with angular speed [latex] {\omega }_{2}=8.0\,\text{rad}\text{/}\text{s} [/latex]. If the cylinders couple so they have the same rotational axis what is the angular speed of the combination? What percentage of the original kinetic energy is lost to friction?
A diver off the high board imparts an initial rotation with his body fully extended before going into a tuck and executing three back somersaults before hitting the water. If his moment of inertia before the tuck is [latex] 16.9\,\text{kg}·{\text{m}}^{2} [/latex] and after the tuck during the somersaults is [latex] 4.2\,\text{kg}·{\text{m}}^{2} [/latex], what rotation rate must he impart to his body directly off the board and before the tuck if he takes 1.4 s to execute the somersaults before hitting the water?
An Earth satellite has its apogee at 2500 km above the surface of Earth and perigee at 500 km above the surface of Earth. At apogee its speed is 730 m/s. What is its speed at perigee? Earth’s radius is 6370 km (see below).
A Molniya orbit is a highly eccentric orbit of a communication satellite so as to provide continuous communications coverage for Scandinavian countries and adjacent Russia. The orbit is positioned so that these countries have the satellite in view for extended periods in time (see below). If a satellite in such an orbit has an apogee at 40,000.0 km as measured from the center of Earth and a velocity of 3.0 km/s, what would be its velocity at perigee measured at 200.0 km altitude?
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When a figure skater is in a spin they pull their arms together this makes them spin faster because of which concept?
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