IB+DP+Topic+6+Circular+Motion+and+Gravitation

//**Discover Centripetal Force**//
Experiment: [|Jell-O Motion] from www.crosswalk.com This experiment takes a little bit more time and a few more ingredients, but it can be really fun. You even get to eat the results. Items Needed: Directions: Begin making your centripetal force test chambers by setting out 3-5 paper cups. Mix one of the boxes of Jell-O according to the directions on the box. Fill each paper cup half full with the mixed Jell-O. Place the cups in the refrigerator till they set—about 2-3 hours. Once the Jell-O is set, gently place a grape on top of the Jell-O in the center of each cup. Then, mix the second box of Jell-O according to the directions. Pour the second batch onto the first. Be sure to leave about one inch from the top of each cup empty. Place the test chambers into the refrigerator till set. In the meantime, you can create the centripetal force machine. Cut a length of string 12-18 inches long. Staple the ends of the sting to opposite sides of the outside of a paper cup. Make sure the ends are secure by stapling them at least 2-4 times each. When the remaining Jell-O has set, place one test chamber into the centripetal force machine. Hold the string in the middle and spin the machine in a circular motion 20 times. Carefully remove the test chamber from the centripetal force machine. Place a paper plate on top of the test chamber. Flip it carefully so the test chamber's ingredients come out onto the plate. You may need to carefully slide the butter knife between the Jell-O and the inside of the cup to help it release onto the plate more easily. Observe where the grape is now. Is it in the same place? Why do you think this happened? Repeat the experiment with the rest of the test chambers to see if your results all agree with one another. Then eat your results. media type="youtube" key="O6yJG_fnJRc" width="560" height="315" MIT Physics Demo -- [|Centrifugal versus Centripetal Motion] [|mittechtv] Uploaded on 6 Nov 2008
 * Paper cups
 * Sturdy string or yarn
 * Grapes
 * Two small packages of very different colored Jell-Os (such as yellow and red)
 * Butter knife
 * Paper plates
 * Stapler

6.1 Circular motion Observable universe: Observations and subsequent deductions led to the realization that the force must act radially inwards in all cases of circular motion. (1.1)
 * // Nature of science: //**

• Period, frequency, angular displacement and angular velocity • Centripetal force • Centripetal acceleration
 * Understandings: **

//** Applications and skills: **// • Identifying the forces providing the centripetal forces such as tension, friction, gravitational, electrical, or magnetic • Solving problems involving centripetal force, centripetal acceleration, period, frequency, angular displacement, linear speed and angular velocity • Qualitatively and quantitatively describing examples of circular motion including cases of vertical and horizontal circular motion

**• Aim 6:** experiments could include (but are not limited to): mass on a string; observation and quantification of loop-the-loop experiences; friction of a mass on a turntable **• Aim 7:** technology has allowed for more accurate and precise measurements of circular motion, including data loggers for force measurements and video analysis of objects moving in circular motion
 * // Aims: //**

//** Data booklet reference: **//

Orbital motion
A force applied perpendicular to its displacement can result in circular motion.

[|Speed and velocity] from physicsclassroom.com [|Circular motion and inertia] from physicsclassroom.com [|Acceleration and circular motion] from physicsclassroom.com [|Centripetal force requirement] from physicsclassroom [|physicsclassroom.com worksheet answers]

UNIFORM CIRCULAR MOTION ACTIVITY: Investigate the variables of a ladybug undergoing circular motion on a carousel: LADYBUG REVOLUTION media type="youtube" key="CzcmUiD39VI" width="560" height="315" [|Circular Motion | A-Level Physics] The acceleration of a particle moving with constant speed in a circle is directed towards the centre of the circle. CENTRIPETAL ACCELERATION: The acceleration of a body moving in a circle. It is directed towards the centre of the circle. A body with a centripetal acceleration must be under the influence of a centripetal force.  v= wr so the centripetal acceleration is: Angular speed: How many radians in a second. Speed does not change but velocity changes so there is an acceleration. This is called centripetal acceleration and the direction of it is always toward the centre of the circle. Quiz) radius = 10 cm, speed of revolution is 600 rpm(revolutions per minute) 1. What is the frequency? 10Hz, 10 revolutions per second 2. What is the period? T = 1/f = 1/10 = 0.1 s 3. What is the angular speed? w=2pi/T = (2 x 3.14) / 0.1 = 62.8 rad/s 4. What is the velocity? v = wr = 6.28 m/s 5. What is the centripetal acceleration? a = w 2 r = (62.8 x 62.8) x 0.1 = 394.3 = 400 m/s 2

Identify the force producing circular motion in various situations. 1. Examples include gravitational force acting on the Moon 2. Friction acting sideways on the tyres of a car turning a corner.

media type="youtube" key="KuNsQJQ-NJY" width="448" height="251"
[|Uniform Circular Motion - Angular Speed, Period and Frequency] [|Circular Motion Lab] from creery.org

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[|Circular motion #6: Calculating orbital velocity] MrDGenova

Q1) A car goes round a curve in a road at constant speed. Explain why, although its speed is constant, it is accelerating.

Escape speed
[|Newton's mountain Applet] from [|Galileoandeinstein.physics.virginia.edu]

media type="youtube" key="-rOKCXir_xE" width="448" height="251" [|Circular motion #5: Calculating orbital acceleration] MrDGenova

Q2) A marble rolls down a track, the bottom part of which has been bent into a loop. The end A of the track, the bottom part of which has been bent into a loop. The end A of the track, from which the marble is released, is at a height of 0.80 m above the ground. Point B is the lowest point and point C the highest point of the loop. The diameter of the loop is 0.35 m. The mass of the marble is 50 g. Friction forces and any gain in kinetic energy due to the rotating of the marble can be ignored. The acceleration due to gravity, g=10 m/s 2 . a) Consider the marble when it is at point C. Draw an arrow to show the direction of the resultant force acting on the marble. [1mark]

b) State the names of the two forces acting on the marble at position B and C .[4marks]

c) Deduce that the speed of the marble is 3.0 m/s at position C. [3marks]

d) Determine the resultant force acting on the marble and hence determine the reaction force of the track on the marble. [ 4marks]

media type="youtube" key="p9dtW7maVGA" width="448" height="251" [|Circular Motion Loop The Loop Problem] Mike Spalding

6.2 – Newton’s law of gravitation Felix Baumgartner's supersonic freefall from 128k' - Mission Highlights || //** Nature of science: **// Laws: Newton’s law of gravitation and the laws of mechanics are the foundation for deterministic classical physics. These can be used to make predictions but do not explain why the observed phenomena exist. (2.4)
 * media type="youtube" key="9zso7ChaQXQ" width="560" height="315" || media type="youtube" key="r8soMU5cKYE" width="560" height="315" || media type="youtube" key="MTY1Kje0yLg" width="560" height="315" || media type="youtube" key="Jsc-pQIMxt8" width="560" height="315" Turning Gravity Into Light - Smarter Every Day 146 by [|SmarterEveryDay] Published on 8 Dec 2015 || media type="youtube" key="FHtvDA0W34I" width="560" height="315"

• Newton’s law of gravitation • Gravitational field strength
 * // Understandings: //**

//** Applications and skills: **// • Describing the relationship between gravitational force and centripetal force • Applying Newton’s law of gravitation to the motion of an object in circular orbit around a point mass • Solving problems involving gravitational force, gravitational field strength, orbital speed and orbital period • Determining the resultant gravitational field strength due to two bodies

• The laws of mechanics along with the law of gravitation create the deterministic nature of classical physics. Are classical physics and modern physics compatible? Do other areas of knowledge also have a similar division between classical and modern in their historical development?
 * // Theory of knowledge: //**

• The law of gravitation is essential in describing the motion of satellites, planets, moons and entire galaxies • Comparison to Coulomb’s law (see Physics sub-topic 5.1)
 * // Utilization: //**

**• Aim 4:** the theory of gravitation when combined and synthesized with the rest of the laws of mechanics allows detailed predictions about the future position and motion of planets
 * // Aims: //**

**Gravitational force and field** Newton’s universal law of gravitation. (Students should be aware that the masses in the force law are point masses. The force between two spherical masses whose separation is large compared to their radii is the same as if the two spheres were point masses with their masses concentrated at the centres of the spheres.)
 * // Data Booklet reference: //**

NEWTON’S LAW OF UNIVERSAL GRAVITATION: The gravitational force between any two point objects is always attractive; proportional to each mass and inversely proportional to the square of their distance apart. Newton, Einstein and Gravitation

[|NEWTON'S LAW OF GRAVITY EXPLAINED] - short audio slideshow from The Guardian newspaper

UNIVERSAL GRAVITATIONAL CONSTANT: The constant of proportionality in the Universal Law of Gravitation. G = 6.67 x 10 –11 Nm 2 kg –2 =Newton's Law of Universal Gravitation=

Experiment to calculate the Law is Cavendish's Experiment

=Cavendish's experiment= __[|The Cavendish Experiment] - [|Sixty Symbols]__ Published on 6 Jul 2011 || Determine the gravitational field due to one or more **point masses**. Gravity Force Lab media type="custom" key="29005621" Gravitational field strength. GRAVITATIONAL FIELD STRENGTH: The force per unit mass on a small test mass placed in the field **gravitational field strength** Derive an expression for gravitational field strength at the surface of a planet, assuming that all its mass is concentrated at its centre. g = F/m and F = GMm/r 2 so, g = GM/r 2
 * [[image:http://upload.wikimedia.org/wikipedia/commons/thumb/9/91/Cavendish_Torsion_Balance_Diagram.svg/548px-Cavendish_Torsion_Balance_Diagram.svg.png width="310" height="339"]] || media type="youtube" key="2PdiUoKa9Nw" width="560" height="315"

The field view eliminates the need for long distance signaling between two masses. Rather, it distorts the space about one mass. – the gravitational field distorts the space around the mass, gravitational curvature, that is causing it so that any other mass placed at any position in the field will “know” how to respond immediately. Bigger masses “curve” the fabric more than smaller masses.

Note that each mass “feels” a different “slope” and must travel at a particular speed to stay in orbit.

In the space surrounding the mass M which sets up the field we can release “test masses” m1 and m2 to determine the strength of the field. To simplify field drawings even more we take the convention of drawing “field lines” as a single arrow. PRACTICE 1: Sketch the gravitational field about the earth (a) as viewed from far away, and (b) as viewed “locally” (at the surface). PRACTICE 2: What would Ms. Lee observe the dropped ball’s acceleration to be, if the elevator were to accelerate downward at 10 ms-2? SOLUTION: Ms. Lee would observe the acceleration of the ball to be zero! She would think that the ball was “weightless!” The ball is NOT weightless, obviously. It is merely accelerating at the same rate as Ms. Lee!
 * (a) || (b) Text page 261 ||
 * [[image:Gravitational field on Earth.PNG width="174" height="170"]] || [[image:Gravitational field on Earth local 2.PNG width="394" height="169"]] ||

PRACTICE 3: The mass of the earth is M = 5.98 x 10 24 kg and the radius of the earth is R = 6.37 x 10 6 m. Find the gravitational field strength at the surface of the earth, and at a distance of one earth radius above its surface. SOLUTION: For r = R: g = GM / R^2 g = (6.67×10^−11)(5.98 x 10^24)/(6.37 x 10^6)^2 g = 9.83 Nkg-1 (ms^-2) For r = 2R: Since r is squared…just divide by 22 = 4. Thus g = 9.83 / 4 = 2.46 ms^-2

PRACTICE 4: A 525-kg satellite is launched from the earth’s surface to a height of one earth radius above the surface. What is its weight (a) at the surface, and (b) at altitude? SOLUTION: Use information from above: (a) AT SURFACE: g surface = 9.83 ms -2 F = mg we get F = (525)(9.83) = 5160 N (b) AT ALTITUDE: g surface+R = 2.46 ms -2 F = mg we get F = (525) x (2.46) = 1290 N

[|VARIATION IN EARTH'S GRAVITATIONAL FIELD] - article on BBC website

Orbital motion
When a particle of mass m orbiting a larger body of mass M in a circular orbit of radius r, there must be the centripetal force on the particle to maintain a constant orbit(Friction is negligible). F C = F G Therefore, mv 2 /r = GMm/r 2 v= sroot(GM/r) and thus, kinetic energy of the satellite = ½ GMm/r As potential energy of the satellite = - GMm/r Total energy of satellite in orbit = - GMm/2r [|Kinetic energy in an orbit] from www.schoolphysics.co.uk

PRACTICE 5: The moon has a mass of m = 7.36 x 10 22 kg. The mean distance between the earth and the moon is 3.82 x 10 8 m. 1. What is the speed of the moon in its orbit about earth? SOLUTION: For circular orbits, the gravitational force is the centripetal force. Thus F C = F G Use FC = FG = mv ^2 /r.

FG = 2.01 x 10^20 N. Then

2.01 x10^20 = ( 7.36 x 10^22) v 2 / 3.82 x 10^8

Then v = 1.02 x 10 3 ms-1

2. What is the period of the moon (in days) in its orbit about earth?

SOLUTION: Use v = d / t = 2pi x r/ T

v = 1.02 x 10 3 ms-1. Then T = 2pi x r/ v = 2pi( 3.82 x 10 8 ) / 1.02 x 10 3 = (2.35 x 10 6 s) = (2.35 x 10 6 s)(1 h / 3600 s)(1 d / 24 h) 27.2 days

media type="custom" key="29043501" Gravity and Orbits Phet

Derive Kepler's 3rd law T 2 is proportional to r 3 media type="youtube" key="afpYBsfbGqc" width="560" height="315" [|A2 Physics Exam Questions] Fields [|DrPhysicsA] Published on 16 May 2013 youtube.com  Examples of exam questions at Physics A2 level for Gravitational and Electric Fields covering Edexcel, AQA and OCR material