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CHAPTER 7 Rotational Motion and the Law of Gravity Angular Speed and Angular Acceleration s = arc length Θ = arc angle (radians) Θ = s ; where r = radius r Term Symbol Units Angular Displacement ∆Θ radians Angular Speed ω radians/sec Average Angular Speed ω radians/sec (Average) Angular Acceleration radians/sec2 Θ2 – Θ1 ω= t2 – t1 ∆Θ Average Angular Speed ω= Average rate at which the arc ∆t angle is changing (Instantaneous) Angular Speed ω = lim ∆Θ The rate at which the arc angle is ∆t0 ∆t changing at a particular instant in time ω2 – ω1 = t2 – t1 Average Angular Acceleration ∆ω = The average rate at which the angular ∆t speed is changing NOTE: You are not required to work with changing rates of acceleration. Therefore, = Linear/ Angular Comparison Term Linear Angular Displacement ∆x m ∆Θ rad Velocity v m/s ω rad/sec Acceleration a m/s2 rad/sec2 Formulas: ∆x = ½(vo+v)t ∆Θ = ½(ωo+ω)t ∆x = vot + ½at2 ∆Θ = ωot + ½ t2 v = vo + at ω = ωo + t v2 = vo2 + 2a∆x ω2 = ωo2 + 2∆Θ N/A s = rΘ NOTE: All points on a rotating disc have the same values for ω and . Example Problem (Rotating Disc) A disc of radius 2.50m accelerates to an angular velocity of 1.35 radians/sec in 12.5 seconds. Calculate (a) the disc’s angular acceleration; (b) the number of revolutions completed during the acceleration; (c) the distance traveled by a point on the edge of the disc. Identify the variables ωo = 0.00 rad/sec = ? ω = 1.35 rad/sec t = 12.5 sec ∆Θ = ? Choose Appropriate Formulas ω = ωo + t = .108 rad/sec2 ∆Θ = [ ]ωo + ω 2 t ∆Θ = ωot + ½ t2 ∆Θ = 8.44 rad ∆Θ = 8.44 rad 1 revolution = 2 rad = 6.2 rad 8.44 rad 1 rev 6.28 rad ∆s = r∆Θ # rev = 1.34rev ∆s = 21.1m Tangential Speed (vt) • arc length of the movement time to complete the movement • vt = ∆s ∆t ∆s = r ∆Θ vt = r ∆Θ ∆t vt = r ω Tangential Acceleration (t) at = r NOTE: Not every point on a rotating disc has the same tangential speed or tangential acceleration. Centripetal Acceleration (center-seeking) Question: The tangential velocity vt illustrated above is clearly changing. How can you tell it is changing? Answer: Its direction is changing. a = ∆v = vf – vi = vf + -vi ∆t ∆t ∆t The illustration on the right clearly shows that ∆v points towards the center of rotation. Centripetal Acceleration (ac) ∆v = ∆s from similar triangles v r ∆v = v ∆s and ∆v = ac ∆t r v ac ∆t = ∆s r v ∆s and ∆s = v ac = r ∆t ∆t v2 ac = = Centripetal Acceleration r Centripetal Force: • The net force that causes an object to accelerate towards its center of rotation (i.e. go in a circle) • Fc = mac • Fc is not a new force • Fc will be either one or a combination of more than one of the following forces: Fg FN Ff Fa (T) Example Problem (Vertical loop) A rollercoaster contains a loop 20m in diameter. a) What speed will cause a passenger to feel “weightless” at the top of the loop? b) If this same person has a mass of 90kg what weight will they experience at the bottom of the loop? Sketch the Force Diagrams Analysis: The only forces Analysis: The person is acting on the person are the “weightless” if no forces other normal force and force of than gravity are acting upon gravity. them (i.e. FN=0). Fc = Fnet(center) = FN – Fg Hence, Fg is Fc FN = Fc + Fg mg = mv2 = mv2 + mg r r v = √rg v = 9.9m/s FN = 1764 Newtons Example Problem (Banked Curve) a) For a car traveling with a speed v around a curve of radius r, determine the formula for the angle at which the road should be bank so that no friction is required. b) what is this angle for an expressway off-ramp of radius 50m at a design speed of 50km/hr? Sketch the Force Diagram Part a: Analysis • Since there is no friction, there is no component of friction directed to the center (right). • Only the component FNsinΘ of the normal force is directed to the center. • FN sinΘ is the net force towards the center = Fc FN sinΘ = Fc FN sinΘ = mv2 r mg = FN cosΘ mg · sinΘ = mv2 cosΘ r mg tanΘ = mv2 r Part b: tanΘ = v2 gr Θ = tan -1 () v2 gr r = 50m v = 50km 1000m 1hr = 14m/s hr km 3600s g = 9.8 Θ = 22° Newton’s Universal Law of Gravitation Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Fg = G m1 m2 r2 G = universal gravitational constant G = 6.67 x 10-11 N·m2/kg2 Force of Earth’s Gravity (on Earth’s Surface) Fg = G mearth mobject rearth2 Fg = [ ] G mearth rearth 2 · massobject (6.67 x10-11 N·m2/kg2)(5.98x1024 kg) Fg = · massobject (6.37x10 6 m)2 Fg = 9.8 x massobject Fg = mobect g Velocity of an Earth Satellite Analysis: What is the only Force acting on the satellite? Answer: Fg If satellite is in circular orbit, then… Fg = Fc mg = mvs2 vs = velocity of satellite r vs = √gr g = GMearth r2 vs = √ GMearth · r r2 √ vs = GMearth r Gravitational Potential Energy PE = -GMEm r = distance of object from r Earth’s center • from Calculus • PE = zero at infinite distance from Earth • PE = some negative value closer to the Earth Escape Speed • Speed at surface of Earth where the object’s KE1 + PE1 is sufficient to reach infinite distance from Earth (PE2=0J) with a final speed of 0m/s (KE2=0J) • From Conservation of Energy: KE1 + PE1 = KE2 + PE1 mv2 + -GMearthm =0+0 [ ] 2 Rearth 2GMearth 1/2 mv2 + -GMearthm vescape = Rearth 2 Rearth