newton's second law - revisited

6/07/12 9:55 PMNewton's Second Law - Revisited

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Applications of Circular Motion

Circular Motion and Satellite Motion - Lesson 2

Newton's Second Law - Revisited

Newton's second law states that the acceleration of an object is directly proportionalto the net force acting upon the object and inversely proportional to the mass of theobject. The law is often expressed in the form of the following two equations.

In Unit 2 of The Physics Classroom, Newton's secondlaw was used to analyze a variety of physical situa-tions. The idea was that if any given physical situa-

tion is analyzed in terms of the individual forces that are acting upon an object, thenthose individual forces must add up as vectors to the net force. Furthermore, the netforce must be equal to the mass times the acceleration. Subsequently, the accelerationof an object can be found if the mass of the object and the magnitudes and directionsof each individual force are known. And the magnitude of any individual force canbe determined if the mass of the object, the acceleration of the object, and the magni-tude of the other individual forces are known. The process of analyzing such physicalsituations in order to determine unknown information is dependent upon the abilityto represent the physical situation by means of a free-body diagram. A free-body dia-gram is a vector diagram that depicts the relative magnitude and direction of all theindividual forces that are acting upon the object.

In this Lesson, we will use Unit 2 principles (free-body diagrams, Newton's secondlaw equation, etc.) and circular motion concepts in order to analyze a variety of phys-ical situations involving the motion of objects in circles or along curved paths. Themathematical equations discussed in Lesson 1 and the concept of a centripetal forcerequirement will be applied in order to analyze roller coasters and other amusementpark rides and various athletic movements.

To illustrate how circular motion principles can be combined with Newton's secondlaw to analyze a physical situation, consider a car moving in a horizontal circle on alevel surface. The diagram below depicts the car on the left side of the circle.

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Applying the concept of a centripetalforce requirement, we know that thenet force acting upon the object is di-rected inwards. Since the car is posi-tioned on the left side of the circle,

the net force is directed rightward. An analysis of the situa-tion would reveal that there are three forces acting upon theobject - the force of gravity (acting downwards), the normalforce of the pavement (acting upwards), and the force of fric-tion (acting inwards or rightwards). It is the friction forcethat supplies the centripetal force requirement for the car tomove in a horizontal circle. Without friction, the car would

turn its wheels but would not move in a circle (as is the case on an icy surface). Thisanalysis leads to the free-body diagram shown at the right. Observe that each force isrepresented by a vector arrow that points in the specific direction that the force acts;also notice that each force is labeled according to type (Ffrict, Fnorm, and Fgrav). Suchan analysis is the first step of any problem involving Newton's second law and a cir-cular motion.

Now consider the following two problems pertaining to this physical scenario of thecar making a turn on a horizontal surface.

Sample Problem #1

The maximum speed with which a 945-kg car makes a 180-degree turn is 10.0 m/s. The radiusof the circle through which the car is turning is 25.0 m. Determine the force of friction and thecoefficient of friction acting upon the car.

Sample Problem #2

The coefficient of friction acting upon a 945-kg car is 0.850. The car is making a 180-degreeturn around a curve with a radius of 35.0 m. Determine the maximum speed with which thecar can make the turn.

Sample problem #1 provides kinematic information (v and R) and requests the valueof an individual force. As such the solution of the problem will demand that the ac-celeration and the net force first be determined; then the individual force value can befound by use of the free-body diagram. Sample problem #2 provides informationabout the individual force values (or at least information that allows for the determi-



nation of the individual force values) and requests the value of the maximum speedof the car. As such, its solution will demand that individual force values be used todetermine the net force and acceleration; then the acceleration can be used to deter-mine the maximum speed of the car. The two problems will be solved using the samegeneral principles. Yet because the given and requested information is different ineach, the solution method will be slightly different.

Solution to Sample Problem #1

The known information and requested information in sample problem #1 is:

Known Information:

m = 945 kg

v = 10.0 m/s

R = 25.0 m

Requested Information:

Ffrict = ???

! = ????

(! - coefficient of friction)

The mass of the object can be used to determine the force of gravity acting in thedownward direction. Use the equation

Fgrav = m * g

where g is 9.8 m/s/s. Knowing that there is no vertical acceleration of the car, it canbe concluded that the vertical forces balance each other. Thus, Fgrav = Fnorm= 9261 N.This allows us to determine two of the three forces identified in the free-body dia-gram. Only the friction force remains unknown.

Since the force of friction is the only horizontal force, itmust be equal to the net force acting upon the object. So ifthe net force can be determined, then the friction force isknown. To determine the net force, the mass and the kine-matic information (speed and radius) must be substitutedinto the following equation:

Substituting the given values yields a net force of 3780 Newton. Thus,the force of friction is 3780 N.



the force of friction is 3780 N.

Finally the coefficient of friction (!) can be determined using the equation that relatesthe coefficient of friction to the force of friction and the normal force.

Substituting 3780 N for Ffrict and 9261 N for Fnorm yields a coeffi-cient of friction of 0.408.

Solution to Sample Problem #2

Once again, the problem begins by identifying the known and requested information.The known information and requested information in the sample problem #2 is:

Known Information:

m = 945 kg

! = 0.85(coefficientof friction)

R = 35.0 m

Requested Information:

v = ???

(the minimum speed would be the speedachieved with the given friction coefficient)

The mass of the car can be used to determine the force of gravity acting in the down-ward direction. Use the equation

Fgrav = m * g

where g is 9.8 m/s/s. Knowing that there is no vertical acceleration of the car, it canbe concluded that the vertical forces balance each other. Thus, Fgrav = Fnorm= 9261 N.Since the coefficient of friction (!) is given, the force of friction can be determined us-ing the following equation:

This allows us to determine all three forces identified in the free-body diagram.



The net force acting upon any object is the vector sumof all individual forces acting upon that object. So if allindividual force values are known (as is the case here),the net force can be calculated. The vertical forces addto 0 N. Since the force of friction is the only horizontalforce, it must be equal to the net force acting upon the

object. Thus, Fnet = 7872 N.

Once the net force is determined, the acceleration can be quickly calculated using thefollowing equation.

Fnet = m*a

Substituting the given values yields an acceleration of 8.33 m/s/s. Finally, the speed atwhich the car could travel around the turn can be calculated using the equation forcentripetal acceleration:

Substituting the known values for a and R into this equation and solving al-gebraically yields a maximum speed of 17.1 m/s.

Each of the two sample problems above was solved using the same basic problem-solving approach. The approach can be summarized as follows.

Suggested Method of Solving Circular Motion Problems

1. From the verbal description of the physical situation, construct a free-bodydiagram. Represent each force by a vector arrow and label the forcesaccording to type.

2. Identify the given and the unknown information (express in terms ofvariables such as m= , a= , v= , etc.).

3. If any of the individual forces are directed at angles to the horizontal andvertical, then use vector principles to resolve such forces into horizontal andvertical components.

4. Determine the magnitude of any known forces and label on the free-bodydiagram.(For example, if the mass is given, then the Fgrav can be determined. And asanother example, if there is no vertical acceleration, then it is known that thevertical forces or force components balance, allowing for the possibledetermination of one or more of the individual forces in the verticaldirection.)

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5. Use circular motion equations to determine any unknown information.(For example, if the speed and the radius are known, then the acceleration canbe determined. And as another example, if the period and radius are known,then the acceleration can be determined.)

6. Use the remaining information to solve for the requested information.a. If the problem requests the value of an individual force, then use the

kinematic information (R, T and v) to determine the acceleration and theFnet ; then use the free-body diagram to solve for the individual forcevalue.If the problem requests the value of the speed or radius, then use thevalues of the individual forces to determine the net force andacceleration; then use the acceleration to determine the value of thespeed or radius.

The method prescribed above will serve you well as you approach circular motionproblems. However, one caution is in order. Every physics problem differs from theprevious problem. As such, there is no magic formula for solving every one. Using anappropriate approach to solving such problems (which involves constructing a FBD,identifying known information, identifying the requested information, and usingavailable equations) will never eliminate the need to think, analyze and problem-solve. For this reason, make an effort to develop an appropriate approach to everyproblem; yet always engage your critical analysis skills in the process of the solution.If physics problems were a mere matter of following a foolproof, 5-step formula orusing some memorized algorithm, then we wouldn't call them "problems."

Check Your Understanding

Use your understanding of Newton's second law and circular motion principles todetermine the unknown values in the following practice problems. Click the buttonto check your answers.

1. A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of1.00 m. At the top of the circular loop, the speed of the bucket is 4.00 m/s. Determinethe acceleration, the net force and the individual force values when the bucket is atthe top of the circular loop.

m = 1.5 kg



a = ________ m/s/s

Fnet = _________ N

See Answer

2. A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of1.00 m. At the bottom of the circular loop, the speed of the bucket is 6.00 m/s. Deter-mine the acceleration, the net force and the individual force values when the bucketis at the bottom of the circular loop.

m = 1.5 kg

a = ________ m/s/s

Fnet = _________ N

See Answer


Gravity is More Than a Name

Nearly every child knows of the word gravity. Gravity is the name associated with



the mishaps of the milk spilled from the breakfast table to thekitchen floor and the youngster who topples to the pavementas the grand finale of the first bicycle ride. Gravity is thename associated with the reason for "what goes up, mustcome down," whether it be the baseball hit in the neighbor-hood sandlot game or the child happily jumping on the back-yard mini-trampoline. We all know of the word gravity - it isthe thing that causes objects to fall to Earth. Yet the role of

physics is to do more than to associate words with phenomenon. The role of physicsis to explain phenomenon in terms of underlying principles. The goal is to explainphenomenon in terms of principles that are so universal that they are capable of ex-plaining more than a single phenomenon but a wealth of phenomenon in a consistentmanner. Thus, a student's conception of gravity must grow in sophistication to thepoint that it becomes more than a mere name associated with falling phenomenon.Gravity must be understood in terms of its cause, its source, and its far-reaching im-plications on the structure and the motion of the objects in the universe.

Certainly gravity is a force that exists between the Earth and the objects that are nearit. As you stand upon the Earth, you experience this force. We have become accus-tomed to calling it the force of gravity and have even represented it by the symbolFgrav. Most students of physics progress at least to this level of sophistication con-cerning the notion of gravity. This same force of gravity acts upon our bodies as wejump upwards from the Earth. As we rise upwards after our jump, the force of gravi-ty slows us down. And as we fall back to Earth after reaching the peak of our motion,the force of gravity speeds us up. In this sense, the force gravity causes an accelera-tion of our bodies during this brief trip away from the earth's surface and back. Infact, many students of physics have become accustomed to referring to the actual ac-celeration of such an object as the acceleration of gravity. Not to be confused withthe force of gravity (Fgrav), the acceleration of gravity (g) is the acceleration experi-enced by an object when the only force acting upon it is the force of gravity. On andnear Earth's surface, the value for the acceleration of gravity is approximately 9.8m/s/s. It is the same acceleration value for all objects, regardless of their mass (andassuming that the only significant force is gravity). Many students of physicsprogress this far in their understanding of the notion of gravity.

In Lesson 3, we will build on this understanding of gravitation, making an attempt tounderstand the nature of this force. Many questions will be asked: How and by



understand the nature of this force. Many questions will be asked: How and bywhom was gravity discovered? What is the cause of this force that we refer to with

the name of gravity? What variables affect the actual value of theforce of gravity? Why does the force of gravity acting upon an ob-ject depend upon the location of the object relative to the Earth?How does gravity affect objects that are far beyond the surface ofthe Earth? How far-reaching is gravity's influence? And is the

force of gravity that attracts my body to the Earth related to the force of gravity be-tween the planets and the Sun? These are the questions that will be pursued. And ifyou can successfully answer them, then the sophistication of your understanding hasextended beyond the point of merely associating the name "gravity" with falling phe-nomenon.


Kepler's Three Laws

In the early 1600s, Johannes Kepler proposed three laws of planetary motion. Keplerwas able to summarize the carefully collected data of his mentor - Tycho Brahe - withthree statements that described the motion of planets in a sun-centered solar system.Kepler's efforts to explain the underlying reasons for such motions are no longer ac-cepted; nonetheless, the actual laws themselves are still considered an accurate de-scription of the motion of any planet and any satellite.

Kepler's three laws of planetary motion can be described as follows:

The path of the planets about the sun is elliptical in shape, with the center of thesun being located at one focus. (The Law of Ellipses)An imaginary line drawn from the center of the sun to the center of the planetwill sweep out equal areas in equal intervals of time. (The Law of Equal Areas)The ratio of the squares of the periods of any two planets is equal to the ratio ofthe cubes of their average distances from the sun. (The Law of Harmonies)

Kepler's first law - sometimes referred to as the law of ellipses - explains that planetsare orbiting the sun in a path described as an ellipse. An ellipse can easily be con-structed using a pencil, two tacks, a string, a sheet of paper and a piece of cardboard.Tack the sheet of paper to the cardboard using the two tacks. Then tie the string into



a loop and wrap the loop around the two tacks. Take your pencil and pull the stringuntil the pencil and two tacks make a triangle (see diagramat the right). Then begin to trace out a path with the pencil,keeping the string wrapped tightly around the tacks. The re-sulting shape will be an ellipse. An ellipse is a special curvein which the sum of the distances from every point on thecurve to two other points is a constant. The two other points

(represented here by the tack locations) are known as the foci of the ellipse. The clos-er together that these points are, the more closely that the ellipse resembles the shapeof a circle. In fact, a circle is the special case of an ellipse in which the two foci are atthe same location. Kepler's first law is rather simple - all planets orbit the sun in apath that resembles an ellipse, with the sun being located at one of the foci of that el-lipse.

Kepler's second law - sometimes referred to as the law of equal areas - describes thespeed at which any given planet will move while orbiting the sun. The speed atwhich any planet moves through space is constantly changing. A planet movesfastest when it is closest to the sun and slowest when it is furthest from the sun. Yet,if an imaginary line were drawn from the center of the planet to the center of the sun,that line would sweep out the same area in equal periods of time. For instance, if animaginary line were drawn from the earth to the sun, then the area swept out by theline in every 31-day month would be the same. This is depicted in the diagram be-low. As can be observed in the diagram, the areas formed when the earth is closest tothe sun can be approximated as a wide but short triangle; whereas the areas formedwhen the earth is farthest from the sun can be approximated as a narrow but long tri-angle. These areas are the same size. Since the base of these triangles are shortestwhen the earth is farthest from the sun, the earth would have to be moving moreslowly in order for this imaginary area to be the same size as when the earth is closestto the sun.



Kepler's third law - sometimes referred to as the law of harmonies - compares the or-bital period and radius of orbit of a planet to those of other planets. Unlike Kepler'sfirst and second laws that describe the motion characteristics of a single planet, thethird law makes a comparison between the motion characteristics of different planets.The comparison being made is that the ratio of the squares of the periods to the cubesof their average distances from the sun is the same for every one of the planets. As anillustration, consider the orbital period and average distance from sun (orbital radius)for Earth and mars as given in the table below.

PlanetPeriod

(s)Average

Dist. (m)

T2/R3

(s2/m3)

Earth 3.156 x 107s

1.4957 x1011

2.977 x 10-19

Mars 5.93 x 107s

2.278 x1011

2.975 x 10-19

Observe that the T2/R3 ratio is the same for Earth as it is for mars. In fact, if the sameT2/R3 ratio is computed for the other planets, it can be found that this ratio is nearly



T2/R3 ratio is computed for the other planets, it can be found that this ratio is nearlythe same value for all the planets (see table below). Amazingly, every planet has thesame T2/R3 ratio.

PlanetPeriod

(yr)

Ave.

Dist. (au)

T2/R3

(yr2/au3)

Mercury 0.241 0.39 0.98

Venus .615 0.72 1.01

Earth 1.00 1.00 1.00

Mars 1.88 1.52 1.01

Jupiter 11.8 5.20 0.99

Saturn 29.5 9.54 1.00

Uranus 84.0 19.18 1.00

Neptune 165 30.06 1.00

Pluto 248 39.44 1.00

(NOTE: The average distance value is given in astronomical units where 1a.u. is equal to the distance from the earth to the sun - 1.4957 x 1011 m. Theorbital period is given in units of earth-years where 1 earth year is the time

required for the earth to orbit the sun - 3.156 x 107 seconds. )

Kepler's third law provides an accurate description of the period and distance for aplanet's orbits about the sun. Additionally, the same law that describes the T2/R3 ra-tio for the planets' orbits about the sun also accurately describes the T2/R3 ratio forany satellite (whether a moon or a man-made satellite) about any planet. There issomething much deeper to be found in this T2/R3 ratio - something that must relateto basic fundamental principles of motion. In the next part of Lesson 4, these princi-ples will be investigated as we draw a connection between the circular motion princi-ples discussed in Lesson 1 and the motion of a satellite.

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How did Newton Extend His Notion of

Newton's comparison of the acceleration of the moon to the acceleration of objectson earth allowed him to establish that the moon is held in a circular orbit by theforce of gravity - a force that is inversely dependent upon the distance between thetwo objects' centers. Establishing gravity as the cause of the moon's orbit does notnecessarily establish that gravity is the cause of the planet's orbits. How then didNewton provide credible evidence that the force of gravity is meets the centripetalforce requirement for the elliptical motion of planets?

Recall from earlier in Lesson 3 that Johannes Kepler proposed three laws ofplanetary motion. His Law of Harmonies suggested that the ratio of the period oforbit squared (T2) to the mean radius of orbit cubed (R3) is the same value k for allthe planets that orbit the sun. Known data for the orbiting planets suggested thefollowing average ratio:

k = 2.97 x 10-19 s2/m3 = (T2)/(R3)

Newton was able to combine the law of universal gravitation with circular motionprinciples to show that if the force of gravity provides the centripetal force for theplanets' nearly circular orbits, then a value of 2.97 x 10-19 s2/m3 could be predictedfor the T2/R3 ratio. Here is the reasoning employed by Newton:

Consider a planet with mass Mplanet to orbit in nearly circular motion about thesun of mass MSun. The net centripetal force acting upon this orbiting planet isgiven by the relationship

Fnet = (Mplanet * v2) / R

This net centripetal force is the result of the gravitational force that attracts theplanet towards the sun, and can be represented as

Fgrav = (G* Mplanet * MSun ) / R2

Since Fgrav = Fnet, the above expressions for centripetal force and gravitationalforce are equal. Thus,

(Mplanet * v2) / R = (G* Mplanet * MSun ) / R2

Since the velocity of an object in nearly circular orbit can be approximated as v =(2*pi*R) / T,

v2 = (4 * pi2 * R2) / T2

Substitution of the expression for v2 into the equation above yields,

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Substitution of the expression for v2 into the equation above yields,

(Mplanet * 4 * pi2 * R2) / (R • T2) = (G* Mplanet * MSun ) / R2

By cross-multiplication and simplification, the equation can be transformed into

T2 / R3 = (Mplanet * 4 * pi2) / (G* Mplanet * MSun )

The mass of the planet can then be canceled from the numerator and thedenominator of the equation's right-side, yielding

T2 / R3 = (4 * pi2) / (G * MSun )

The right side of the above equation will be the same value for every planetregardless of the planet's mass. Subsequently, it is reasonable that the T2/R3 ratiowould be the same value for all planets if the force that holds the planets in theirorbits is the force of gravity. Newton's universal law of gravitation predicts resultsthat were consistent with known planetary data and provided a theoreticalexplanation for Kepler's Law of Harmonies.

Scientists know much more about the planets than they did in Kepler's days. Use ThePlanets widget bleow to explore what is known of the various planets.

Check Your Understanding

1. Our understanding of the elliptical motion of planets about the Sun spanned sever-al years and included contributions from many scientists.

a. Which scientist is credited with the collection of the data necessary tosupport the planet's elliptical motion?

b. Which scientist is credited with the long and difficult task of analyzingthe data?

c. Which scientist is credited with the accurate explanation of the data?



See Answer

2. Galileo is often credited with the early discovery of four of Jupiter's many moons.The moons orbiting Jupiter follow the same laws of motion as the planets orbiting thesun. One of the moons is called Io - its distance from Jupiter's center is 4.2 units and itorbits Jupiter in 1.8 Earth-days. Another moon is called Ganymede; it is 10.7 unitsfrom Jupiter's center. Make a prediction of the period of Ganymede using Kepler'slaw of harmonies.

See Answer

3. Suppose a small planet is discovered that is 14 times as far from the sun as theEarth's distance is from the sun (1.5 x 1011 m). Use Kepler's law of harmonies to pre-dict the orbital period of such a planet. GIVEN: T2/R3 = 2.97 x 10-19 s2/m3

See Answer

4. The average orbital distance of Mars is 1.52 times the average orbital distance ofthe Earth. Knowing that the Earth orbits the sun in approximately 365 days, use Ke-pler's law of harmonies to predict the time for Mars to orbit the sun.



See Answer

Orbital radius and orbital period data for the four biggest moons of Jupiter are listedin the table below. The mass of the planet Jupiter is 1.9 x 1027 kg. Base your answersto the next five questions on this information.

Jupiter'sMoon Period (s) Radius (m) T2/R3

Io 1.53 x 105 4.2 x 108 a.Europa 3.07 x 105 6.7 x 108 b.

Ganymede 6.18 x 105 1.1 x 109 c.Callisto 1.44 x 106 1.9 x 109 d.

5. Determine the T2/R3 ratio (last column) for Jupiter's moons.

See Answer

6. What pattern do you observe in the last column of data? Which law of Kepler'sdoes this seem to support?

See Answer

7. Use the graphing capabilities of your TI calculator to plot T2 vs. R3 (T2 should beplotted along the vertical axis) and to determine the equation of the line. Write theequation in slope-intercept form below.

See Answer



See graph below.

8. How does the T2/R3 ratio for Jupiter (as shown in the last column of the data table)compare to the T2/R3 ratio found in #7 (i.e., the slope of the line)?

See Answer

9. How does the T2/R3 ratio for Jupiter (as shown in the last column of the data table)compare to the T2/R3 ratio found using the following equation? (G=6.67x10-11

N*m2/kg2 and MJupiter = 1.9 x 1027 kg)

T2 / R3 = (4 * pi2) / (G * MJupiter )

See Answer

Graph for question #6

Return to Question #6



Return to Question #6

newton's second law - revisited

Documents

centripetal force requirement

johannes kepler proposed

vertical forces balance

individual force values

circular motion principles

average orbital distance

net force acting

t2 r3 ratio found