PHY 101 1nd Edition Lecture 20 Outline of Last Lecture I. 7.2 Rotational Motion Under Constant Angular AccelerationII. 7.3 Angular & Linear Quantitiesa. Tangential Speed & AccelerationIII. 7.4 Centripetal Accelerationa. Centripetal & Tangential AccelerationOutline of Current Lecture IV. Summary: Angular & Linear QuantitiesV. Forces Causing Centripetal AccelerationVI. Problem Solving StrategiesVII. 7.5 Newtonian Gravitation RevisitedCurrent LectureExample: Discus Throw- A discus thrower turns with an angular acceleration of a = 50rad/s2. Find the magnitude of the total acceleration and tangential speed when he reaches an angular speed of 10 rad/s.- Note the total acceleration a = (at2 + ac2)1/2 - Solution: find centripetal accelerationo At = rα = (0.8)(50) = 40m/s2o A = (40^2 + 80^2)^1/2 = 89 m/s2Summary: Angular & Linear Quantities- When a rigid object rotates:o The arc length (linear distance): Δs = rΔΘo The velocity is ltangential to its trajectoryo Tangential speed: v = rωo Tangential acceleration: at = rαo Centripetal acceleration: ac = vt2/rForces Causing Centripetal AccelerationThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- Even an object is doing a constant speed circular motion, it experiences a centripetal acceleration ac- According to Newton’s 2nd law, any acceleration is accompanied by a force in this caseo Fc = mac ac = v2/r = rω- The force is also called centripetal force. Examples of centripetal force:o Tension in a stringo Gravityo Force of frictionCentripetal Force Examples- Whirling a ball of mass m with a string, the tension T of the string provides the centripetal forceo The faster the ball moves, the greater the tensiono T = mac = mv2/r- Car travelling along a circular track: friction provides centripetal forceo Fs = mac = mv2/ro dangerous to make a sudden turn while the car is moving at high speeds, especially on icy roadso the friction may not be enough to provide the needed centripetal force. The car may be out of control- example o the gravitational forces between planets & the sun provide the centripetal force for their circular motions around the suno Fg = Fco Fg = Gm1m2/r2; Fc = mv2/ro Gm1m2/r2 = mv2/ro Gm2/r = v2 - The closer the planet is to the sun, the faster it is orbitingProblem Solving Strategy- Draw free body diagram, showing and labeling all the forces acting on the object(s)- Choose a coordinate system that has one axis perpendicular to the circular path and the other axis tangent to the circular path- Find the net force toward the center of the circular path (this is the force that causes thecentripetal acceleration, Fc)- Use Newton’s 2nd lawo The directions will be radial, normal and tangentialo The acceleration in the centripetal acceleration- Solve for the unknowns7.5 Newtonian Gravitation Revisited- Newton’s law of universal gravitationo Fg = Gm1m2/r2- On earth’s surface, the weight of an object with a mass m iso Fg = mg = GmMg/Re2 g = GMe/Re2- The gravitational acceleration g will vary with altitudeGravitational PE- Pe = mgy is only valid near the Earth’s surfaceo In this case, the reference point is chosen as Earth’s surface- For object’s high above the Earth’s surface, an alternate expression is neededo PE = - GMem/ro Zero reference level is infinitely far from the Earth’s surface- On Earth’s surface, the PE of the Earth – object system iso PE = -
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