Circular Motion & Angular Quantities in Physics I Lecture Notes by Dr. Donald Luttermoser, Study notes of Physics

Download Circular Motion & Angular Quantities in Physics I Lecture Notes by Dr. Donald Luttermoser and more Study notes Physics in PDF only on Docsity! PHYS-2010: General Physics I Course Lecture Notes Section VIII Dr. Donald G. Luttermoser East Tennessee State University Edition 2.3 Abstract These class notes are designed for use of the instructor and students of the course PHYS-2010: General Physics I taught by Dr. Donald Luttermoser at East Tennessee State University. These notes make reference to the College Physics, 7th Edition (2005) textbook by Serway and Faughn. Donald G. Luttermoser, ETSU VIII–3 e) A body revolves around another object if the first body is in orbit about the second object (e.g., the Earth “re- volves” about the Sun). Sun Earth "orbits" Sun Example VIII–1. Problem 7.2 (Page 218) from the Ser- way & Faughn textbook: A wheel of radius 4.1 meters rotates at a constant velocity. How far (path length) does a point on the circumference travel if the wheel is rotated through angles of 30◦, 30 rad, and 30 rev? Let r be the radius of the wheel, s be the path length along the circumference, and θ be the angle that this point subtends from a reference point as the wheel rotates. As such, r = 4.1 m, and θ1 = 30 ◦, θ2 = 30 rad, and θ3 = 30 rev. We will use Eq. (VI-1), s = θ r , except that all of these angles must be expressed in radians, so θ1 = 30 ◦ × π rad 180◦ = 0.5236 rad, θ2 = 30 rad (which is what we want), θ3 = 30 rev × 2π rad 1 rev = 188.5 rad. VIII–4 PHYS-2010: General Physics I Plugging these values of θ into Eq. (VI-1) gives s1 = θ1 r = 0.5236 (4.1 m) = 2.1 m s2 = θ2 r = 30 (4.1 m) = 120 m s3 = θ3 r = 188.5 (4.1 m) = 770 m 2. The length s in Eq. (VIII-1) is actually an arc length (curved length). However, if θ is small (i.e., θ  1), s → D, the actual linear size (or diameter if the object is round). a) As such, by measuring the angle that a distance object subtends, we can calculate its actual size (or diameter) D by knowing the distance d to the object. b) Since θ is small here, it is often more convenient to express this angle in terms of arcseconds instead of radians. We can rewrite Eq. (VIII-1) by setting s = D, r = d, and expressing θ in arcseconds and relabeling it as α, then θ(rad) = α(arcseconds) × π rad 180◦ × 1 ◦ 3600 arcsec = α arcsec 206, 265 arcsec/rad . c) Using α instead of θ in Eq. (VIII-1) gives D = α d 206, 265 , (VIII-4) which is called the small-angle formula. Here, D is the linear size (i.e., diameter) of the object at a distance d which subtends an angle α measured in arcseconds (the Donald G. Luttermoser, ETSU VIII–5 206,265 is the conversion factor between arcseconds and radians). The lengths d and D will be in the same units (e.g., if d is measured in km, then D will be in km). Note that this formula is only valid when θ  1 (typically, one would want α < 1000 arcseconds in order to use this small-angle approximation). Example VIII–2. The star Betelguese (α Ori) has an angular size of 0.040 arcsec (40 milliarcseconds = 40 mas) and it is at a distance of 200 pc. What is the linear size of Betelgeuse? How does this size compare to the planet’s distances from the Sun in our solar system? d = 200 pc x 3.09 × 1016 m/pc = 6.18 × 1018 m D = 0.040 x 6.18 × 1018 m / 206,265 = 1.20 × 1012 m. The Sun is 1.39×109 m in diameter which means that Betelgeuse is D = 1.20 × 1012 m 1.39 × 109 m/D = 860 D = 8.0 AU, where an Astronomical Unit (AU) is the size of the Earth’s orbital semimajor axis (note that 1 AU = 1.496 × 1011 m). Betelgeuse is 860 times bigger than the Sun! This diameter gives a stellar radius of 4.0 AU for Betelgeuse. If it was put in the place of the Sun, then Mercury (0.39 AU), Venus (0.72 AU), Earth (1.0 AU), and Mars (1.5 AU) would be inside Betelgeuse! The planet Jupiter (5.2 AU) would be close (1.2 AU) from the photosphere (“surface”) of Betelgeuse. 3. Angular speed, ω, is the change of angular displacement di- vided by the time interval that the angular displacement took place. VIII–8 PHYS-2010: General Physics I where vt is the “tangential” velocity tangent to the curved path of motion the particle is taking. r s vt ω θ 3. The tangential acceleration, a t, of a point on a rotating object equals the distance of that point from the axis of rotation (r) multiplied by the angular acceleration (α): ∆vt = r ∆ω , ∆vt ∆t︸ ︷︷ ︸ a = r ∆ω ∆t︸ ︷︷ ︸ α lim ∆t→0 ∆vt ∆t = r lim ∆t→0 ∆ω ∆t , a t = r α . (VIII-13) C. Centripetal Acceleration and Force. 1. If an object travels at constant speed in a curved path, it is accelerating since the velocity vector is continuously changing direction. 2. Besides, the tangential acceleration (i.e., acceleration tangent to the curved path), there must be an acceleration perpendicular (⊥) to the tangent line pointing toward the center of the curved path. Donald G. Luttermoser, ETSU VIII–9 a) Otherwise, the object would fly off in a straight path in the direction of a t as per Newton’s 1st law of motion. b) This center-seeking acceleration is called the centripetal acceleration, a c = r ω 2 = v2t r . (VIII-14) i) Gravity is a centripetal force (hence acceleration) for planets orbiting the Sun. ii) A string’s tension is a centripetal force (hence ac- celeration) for an object attached to that string be- ing rotated in circular motion. ac at ω stri ng rock c) The total acceleration of an object in circular motion is ~a = ~at + ~ac , (VIII-15) a = √ a2t + a 2 c . (VIII-16) VIII–10 PHYS-2010: General Physics I Example VIII–3. Problem 7.18 (Page 219) from the Ser- way & Faughn textbook: A race car starts from rest on a circular track of radius 400 m. The car’s speed increases at a constant rate of 0.500 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, determine (a) the speed of the race car, (b) the distance traveled, and (c) the elapsed time. Solution (a): From Eq. (VIII-14), the centripetal acceleration is a c = v2t r . Thus, when a c = a t = 0.500 m/s 2, we have vt = √ r a c = √ (400 m)(0.500 m/s2) = 14.1 m/s . Solution (b): The total distance traveled is just the total arc length traveled on the circular track, s, which can be found from the linear 1-D equation of motion: v2 = v2◦ + 2 a s , where v = vt = 14.1 m/s, v◦ = 0 (starts from rest), and a = a t = 0.500 m/s2. Hence, the total arc length (i.e., distance) traveled is s = v2 − v2◦ 2 a t = (14.1 m/s)2 − 0 2 · 0.500 m/s2 = 200 m . Solution (c): We can use another linear 1-D equation of motion to find the elapse time: v = v◦ + a t , where v = vt = 14.1 m/s, v◦ = 0 (starts from rest), and a = a t = 0.500 m/s2. Hence, the time it take to reach the tangential Donald G. Luttermoser, ETSU VIII–13 Solution (b): Lθmax vi = 3.00 m/s yi = 0 yf vf = 0 L - yf yf The diagram above shows the bob at its maximum height. To find this height, hence θmax, we only need to use the conservation of mechanical energy. We will set the lowest position of the bob at the “ground” level, y i = 0. At its highest position, y f, the velocity changes from the counter-clockwise to the clockwise direction, hence goes through zero at this point, v f = 0. From the diagram above, we can solve for y f in terms of θmax: cos θmax = L − y f L L cos θmax = L − y f y f = L − L cos θmax y f = L (1 − cos θmax) . We can now determine θmax from the conservation of mechanical energy by setting the lowest point as the initial position, so v i = vt [from part (a)], and y i = 0, and the highest point as the final position, so v f = 0 and y f is given in the equation above: (KE + PEg) i = (KE + PEg) f 1 2 mv2i + mgy i = 1 2 mv2f + mgy f VIII–14 PHYS-2010: General Physics I 1 2 mv2t + 0 = 0 + mgL (1 − cos θmax) mgL (1 − cos θmax) = 1 2 mv2t 1 − cos θmax = v2t 2gL cos θmax = 1 − v2t 2gL θmax = cos −1  1 − v2t 2gL   = cos−1  1 − (3.00 m/s)2 2(9.80 m/s2)(0.800 m)   = cos−1(0.4260) = 64.8◦ . Solution (c): θmax Fg T y x θmax The diagram above shows the forces acting on the bob at the highest point with respect to an arbitrary coordinate system cho- sen such that the tension and total centripetal force points in the positive y direction. From this, we now construct a free-body diagram: Donald G. Luttermoser, ETSU VIII–15 y x T Fg θmax Since the tangential velocity at this highest point is zero [as men- tioned in part (b)], the centripetal force must be zero: Fc = mac = m v2t r = 0 . Using this equilibrium equation in conjunction with the free-body diagram above, we get Fc = ∑ Fy = T − Fg-y 0 = T − mg cos θmax T = mg cos θmax = (0.400 kg)(9.80 m/s2) cos(64.8◦) = 1.67 N .