Astronomy

Rotational inertia, angular momentum of the Solar system

Rotational inertia, angular momentum of the Solar system


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Is there Rotational inertia of the Solar system as one entity "disk"? Is this even possible?

  1. The angular momentum of a multi-point system in the Solar system - Since the different planets have different mass and orbital periods then does create angular momentum, inertia in the system?

  2. How can one planet affect another planet in the system from its individual rotation?

  3. Does the solar system exhibit rigid body rotation from the mass distributed over planets?


Solar system planetary alignment triggers tides and earthquakes

This research hypothesizes that tidal and earthquakes are induced by solar system planet positions, as the planetary attraction act as a trigger force change the speed of the Earth rotation. The occurrence of a sea tide is only a consequence of a relative slowdown of the rotational/revolving speed of the Earth which urges the Earth’s plates to move. The research included analyzing earthquake data for the whole Earth over July, 2019 with a case study of the Arabian Plate (AP) seismicity included the Zagros Folded Belt (ZFB) and Zagros Thrust Zone (ZTZ) as a seismic active belt in the northern hemisphere. The rotational velocity of the Earth has been calculated for eight seismic events, and it turns out that the velocity was different for each case. A negative proportional was found between earthquake and the Earth rotational speed. During the configuration of the Jupiter and Saturn in a straight line with the Earth over July 2019, one thousand and thirty-seven of earthquakes occurred around the world were statistically analyzed having 2–6 magnitudes. Rotational/ revolving speed, angular momentum and rotational inertia kinetic energy gravitational potential energy of the Earth at equator and at 45 degrees were computed to show how rotational speed triggering plates. Planets interact with each other influencing earthquakes via the gravitational stresses arising from the configuration of the solar system planets that cause a slowdown of the rotational/revolving speed of the Earth. This stimulates the Earth’s plate to move generating earthquake due to the activation of faults.

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The angular momenta of solar system bodies: Implications for asteroid strengths

The angular momentum H is plotted versus mass M for the planets and for all asteroids with known rotation rates and shapes, primarily taken from D. C. McAdoo and J. A. Burns [Icarus 18, 285–293 (1973)]. An asteroid's angular momentum is derived from its rotation rate as determined by the period of its lightcurve, its shape as indicated by the lightcurve amplitude, and where possible its size as given by polarimetry or radiometry. The asteroid is assumed to be rotating about its axis of maximum moment of inertia. As previously found by F. F. Fish [Icarus 7, 251–256 (1967]) and W. K. Hartmann and S. M. Larson [Icarus 7, 257–260 (1967)], H is approximately proportional to M 5 3 , which shows that the asteroids and most planets spin with nearly the same rate. The very smallest asteroids on the plot deviate from the above reaction, usually containing excess angular momentum. This suggests that collisions have transferred substantial angular momentum to the smallest asteroids, perhaps causing their internal stress states to be substantially modified by centrifugal effects.

The forces produced by gravitation are then compared to centrifugal effects for a rotating, triaxial ellipsoid of density 3 g cm −3 . For all asteroids with known properties the gravitational attraction is shown to be larger than the centrifugal acceleration of a particle on the surface: thus the observed asteroid regoliths are gravitationally bound. Poisson's equation for the gravitational potential is investigated and it is shown by mathematical and physical arguments that any arbitrarily shaped ellipsoid with the attractive surface force boundary condition found above will have only attractive internal forces. Thus the internal stress states in asteroids are always compressive so that asteroids could be internally fractured without losing their integrity.


Evolution of the Earth

9.09.1 Polar Motion and Length-of-Day Variations Through (Geologic) Time

Observations of the evolving state of planetary rotation reveal the occurrence of variability on timescales ranging from daily, to annual, to interannual, decadal, and millennial and extending even to the timescale of hundreds of millions of years on which the process of mantle convection governs planetary evolution and to the 4.56 billion-year age of the planet itself. These variations in the state of rotation are most usefully discussed in terms of changes in the rate of planetary rotation about the instantaneous spin axis and thus variations in the length of the day, on the one hand, or in terms of the wobble of the spin axis as observed in a body-fixed frame of reference on the other. On relatively short subannual to annual timescales, variations in the length of day have been clearly shown to arise primarily as a consequence of the exchange of angular momentum between the solid Earth and its overlying atmosphere (e.g., Hide et al., 1980 ) and oceans. On the interannual timescale, an important recent discovery concerning length-of-day (l.o.d.) variability has been the documentation of a significant excitation associated with El Niño Southern Oscillation (ENSO) events ( Cox and Chao, 2002 Dickey et al., 2002 ). Concerning the sources of wobble excitation, it is clear that on the timescale of the seasonal cycle of climate change, the excitation of the annual component of wobble variability is due to the interhemispheric exchange of atmospheric mass. The Chandler wobble, however, a free oscillation of the Earth's spin axis in a body-fixed frame of reference with a period close to 14 months, is apparently significantly forced by the dynamic state of the oceans (e.g., Gross, 2000 ) and by the atmosphere. These most recent analyses of the problem of Chandler wobble excitation, to be reviewed in what follows, appear to have finally resolved what had remained an unresolved problem for decades.

On the longer timescale of millennia, both of these ‘anomalies’ in the Earth's rotation exhibit apparent secular variations that are caused primarily by the Late Pleistocene cycle of glaciation and deglaciation ( Peltier, 1982 ) that has been an enduring feature of climate system variability for the past 900 000 years of Earth history (e.g., Deblonde and Peltier, 1991 ). A primary focus of the discussion to follow in this chapter will be upon the manner in which, through the process of glacial isostatic adjustment (GIA), these ice-age engendered variations in the Earth's rotation feed back upon postglacial relative sea-level history, thus enabling detailed tests to be performed on the quality of the theory that has been developed to compute the rotational response to the GIA process ( Peltier, 2005 ). The importance of an accurate attribution of the source of excitation of the observed secular changes in the l.o.d. and polar motion to the GIA process concerns the important role that these observations may be invoked to play in the inference of the viscosity of the deep Earth, a parameter that is required in the construction of models of the mantle convection and continental drift processes.

An interesting additional aspect of the history of the Earth's rotation on the timescale of the Late Pleistocene ice-age cycle concerns the way in which temporal variations in the precession and obliquity components of the evolving geometry of the Earth's orbit around the Sun, forced by gravitational n-body effects in the solar system, have been employed to refine the timing of the ice-age cycle itself ( Shackleton et al., 1990 ).

On the very longest timescales on which the thermal evolution of the planet is governed by the mantle convection process, there also exists the distinct possibility that relatively rapid and large amplitude changes in the rotational state could have occurred in association with an ‘avalanche effect’ during which the style of the mantle convective circulation switches from one characterized by significant radial layering of the thermally forced flow to one of ‘whole mantle’ form (e.g., Solheim and Peltier, 1994a,b ). This process could conceivably act so as to induce the inertial interchange true polar wander (IITPW) instability that was suggested initially by Gold (1955) and that has recently been invoked by Kirschvink et al. (1997) as plausibly having occurred in the early Cambrian period of Earth history.

Figure 1 provides a schematic depiction of the extremely broad range of processes that contribute to the excitation of variations in the Earth's rotation on all timescales. The centipedes in the sketch, following the colorful analogy by Gold (1955) , are intended to represent, by their ability to move over the surface and thereby slowly (?) modify the moment of inertia tensor of the planet, the excitation of a rotational response due to the action of the internal mantle convective mixing process. How this component of the rotational excitation might be most accurately described is still a matter of considerable debate, as will be discussed in the final section of this chapter.

Figure 1 . Schematic diagram illustrating the range of processes that contribute to the excitation of variations in the rotational state of the planet. The centipedes in the figure, which are a modification of the beetles employed by Gold (1955) for the same illustrative purpose, are intended to represent the contribution to rotational excitation due to the mantle convection process. This schematic is a modification of that in Lambeck (1980a) paper commenting upon the important paper of Hide et al. (1980) .


11.4: Conservation of Angular Momentum

  • Contributed by OpenStax
  • General Physics at OpenStax CNX
  • Apply conservation of angular momentum to determine the angular velocity of a rotating system in which the moment of inertia is changing
  • Explain how the rotational kinetic energy changes when a system undergoes changes in both moment of inertia and angular velocity

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. 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&rsquos rotation over millions of years due to frictional forces exerted on tidal deformations.

However, suppose there is no net external torque on the system, (sum vec< au>) = 0. In this case, we can introduce the law of conservation of angular momentum.

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:

[vec = vec_ <1>+ vec_ <2>+ cdots + vec_ = constant ldotp label<11.11>]

Note that the total angular momentum (vec) 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 (PageIndex<1>) 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 (|vec|) and (|vec|) are small, so (|vec< au>|) 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

where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because I&prime is smaller, the angular velocity (omega)&prime must increase to keep the angular momentum constant.

Figure (PageIndex<1>): (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

whereas her final rotational energy is

Since I&prime(omega)&prime = I(omega), we can substitute for (omega)&prime and find

[K'_ = frac<1> <2>I' (omega')^ <2>= frac<1> <2>I' left(dfrac omega ight)^ <2>= frac<1> <2>I omega^ <2>left(dfrac ight) = K_ left(dfrac ight) ldotp]

Because her moment of inertia has decreased, (I&prime < I), 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&rsquos arms do not move in a perfect circle&mdashthey 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 (PageIndex<2>)).

Figure (PageIndex<2>): 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.

Example (PageIndex<1>): Coupled Flywheels

A flywheel rotates without friction at an angular velocity (omega_<0>) = 600 rev/min 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 (PageIndex<3>)). Because friction exists between the surfaces, the flywheels very quickly reach the same rotational velocity, after which they spin together.

  1. Use the law of conservation of angular momentum to determine the angular velocity (omega) of the combination.
  2. What fraction of the initial kinetic energy is lost in the coupling of the flywheels?

Part (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).

  1. 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 [I_ <0>omega_ <0>= (I_ <0>+ 3I_<0>) omega, onumber] [omega = frac<1><4>omega_ <0>= 150 rev/min = 15.7 rad/s ldotp onumber]
  2. Before contact, only one flywheel is rotating. The rotational kinetic energy of this flywheel is the initial rotational kinetic energy of the system, (frac<1><2>I_ <0>omega_<0>^<2>). The final kinetic energy is [frac<1><2>(4I_<0>) omega^ <2>= frac<1><2>(4I_<0>) left(dfrac><4> ight)^ <2>= frac<1><8>I_ <0>omega_<0>^<2>. onumber] Therefore, the ratio of the final kinetic energy to the initial kinetic energy is [frac<8>I_ <0>omega_<0>^<2>><2>I_ <0>omega_<0>^<2>> = frac<1><4>ldotp onumber] Thus, 3/4 of the initial kinetic energy is lost to the coupling of the two flywheels.

Significance

Since 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 (PageIndex<1>) 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 25%. What is the new rotation rate?

Example (PageIndex<2A>): Dismount from a High Bar

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 (PageIndex<4>).

Figure (PageIndex<4>): A gymnast dismounts from a high bar and executes a number of revolutions in the tucked position before landing upright.

Using 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&rsquot need to convert to radians/s.

The moment of inertia at full extension is [I_ <0>= frac<1> <12>mL^ <2>= frac<1> <12>(80.0 kg)(1.8 m)^ <2>= 21.6 kg cdotp m^ <2>ldotp onumber]

The moment of inertia in the tuck is [I_ = frac<1> <12>mL_^ <2>= frac<1> <12>(80.0 kg)(0.9 m)^ <2>= 5.4 kg cdotp m^ <2>ldotp onumber]

Conservation of angular momentum: [I_ omega_ = I_ <0>omega_ <0>Rightarrow omega_ = frac omega_<0>><>> = frac<(21.6 kg cdotp m^<2>)(1.0 rev/s)><5.4 kg cdotp m^<2>> = 4.0 rev/s ldotp onumber]

Time interval in the tuck: [t = sqrt> = sqrt<9.8 m/s>> = 0.5 s ldotp onumber]

In 0.5 s, he will be able to execute two revolutions at 4.0 rev/s.

Significance

Note 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.

Example (PageIndex<2B>): Conservation of Angular Momentum of a Collision

A bullet of mass m = 2.0 g is moving horizontally with a speed of 500.0 m/s. The bullet strikes and becomes embedded in the edge of a solid disk of mass M = 3.2 kg and radius R = 0.5 m. The cylinder is free to rotate around its axis and is initially at rest (Figure (PageIndex<5>)). What is the angular velocity of the disk immediately after the bullet is embedded?

Figure (PageIndex<5>): A bullet is fired horizontally and becomes embedded in the edge of a disk that is free to rotate about its vertical axis.

For 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 mvR, 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 mvR. 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.

The initial angular momentum of the system is

[L_ = mvR ldotp onumber]

The moment of inertia of the system with the bullet embedded in the disk is

[I = mR^ <2>+ frac<1> <2>MR^ <2>= left(m + dfrac<2> ight) R^ <2>ldotp onumber]

The final angular momentum of the system is

[L_ = I omega_ ldotp onumber]

Thus, by conservation of angular momentum, Li = Lf and

[mvR = left(m + dfrac<2> ight) R^ <2>omega_ ldotp onumber]

Significance

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.


Angular Momentum

A concept that is a bit more complex, but important for understanding many astronomical objects, is angular momentum, which is a measure of the rotation of a body as it revolves around some fixed point (an example is a planet orbiting the Sun). The angular momentum of an object is defined as the product of its mass, its velocity, and its distance from the fixed point around which it revolves.

If these three quantities remain constant—that is, if the motion of a particular object takes place at a constant velocity at a fixed distance from the spin center—then the angular momentum is also a constant. Kepler’s second law is a consequence of the conservation of angular momentum. As a planet approaches the Sun on its elliptical orbit and the distance to the spin center decreases, the planet speeds up to conserve the angular momentum. Similarly, when the planet is farther from the Sun, it moves more slowly.

The conservation of angular momentum is illustrated by figure skaters, who bring their arms and legs in to spin more rapidly, and extend their arms and legs to slow down (Figure 3). You can duplicate this yourself on a well-oiled swivel stool by starting yourself spinning slowly with your arms extended and then pulling your arms in. Another example of the conservation of angular momentum is a shrinking cloud of dust or a star collapsing on itself (both are situations that you will learn about as you read on). As material moves to a lesser distance from the spin center, the speed of the material increases to conserve angular momentum.

Figure 3: Conservation of Angular Momentum. When a spinning figure skater brings in her arms, their distance from her spin center is smaller, so her speed increases. When her arms are out, their distance from the spin center is greater, so she slows down.

Key Concepts and Summary

In his Principia, Isaac Newton established the three laws that govern the motion of objects: (1) objects continue to be at rest or move with a constant velocity unless acted upon by an outside force (2) an outside force causes an acceleration (and changes the momentum) for an object and (3) for every action there is an equal and opposite reaction. Momentum is a measure of the motion of an object and depends on both its mass and its velocity. Angular momentum is a measure of the motion of a spinning or revolving object and depends on its mass, velocity, and distance from the point around which it revolves. The density of an object is its mass divided by its volume.


11.3 Conservation of Angular Momentum

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,

In this case, (Figure) becomes the law of conservation of angular momentum.

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:

Note that the total angular momentum

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

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

where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because

is smaller, the angular velocity

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

whereas her final rotational energy is

Because her moment of inertia has decreased,

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.

Example

Coupled Flywheels

A flywheel rotates without friction at an angular velocity

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

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.

Strategy

Part (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).

Solution

a. 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

b. Before contact, only one flywheel is rotating. The rotational kinetic energy of this flywheel is the initial rotational kinetic energy of the system,

. The final kinetic energy is

Therefore, the ratio of the final kinetic energy to the initial kinetic energy is

Thus, 3/4 of the initial kinetic energy is lost to the coupling of the two flywheels.

Significance

Since 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.

Check Your Understanding

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

. What is the new rotation rate?

[reveal-answer q=”fs-id1165037935148″]Show Solution[/reveal-answer]

Using conservation of angular momentum, we have

Example

Dismount from a High Bar

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.

Strategy

Using 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.

Solution

The moment of inertia at full extension is

The moment of inertia in the tuck is

.
Conservation of angular momentum:

Time interval in the tuck:

In 0.5 s, he will be able to execute two revolutions at 4.0 rev/s.

Significance

Note 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.

Example

Conservation of Angular Momentum of a Collision

is moving horizontally with a speed of

The bullet strikes and becomes embedded in the edge of a solid disk of mass

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.

Strategy

For 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

, 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

. 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.

Solution

The initial angular momentum of the system is

The moment of inertia of the system with the bullet embedded in the disk is

The final angular momentum of the system is

Thus, by conservation of angular momentum,

Significance

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

  • In the absence of external torques, a system’s total angular momentum is conserved. This is the rotational counterpart to linear momentum being conserved when the external force on a system is zero.
  • For a rigid body that changes its angular momentum in the absence of a net external torque, conservation of angular momentum gives

Conceptual Questions

What is the purpose of the small propeller at the back of a helicopter that rotates in the plane perpendicular to the large propeller?

[reveal-answer q=”fs-id1165038199297″]Show Solution[/reveal-answer]

Without the small propeller, the body of the helicopter would rotate in the opposite sense to the large propeller in order to conserve angular momentum. The small propeller exerts a thrust at a distance R from the center of mass of the aircraft to prevent this from happening.

As the rope of a tethered ball winds around a pole, what happens to the angular velocity of the ball?

[reveal-answer q=”fs-id1165038356984″]Show Solution[/reveal-answer]

The angular velocity increases because the moment of inertia is decreasing.

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.

[reveal-answer q=�″]Show Answer[/reveal-answer]
[hidden-answer a=�″]More mass is concentrated near the rotational axis, which decreases the moment of inertia causing the star to increase its angular velocity.[/hidden-answer]

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.

Problems

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?

and it has a rotational period of approximately 28 days. If the Sun should collapse into a white dwarf of radius

what would its period be if no mass were ejected and a sphere of uniform density can model the Sun both before and after?

[reveal-answer q=”fs-id1165037231632″]Show Solution[/reveal-answer]

A cylinder with rotational inertia

rotates clockwise about a vertical axis through its center with angular speed

A second cylinder with rotational inertia

rotates counterclockwise about the same axis with angular speed

. 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

and after the tuck during the somersaults is

, 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?

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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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Shown below is a small particle of mass 20 g that is moving at a speed of 10.0 m/s when it collides and sticks to the edge of a uniform solid cylinder. The cylinder is free to rotate about its axis through its center and is perpendicular to the page. The cylinder has a mass of 0.5 kg and a radius of 10 cm, and is initially at rest. (a) What is the angular velocity of the system after the collision? (b) How much kinetic energy is lost in the collision?

A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk

rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 10.0 rad/s. The bug crawls to the center of the disk. (a) What is the new angular velocity of the disk? (b) What is the change in the kinetic energy of the system? (c) If the bug crawls back to the outer edge of the disk, what is the angular velocity of the disk then? (d) What is the new kinetic energy of the system? (e) What is the cause of the increase and decrease of kinetic energy?

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back to the original value

back to the original value

e. work of the bug crawling on the disk

A uniform rod of mass 200 g and length 100 cm is free to rotate in a horizontal plane around a fixed vertical axis through its center, perpendicular to its length. Two small beads, each of mass 20 g, are mounted in grooves along the rod. Initially, the two beads are held by catches on opposite sides of the rod’s center, 10 cm from the axis of rotation. With the beads in this position, the rod is rotating with an angular velocity of 10.0 rad/s. When the catches are released, the beads slide outward along the rod. (a) What is the rod’s angular velocity when the beads reach the ends of the rod? (b) What is the rod’s angular velocity if the beads fly off the rod?

A merry-go-round has a radius of 2.0 m and a moment of inertia

A boy of mass 50 kg runs tangent to the rim at a speed of 4.0 m/s and jumps on. If the merry-go-round is initially at rest, what is the angular velocity after the boy jumps on?

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A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge? The child is initially at rest.

Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

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(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is

. (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 15.0 s?

Twin skaters approach one another as shown below and lock hands. (a) Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius. (b) Compare the initial kinetic energy and final kinetic energy.

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A baseball catcher extends his arm straight up to catch a fast ball with a speed of 40 m/s. The baseball is 0.145 kg and the catcher’s arm length is 0.5 m and mass 4.0 kg. (a) What is the angular velocity of the arm immediately after catching the ball as measured from the arm socket? (b) What is the torque applied if the catcher stops the rotation of his arm 0.3 s after catching the ball?

In 2015, in Warsaw, Poland, Olivia Oliver of Nova Scotia broke the world record for being the fastest spinner on ice skates. She achieved a record 342 rev/min, beating the existing Guinness World Record by 34 rotations. If an ice skater extends her arms at that rotation rate, what would be her new rotation rate? Assume she can be approximated by a 45-kg rod that is 1.7 m tall with a radius of 15 cm in the record spin. With her arms stretched take the approximation of a rod of length 130 cm with

of her body mass aligned perpendicular to the spin axis. Neglect frictional forces.

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Moment of inertia in the record spin:

A satellite in a geosynchronous circular orbit is 42,164.0 km from the center of Earth. A small asteroid collides with the satellite sending it into an elliptical orbit of apogee 45,000.0 km. What is the speed of the satellite at apogee? Assume its angular momentum is conserved.

A gymnast does cartwheels along the floor and then launches herself into the air and executes several flips in a tuck while she is airborne. If her moment of inertia when executing the cartwheels is

and her spin rate is 0.5 rev/s, how many revolutions does she do in the air if her moment of inertia in the tuck is

and she has 2.0 s to do the flips in the air?

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Her spin rate in the air is:


She can do four flips in the air.

The centrifuge at NASA Ames Research Center has a radius of 8.8 m and can produce forces on its payload of 20 gs or 20 times the force of gravity on Earth. (a) What is the angular momentum of a 20-kg payload that experiences 10 gs in the centrifuge? (b) If the driver motor was turned off in (a) and the payload lost 10 kg, what would be its new spin rate, taking into account there are no frictional forces present?

A ride at a carnival has four spokes to which pods are attached that can hold two people. The spokes are each 15 m long and are attached to a central axis. Each spoke has mass 200.0 kg, and the pods each have mass 100.0 kg. If the ride spins at 0.2 rev/s with each pod containing two 50.0-kg children, what is the new spin rate if all the children jump off the ride?

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Moment of inertia with all children aboard:

An ice skater is preparing for a jump with turns and has his arms extended. His moment of inertia is

while his arms are extended, and he is spinning at 0.5 rev/s. If he launches himself into the air at 9.0 m/s at an angle of

with respect to the ice, how many revolutions can he execute while airborne if his moment of inertia in the air is

A space station consists of a giant rotating hollow cylinder of mass

including people on the station and a radius of 100.00 m. It is rotating in space at 3.30 rev/min in order to produce artificial gravity. If 100 people of an average mass of 65.00 kg spacewalk to an awaiting spaceship, what is the new rotation rate when all the people are off the station?

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from the Sun with an orbital period of 165 years. Planetesimals in the outer primordial solar system 4.5 billion years ago coalesced into Neptune over hundreds of millions of years. If the primordial disk that evolved into our present day solar system had a radius of

km and if the matter that made up these planetesimals that later became Neptune was spread out evenly on the edges of it, what was the orbital period of the outer edges of the primordial disk?


A tensor can not be equal to zero in one frame and not be equal to zero in another frame.

If you consider an angular velocity ##omega## relative an inertial frame then the phrase "relative an inertial frame" is a part of the definition of the vector ##omega##. Once you have defined the vector ##omega## you can expand this vector in any frame you wish. If ##omega e 0## then it is so in any frame.

If you consider an angular velocity relativ noninertial frame then this angular velocity is just another vector, not the same as the previously discussed one

Even in the body frame, if the body is rotating, the angular velocity vector is not zero. The choice of a reference frame is essentially just choosing a set of base vectors on which to resolve the velocity vector.

Now there may be some confusion about what it means to rotate. The best understanding is to say that rotation refers to rotation with respect to an inertial frame. If there is rotation with respect to an inertial frame, then the vector exists and is nonzero, and it may be resolved on any convenient frame, including a body frame.

I am not sure I understand what you mean by "exist". A velocity vector has magnitude and direction. A coordinate system is necessary to specify the direction. If you take that reference away, the vector loses one of its two principal properties and becomes a scalar. Doesn't that mean that it loses its "existence" as a vector?

On edit: The exchange between @Dr.D and me has gone off-thread to private messaging in order to keep the main discussion focused.


Conservation of Angular Momentum

We can now understand why Earth keeps on spinning. As we saw in the previous example, ( ext<δ>L=( extphantom< ule<0.25em><0ex>> au ) ext<δ>t) . This equation means that, to change angular momentum, a torque must act over some period of time. Because Earth has a large angular momentum, a large torque acting over a long time is needed to change its rate of spin. So what external torques are there? Tidal friction exerts torque that is slowing Earth’s rotation, but tens of millions of years must pass before the change is very significant. Recent research indicates the length of the day was 18 h some 900 million years ago. Only the tides exert significant retarding torques on Earth, and so it will continue to spin, although ever more slowly, for many billions of years.

What we have here is, in fact, another conservation law. If the net torque is zero, then angular momentum is constant or conserved. We can see this rigorously by considering ( extphantom< ule<0.25em><0ex>> au =cfrac< ext<δ>L>< ext<δ>t>) for the situation in which the net torque is zero. In that case,

If the change in angular momentum ( ext<δ>L) is zero, then the angular momentum is constant thus,

These expressions are the law of conservation of angular momentum. Conservation laws are as scarce as they are important.

An example of conservation of angular momentum is seen in the figure below, in which an ice skater is 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 and because the friction is exerted very close to the pivot point. (Both (F) and (r) are small, and so ( au ) is negligibly small.) Consequently, she can spin for quite some time. She can do something else, too. She can 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

Expressing this equation in terms of the moment of inertia,

where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because (Iprime ) is smaller, the angular velocity (omega prime ) must increase to keep the angular momentum constant. The change can be dramatic, as the following example shows.

(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. In the next image, 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.

Example: Calculating the Angular Momentum of a Spinning Skater

Suppose an ice skater, such as the one in the figure above, is spinning at 0.800 rev/ s with her arms extended. She has a moment of inertia of (2 ext<.> ext<34>phantom< ule<0.25em><0ex>> extcdot < ext>^<2>) with her arms extended and of (0 ext<.> ext<363>phantom< ule<0.25em><0ex>> extcdot < ext>^<2>) with her arms close to her body. (These moments of inertia are based on reasonable assumptions about a 60.0-kg skater.) (a) What is her angular velocity in revolutions per second after she pulls in her arms? (b) What is her rotational kinetic energy before and after she does this?

In the first part of the problem, we are looking for the skater’s angular velocity (omega prime ) after she has pulled in her arms. To find this quantity, we use the conservation of angular momentum and note that the moments of inertia and initial angular velocity are given. To find the initial and final kinetic energies, we use the definition of rotational kinetic energy given by

Solution for (a)

Because torque is negligible (as discussed above), the conservation of angular momentum given in (mathrm=Iprime omega prime ) is applicable. Thus,

Solving for (omega prime ) and substituting known values into the resulting equation gives

Solution for (b)

Rotational kinetic energy is given by

The initial value is found by substituting known values into the equation and converting the angular velocity to rad/s:

The final rotational kinetic energy is

Substituting known values into this equation gives

In both parts, there is an impressive increase. First, the final angular velocity is large, although most world-class skaters can achieve spin rates about this great. Second, the final kinetic energy is much greater than the initial kinetic energy. The increase in rotational kinetic energy comes from work done by the skater in pulling in her arms. This work is internal work that depletes some of the skater’s food energy.

There are several other examples of objects that increase their rate of spin because something reduced their moment of inertia. Tornadoes are one example. Storm systems that create tornadoes are slowly rotating. When the radius of rotation narrows, even in a local region, angular velocity increases, sometimes to the furious level of a tornado. Earth is another example. Our planet was born from a huge cloud of gas and dust, the rotation of which came from turbulence in an even larger cloud. Gravitational forces caused the cloud to contract, and the rotation rate increased as a result. (See the figure below.)

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.

In case of human motion, one would not expect angular momentum to be conserved when a body interacts with the environment as its foot pushes off the ground. Astronauts floating in space aboard the International Space Station have no angular momentum relative to the inside of the ship if they are motionless. Their bodies will continue to have this zero value no matter how they twist about as long as they do not give themselves a push off the side of the vessel.

Check Your Understanding

Is angular momentum completely analogous to linear momentum? What, if any, are their differences?

Solution

Yes, angular and linear momentums are completely analogous. While they are exact analogs they have different units and are not directly inter-convertible like forms of energy are.


Inertia Tensor

I now deal with a third topic in rather more detail, namely the relation between angular momentum ( f L ) and angular velocity ( oldsymbol ). The reader will be familiar from elementary (and two- dimensional) mechanics with the relation (L = I omega ). What we are going to find in the three- dimensional solid-body case is that the relation is (f = mathbboldsymbol ). Here ( f L ) and ( oldsymbol ) are, of course, vectors, but they are not necessarily parallel to each other. They are parallel only if the body is rotating about a principal axis of rotation. The quantity ( mathbb ) is a tensor known as the inertia tensor. Readers will be familiar with the equation (<f F>= m <f a>). Here the two vectors are in the same direction, and m is a scalar quantity that does not change the direction of the vector that it multiplies. A tensor usually (unless its matrix representation is diagonal) changes the direction as well as the magnitude of the vector that it multiplies. The reader might like to think of other examples of tensors in physics. There are several. One that comes to mind is the permittivity of an anisotropic crystal in the equation ( <f D= oldsymbolE>) and ( f E ) are not parallel unless they are both directed along one of the crystallographic axes.

If there are no external torques acting on a body, ( f L ) is constant in both magnitude and direction. The instantaneous angular velocity vector, however, is not fixed either in space or with respect to the body - unless the body is rotating about a principal axis and the inertia tensor is diagonal.

So much for a preview and a qualitative description. Now down to work.

I am going to have to assume familiarity with the equation for the components of the cross product of two vectors:

I am also going to assume that the reader knows that the angular momentum of a particle of mass ( m ) at position vector ( f r ) (components (( x,y,z ) ) ) and moving with velocity (f v ) (components ( ( dot , dot , dot ))) is ( m < f r imes v>). For a collection of particles, (or an extended solid body, which, I'm told, consists of a collection of particles called atoms), the angular momentum is

I also assume that the relation between linear velocity ( f v )( ( dot , dot , dot ) )and angular velocity ( oldsymbol ) ( ( omega_ , omega_ , omega_ ) ) is understood to be ( f v = oldsymbol imes r ), so that , for example ( dot = omega_ y - omega_x. ) then

[ < f L >= left(eginL_x L_y L_z end ight) = left(eginA -H -G -H quad B -F -G -F quad Cend ight) left(eginomega_x omega_y omega_z end ight) label ]

This is the equation ( f L =mathbb oldsymbol ) referred to above. The inertia tensor is sometimes written in the form

[mathbb I = left( eginI_ I _ I _ I_ I _ I _ I_ I _ I _ end ight) ]

so that, for example, ( I_ = &minus H ). It is a symmetric matrix (but it is not an orthogonal matrix).


Watch the video: More on moment of inertia. Moments, torque, and angular momentum. Physics. Khan Academy (September 2022).