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Explore the principles of circular motion, including uniform and non-uniform motion, Kepler's laws, and centripetal acceleration. Learn about tangential and angular acceleration, apparent weight, and artificial gravity. Practice with angular measurement and conversions. Understand centripetal force and planar acceleration in various scenarios.
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Chapter 5: Circular Motion • Uniform circular motion • Radial acceleration • Unbanked turns (banked) • Circular orbits: Kepler’s laws • Non-uniform circular motion • Tangential & Angular acceleration • (apparent weight, artificial gravity) • Hk: CQ 1, 2. Prob: 5, 11, 15, 19, 39, 49.
angular measurement degrees (arbitrary numbering system, e.g. some systems use 400) radians (ratio of distances) e.g. distance traveled by object is product of angle and radius. 2
Radians r s r s = arc length r = radius 3
Angular Motion radian/second (radian/second)/second 5
angular conversions Convert 30° to radians: Convert 15 rpm to radians/s 6
Angular Equations of Motion Valid for constant-a only 7
Centripetal Acceleration • Turning is an acceleration toward center of turn-radius and is called Centripetal Acceleration • Centripetal is left/right direction • a(centripetal) = v2/r • (v = speed, r = radius of turn) • Ex. V = 6m/s, r = 4m. a(centripetal) = 6^2/4 = 9 m/s/s
Planar Acceleration • Total acceleration = tangential + centripetal • = forward/backward + left/right • a(total) = ra (F/B) + v2/r (L/R) • Ex. Accelerating out of a turn; 4.0 m/s/s (F) + 3.0 m/s/s (L) • a(total) = 5.0 m/s/s
Centripetal Force • required for circular motion • Fc = mac = mv2/r • Example: • 1.5kg moves in r = 2m circle v = 8m/s. • ac = v2/r = 64/2 = 32m/s/s • Fc = mac = (1.5kg)(32m/s/s) = 48N
Rounding a Corner • How much horizontal force is required for a 2000kg car to round a corner, radius = 100m, at a speed of 25m/s? • Answer: F = mv2/r = (2000)(25)(25)/(100) = 12,500N • What percent is this force of the weight of the car? • Answer: % = 12,500/19,600 = 64%
Mass on Spring 1 • A 1kg mass attached to spring does r = 0.15m circular motion at a speed of 2m/s. What is the tension in the spring? • Answer: T = mv2/r = (1)(2)(2)/(.15) = 26.7N
Mass on Spring 2 • A 1kg mass attached to spring does r = 0.15m circular motion with a tension in the spring equal to 9.8N. What is the speed of the mass? • Answer: T = mv2/r, v2 = Tr/m • v = sqrt{(9.8)(0.15)/(1)} = 1.21m/s
Kepler’s Laws of Orbits • Elliptical orbits • Equal areas in equal times (ang. Mom.) • Square of year ~ cube of radius
Elliptical Orbits • One side slowing, one side speeding • Conservation of Mech. Energy • ellipse shape • simulated orbits
s = rq v = rw a(tangential) = ra. a(centripetal) = v2/r F(grav) = GMm/r2 Kepler’s Laws, Energy, Angular Momentum Summary
Centrifugal Force • The “apparent” force on an object, due to a net force, which is opposite in direction to the net force. • Ex. A moving car makes a sudden turn to the left. You feel forced to the right of the car. • Similarly, if a car accelerates forward, you feel pressed backward into the seat.
rotational speeds • rpm = rev/min • frequency “f” = cycles/sec • period “T” = sec/cycle = 1/f • degrees/sec • rad/sec w = 2pf
7-43 • Merry go round: 24 rev in 3.0min. • W-avg: 0.83 rad/s • V = rw = (4m)(0.83rad/s) = 3.3m/s • V = rw = (5m)(0.83rad/s) = 4.2m/s
Rolling Motion v = vcm = Rw
Example: Rolling A wheel with radius 0.25m is rolling at 18m/s. What is its rotational rate?
Example A car wheel angularly accelerates uniformly from 1.5rad/s with rate 3.0rad/s2 for 5.0s. What is the final angular velocity? What angle is subtended during this time?
vt at ac vt ac Rotational Motion r
Convert 50 rpm into rad/s. • (50rev/min)(6.28rad/rev)(1min/60s) • 5.23rad/s