Circular Motion

Centripetal Acceleration Ex: A 5-kg object moves at a constant speed of 10 m/s in a 5.0 m radius circle. What is the object's acceleration? Step 1: Pull out information from question. Vt= 10m/s r= 5m m= 5kg Step 2: Find the appropriate equation to plug in given to find the acceleration. a=V^2/r Step 3: Plug in given and solve. a=10^2/5 a=100/5 a= 20m/s^2

Centripetal Force Ex: Using the last problem we found out that a= 20m/s^2. Now we want to find the centripetal force. Step 1: Pull out information from question. (Same as previous) Vt= 10m/s r= 5m m= 5kg Step 2: Find the appropriate equation to plug in given to find the acceleration. Fnet= ma Step 3: Plug in given and solve. Fnet= (5)(20) Fnet= 100N

Reference Videos Centripetal Acceleration: http://youtu.be/EX5DZ2MHlV4 http://youtu.be/kgtUD71_txE Centripetal Force: http://youtu.be/dN8eTFSfaEg

Forces Present During Circular Motion Tension Normal Force Static Friction Gravity Coulomb Force FYI- During Uniform circular motion the acceleration of the object always points toward the center of radius.

Newton’s Law of Universal Gravitation Newton’s Law of Universal Gravitation- states that any two bodies in the universe attract each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This law involves things such as satellite, planetary orbits, falling objects, and tides.

Applying Newton’s Law of Universal Gravitation Formula= (G)(m1*m2)/(d^2) Let G= (6.673*10^-11 N m^2/kg^2) m1= (5.98*10^24kg) m2= (70kg) d= (6.38*10^6m) Fgrav= (6.673*10^-11)(5.98*10^24)(70)/(6.38*10^6) Fgrav= 686N

Kepler’s Laws of Planetary Motion Kepler’s three Laws The path of the planets about the sun is elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses) An imaginary line drawn from the center of the sun to the center of the planet will 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 of the cubes of their average distances from the sun. (The Law of Harmonies)

Kepler’s Laws References Watch the video below for help! http://youtu.be/uIYRFxF1DjI

Orbital Speed and Period Equations for orbital speed and period. Speed: v= SQRT(g*m/R) Acceleration: a= g*m/R^2 Period t^2/R^3= 4*pi^2/g*m In the chart to the right are life applications of these formulas to find the orbital speeds and periods of the planets.

Applying Orbital Speed Find the orbital speed of a planet with g= (6.673*10^-11), m= (5.98*10^24), and R= (6.47*10^6). Choose a formula from the previous slide. v= SQRT [ (g)*(m)/(R) Plug in given information. v = SQRT [ (6.673 x 10^-11 N m2/kg2) • (5.98 x 10^24 kg) / (6.47 x 10^6 m) ] Solve. v = 7.85 x 103 m/s

Apparent Weightlessness Weightlessness is described by an individual having no external objects touching his/her body at one point in time. This occurs when an individual is in a moment of freefall. During freefall the only force acting upon the individual is gravity. Since gravity can not be felt the individual acquires a sense of weightlessness.

Torque vs. Force Force: Strength or energy as an attribute of physical action or movement. Torque: A twisting force that tends to cause rotation. The major difference between force and torque is that a force is more general and can be many things. Torque is one specific type of force that can take a rotational force and transform it into a lateral force.

MORE REFERENCES Helpful Websites YouTube Videos http://www.physicsclassroom.com/class/circles http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation YouTube Videos http://youtu.be/Ah8jt8k3lAE http://youtu.be/6TGCPXhMLtU