VIEWS: 163 PAGES: 11 CATEGORY: Education POSTED ON: 9/24/2009
1 Rotational Kinematics and Uniform Circular Motion Now we turn to a very common physical motion, that of circular motion. It is every where as we speak, ♦ the Earth revolves in nearly circular motion about the Sun at an astounding linear velocity of 2.6 × 107 km /day (1.6 × 107 miles per day). This translates to about 1o of angle conquered a day (360o, 365 days)—an humble 1.7 x 10-2 rad/day. ♦ The Earth spins about its own axis at rate 360o per day, and, as its radius is ~6.4 x 103 km, this translates to 4.0 x 104 km/day (at the equator) (or 1.1 x 103 km/day at latitude 80o. ♦ The moon revolves about the Earth at approximately 12o per day ♦ Circular motion describes motions of planets, planetary systems, Milky Way ♦ Clocks, pendula ♦ Racing tracks ♦ Negotiating a curve on a road ♦ You are now going to fall in love with the word “angular.” You’ll find that for every term in kinematics that you’re familiar with, there’s an “angular” counterpart: angular displacement, angular velocity, angular acceleration, etc. And you’ll find that, “angular” aside, very little changes when dealing with rotational kinematics. http://www.sparknotes.com/testprep/books/sat2/physics/chapter10section3.rhtml Uniform Circular Motion—where most of our focus will be. By uniform we mean that the rate of change of angle with time is constant and does not change sense of rotation during our observation. By circular motion, we mean that the distance of the object from the origin of our reference frame remains constant as object rotates about the origin. By convention, we measure angles in a circle in a counterclockwise direction from the positive x-axis. The angular position of a particle is the angle θ, made between the line connecting that particle to the origin, O, and the positive x-axis, measured counterclockwise. Let’s take the example of a point P on a counter- clockwise rotating wheel: P θ In this figure, point P has an angular position of θ. Note that every point on the line OP has the same angular position: the angular position of a point does not 2 depend on how far that point is from the origin, O. The green segment from the x–axis to the point P is called an arc. We can relate the angular position of P to the length of the arc of the circle, s. between P and the angular position by: s = Rθ θ in radians In this equation, s is the length of the arc, and R is the radius of the circle. Emphasize difference between angular and linear velocity: suppose you and a friend are eight years old again, and you decide to run in circles around a tree. The tree is your “origin”. The circle your friend maps out has a radius of about 1 meter, but the circle you map out has a radius of 2 meters. If both of you keep up with each other (cover the same number of angles per time), you have a lot more running to do: you (faster linear velocity to “keep up”) friend You and your friend have the same angular velocity (both sweep out the same number of rads per time), but different linear velocities (which scale linearly with your respective radii). 1 Angular Displacement Let the wheel be rotated at a constant—uniform—rate. A complete trip (360o) is called a revolution. The time to complete one revolution is called a period, T. The linear speed of point P is given by: Δs/Δt = (Δθ/Δt)R = ωR where ω, the angular velocity = Δθ/Δt. Thus, in one revolution, Δθ = 2π and the time is the period T so ω = Δθ/Δt = 2π/T. The frequency, f = 1/T and has SI unit s-1, Hertz (Hz) or cycles per second (cps) Rigidity of circular motion assumed. For line to move in the way described above, every point 1 along the line must rotate 90º counterclockwise (⇔ ccw ; clockwise ⇔ cw). By definition, the particles that make up a rigid body must stay in the same relative position to one another. As a result, the angular displacement is the same for every point in a rotating rigid body. 3 So far we have learned: s = Rθ Angular velocity ω = Δθ/Δt = 2π/T = where s is the length of an arc of 2πf also equals v/R units rad/s a circle with radius R and T is the period of motion in s or min θ is the angle it subtends. f is the frequency in s-1 or min-1 Hopefully, you’ve bought into fact that distance and displacement are not the same thing: well, the same distinction applies with angular distance and angular displacement. For example, if you rotate a record 30º clockwise and then 20º counterclockwise, the angular displacement of the record is 10º, although the particles have traveled a total angular distance of 50º. Ex 1. A toy top rotates at 13 rev/min. Determine the period in min and the angular velocity rad/s if the radius of the top is 5 cm. 13 rev/min is a frequency. Then period T = 1/13 min/rev or just 0.0769 min. Angular speed = (2πf) = 82 min-1 Linear Velocity v2 R2 v1 look at positions only: R2 ΔR R1 R1 Change in displacement : In time t, Δθ = ωΔt ΔR = vΔt, By definition, the change in the position of the vector R is proportional to velocity (linear velocity). Angular acceleration, is defined as the rate of change of angular velocity over time. Average angular acceleration, α, is defined by α = Δω/Δt typically units of rad/s2. In general, Kinematic Relation between Linear and Angular variables: Quantity Linear Expression Angular Expression acceleration a = constant α = constant velocity v = (v0 + at) ω = (ω0 + αt) position 1 2 ΔS = at + v0t 1 Δθ = αt2 + ω0t 2 2 Relation between v2- v02 = 2ad ω2- ω02 = 2αΔθ Speed,acceleration and displacement 4 Ex. 2 A CD accelerates uniformly from rest to 200 spins per minute in 8.0 seconds. If the rate of change in speed is constant, determine (a) the instantaneous angular acceleration (rad/s2 ) 2 , (b) how many radians the CD swept out in 8.0 seconds. Solution: Uniform acceleration means that α = Δω/Δt. Thus ω(final) = spin 2π rad 1 min 200 ⋅ 2π 200 × × = = 20.9 rad/s. s spin 60 s 60 s 20.9 rad/s Then we set α = Δω/Δt = = 2.6 rad/s2 answer to (a). 8.0 s (b)The angular formula is exactly analogous to the linear one, ωf2 – ωo2 = 2αΔθ Thus here the final angular speed is 200 rpm = 20.9 rad/s, the initial angular speed is zero; from (a), the angular acceleration α is 2.6 rads-2, find Δθ: (20.9)2 = 2(2.6) Δθ, Δθ = 84 rad Linear acceleration’s relationship to angular velocity and radius: Consider how the velocity vectors change in time Δt. Right diagram concentrates on velocity vectors I translate the initial velocity next to the present velocity so we can more readily compare the change in velocity : r We know that a is proportional to the change in velocity, this change, the velocity difference vector = v1 - v2 (red, at later time Δt) . Graphically we observe that it lies along a line that points back to the center of the circle. In other words, it anti- Uniform acceleration means that α = Δω/Δt is not is not just the average acceleration but also the 2 instantaneous one, dω/dt. Some of us would not be aware of this distinction, but we would have spin 2π rad 1 min 200 ⋅ 2π found ω(final) = 200 × × = = 20.9 rad/s. s spin 60 s 60 s 5 parallel to the position vector R. Let R ˆ be a unit vector pointing (radially) outward from the origin, r Then a = -a R where a is the magnitude of the radial ˆ acceleration. Now Δv/Δt = Δ(ωR)/Δt = ω ΔR/Δt = ω v = ω2R or equivalently, v2/R r Then magnitude a = v2/R and vector a = -(v2/R) R ˆ r The acceleration a , called the centripetal acceleration, points inward along a radial line that “keeps the object turning”. The magnitude of this linear, centripetal acceleration is given in several, equivalent forms, we use the subscript c to remind us that this is a circular acceleration—a centripetal acceleration, ac = v2/R = ω2R We make use of the form that is most convenient for the problem encountered. IMPORTANT POINT: Since there is an acceleration, there must necessarily be a force responsible for it, Fc = mac = mv2/R “centripetal force” In the types of problems that will follow, we will determine which force or which component of a force supplies the required centripetal force. Ex 3. A demo CD, 8.00 cm in diameter, spins 200 revolutions per minute (200 rpm). Determine (a) angular velocity rad/s (b) linear velocity (cm/s) for a point 2.00 cm from the center (c) linear velocity (cm/s) for a point 2.00 cm from the edge (d) centripetal acceleration experienced by a point on the rim (edge) of the CD Solution: (a) 200 rpm ≡ 200(2π)/60 s = 400π (rad)/60 s = 20.9 rad/s = ω (b) v = ωRb = (20.9 rad s-1)(1.00 cm) = 20.9 cm/s (notice how radian dropped) (c) 2.00 cm from edge means 3.00 cm from center, so: v = ωRc = (20.9 rad s-1)(3.00 cm) = 62.8 cm/s (notice how radian dropped) (d) ac = ω2Rd = (20.9 rad s-1)2(4.00 cm) = 1.75 x 103 cm/s2 Example 4. A car (681 kg) negotiates a frictionless banked curve (θ = 20o). Draw a free body diagram and determine the force (or component) that supplies the centripetal force acting on the car. (b) If the radius of the curve corresponds to 14 m, determine the fastest speed the car can go and still safely navigate the curve. Solution: The car’s navigation can be modeled by car 6 following a small arc of a circle. During that time, the car experiences an inward tug, a centripetal force pointing Nsinθ towards center of the imaginary circle. Ncosθ N θ The two forces acting on the car are N and W. Of these, W only the horizontal component of the normal force, Nsinθ, points radially inward. Notice how the green block arrow and the horizontal component of the normal force are parallel. It is this component of force that supplies the required r r centripetal force. In other words, 2nd law, F = m a , yields: horizontal direction: Nsinθ = mac = mv2/R equals the forced asked for in (a) “ (a) Fc = (168 kg)(9.81 ms-2)(sin20o) = 534 N vertical direction : Ncosθ - mg = 0 N = mg/cosθ Inserting expression for N from vertical expression into horizontal expression, we learn that v2 = [(N/m) sinθ]R = rearranging = (g/cosθ)Rsinθ = gR(sinθ/cosθ) = gRtanθ OR v = (gRtanθ )1/2 = 7.07 m/s Practical Interpretation: v = (gRtanθ )1/2 is the fastest speed the car can go and match the force needed to smoothly negotiate the curve. Note dim[gR] = (m/s)2 = dim[v2] Ex 5. A bob consists of a mass m (0.300 kg) connected to a string of length L (15 cm) and connected to a stick. The bob whirls at a uniform rate of 2 cps (cycles per second) θ L Assume string is massless and determine (a) angular velocity ω (rad/s) m (b) which force or component of force provides the centripetal force? Draw an FBD. (c) the tension in the string (N) in terms of m, ω, and L (d) angle θ, (e) the linear velocity (ms-1) (f) the centripetal acceleration (ms-2) (g) An algebraic expression for angular frequency in terms of g, θ and L Solution (a) ω = (2π)2 s-1 = 12.6 rad s-1 7 (b) since a force is sought for, draw an FBD Only two forces acting are tension and Tcosθ θ weight of bob: T Free Body Diagram, FBD Vertical force eqn: Tcosθ - W = 0 W Horizontal force : Tsinθ = mac = mv2/R In other words: The horizontal component of T points radially inward and so provides the centripetal force. (c) from trigonometry, the radius R = Lsinθ. Application of Newton’s 2nd Law yields: Vertical Equation, T = mg/cosθ and Horizontal Equation, T, tension = (mω2R) ÷ sinθ = (mω2Lsinθ) ÷ sinθ = mω2L = 7.11 N (d) Angle can be found by again using the vertical relation: W (0.3 kg)(9.8 N/kg) cosθ = = = 0.411. T 7.11 N Thus taking inverse, cos-1(0.411) = θ = 65.5o. v = ωR where the radius of the circle in the plane is R = Lsinθ. Thus (e) v = ωLsinθ = (12.6 rad/s)(0.20 m)·sin(30o) = 1.26 m/s (radian dropped out) (f) a = ω2R = ω2Lsinθ = (12.6)2(0.15·sin(30o)) = 15.9 ms-2 (rad ‘unit’ dropped out) (g) One can also find a relation between angular velocity, length of string, and angle string makes with respect to the vertical, and gravity: T = W/cosθ also equals mω2L so, dividing both sides by mass: g/cosθ = ω2L or solving for angular speed g ω2 = g/(Lcosθ) ω = is a general expression. Lcosθ Ex 6. From a safe-design point of view, determine the criterion at point A B and again at point B for the cart on the roller coaster at right to just stay on the circular track of radius R. A By criterion, we mean a force condition that arises as a consequence of some R necessary constraint. Thus determine a minimum speed at point B. Solution: 8 As the motion must be circular motion, at A, some force must point to the west t o provide the centripetral force (car is headed north). That force turns out to be the normal force: N = mv2/r. N The FBD diagram at right shows N and W, only N points west. Similarly, at B, the FBD below shows both N and W pointing south: Seemingly both forces could add to provide the centripetal force (let down be (+), then): N W N + W = mv2/r However, from a design point of view, we want to be certain that even in the worst circumstance, the centripetal criterion is met. The worst instance would occur if, for an instant, the car no longer makes contact with the track. The press now gone, the normal force would vanish, but the force due to gravity, of course, would still be there. Thus, this minimum criterion: W = mg = mv2/r which leads to the criterion on the minimum speed to clear point B safely: v2 = gR thus v = gR As a summary of what we have done so far, we have demonstrated the striking similarity between linear and angular kinematics. In fact, if one remembers the coordinate associations: v ⇔ ω a ⇔ α Δs ⇔ Δθ, then the table on the next page displays the useful kinematic relations in angular coordinates, Kinematic Relation between Linear and Angular variables: Quantity Linear Expression Angular Expression acceleration a = constant α = constant velocity v = (v0 + at) ω = (ω0 + αt) position 1 ΔS = at2 + v0t 1 Δθ = αt2 + ω0t 2 2 2 2 Relation between v - v0 = 2ad 2 ω - ω02 = 2αΔθ Speed,acceleration And displacement Ex 7, A merry-go-round (diameter 4.0 m) starts from rest and uniformly accelerates until it rotates at 0.8 rev/minute. This occurs in 11 seconds. Once it has reached this rate, it rotates at 0.8 rev/minute uniformly. Determine (a) angular speed after 1 minute (b) the angular acceleration (rad/s2) (c) speed after 25 seconds (d) linear velocity after 40 s (e) centripetal acceleration at the rim of the merry-go-round. 9 Solution: (a) 0.8 rev/min = (0.8)(2π)/60 s = 0.083 rad/s (b) α = Δω/Δt = (0.083 – 0)/(11) = 7.6 × 10-3 rads-2 (c) ω = (ω0 + αt) = αt = (7.6 × 10-3 rads-2)(25 s) = 0.19 rad/s (d) At 35 s, ω = (7.6 × 10-3 rads-2)(35 s) = 0.27 rad/s ∴ v = ωR =(0.27 rad/s)(2.0 m) = 0.53 m/s (rad “unit” drops) (e) centripetal acceleration ac = v2/R = 0.14 ms2 Kepler’s Laws of Motion Another important application of circular motion is found in the observations of Johannes Kepler (1571-1630) 3 . An protégé of the respected astromologist Tycho Brahe, he is most famous for his three laws of planetary motion— published in 1609 and 1615—although he also contributed to the fields of optics, geometry and pre-calculus in his investigations of logarithms and solids of revolution. The three laws are as follows: For bodies in orbit around a common center, 1. Each body follows an elliptical path with the sun at one foci. 2. Periods squared goes as radius cubed T2 ~ R3 3 Johannes Kepler is now chiefly remembered for discovering the three laws of planetary motion that bear his name published in 1609 and 1619). He also did important work in optics (1604, 1611), discovered two new regular polyhedra (1619), gave the first mathematical treatment of close packing of equal spheres (leading to an explanation of the shape of the cells of a honeycomb, 1611), gave the first proof of how logarithms worked (1624), and devised a method of finding the volumes of solids of revolution that (with hindsight!) can be seen as contributing to the development of calculus (1615, 1616). Moreover, he calculated the most exact astronomical tables hitherto known, whose continued accuracy did much to establish the truth of heliocentric astronomy 10 3. Each areas are swept out in equal times Δ(Area)/Δt = constant The departure of an orbit from being F = GMm/R2. Since Earth moves in truly circular is measured by its circular orbit, the gravitational force eccentricity. Eccentricity varies Johannes Kepler (1571-1630) from 0 (circular) to 1 (very elongated ellipse; a cigar). Fortunately the trajectory of the Earth and most planets is nearly circular enough that the main premises of Kepler’s law can be explored without the complications of elliptical geometry. Thus, aided by Newton’s Law of Gravitational that Mr. Kepler did not have the benefit to know, we set out to prove the 2nd Law, that T2 ~ R3: Force of attraction between Earth and Sun is due to gravitational force, supplies the necessary centripetal motion: mv2/R = GMm/R2 multiply both sides by R, we have v2 = GM/R but the linear speed equals 2πR/T so (4π2R2/T2) = GM/R, multiply both sides by R: 4π2R3/T2 = GM or R3/T2 = GM/(4π2) = a constant Next for the 3rd Kepler Law, equal areas swept out in equal times means that the rate of change of area with change in time should be a constant, For a circle, this is no big deal (think about the 2nd hand of a clock, sweeping out 6o per second, thus the same area per unit time). For an ellipse, it may not seem as obvious: We claim in the same time interval that this area equals this area Proof: ΔA/Δt = Δ(πR2)/Δt = π [R(ΔR/Δt) + R(ΔR/Δt) ] = 2π R(ΔR/Δt) = 2π vR But for uniform circular motion, the linear speed v is constant 4 and of course the radius is constant, so 2π vR = a constant. It will be shown that ΔA/Δt is proportional 4 the linear velocity being a vector, is not 11 to the angular momentum of the planet or object, and that it is conserved in the gravitational force field. The consequence of Kepler’s 3rd law is that for an elliptical orbit, the nearer a planet is to the Sun the faster it must move to sweep out the same area, say in 2 weeks that it sweeps out at the other extreme of its orbit in 2 weeks where it’s distance from the Sun is farther. For your information, the nearest distance of approach is called the perihelion and the furthest distance is called the aphelion. Certainly one of the several triumphs of Newton’s Principia was the verification of these well respected empirical astronomical laws. Ex. 8 For a certain planetary orbit, the aphelion of the planet = 1.50 times its perihelion. If the speed at the perihelion is 4.71 km/s, determine the speed at its aphelion. Kepler’s 3rd Law says R·v = constant, so Ra·va = Rp·vp or as Ra = 1.50·Rp we have 1.50·va = vp or va = vp/1.50 = 3.14 km/s Example 9. Saturn is 1.43 x 109 km from the Sun. Earth is 1.49 x 108 km from the Sun. Assuming circular orbit and any obvious other data, determine the period of Saturn’s orbit without looking it up. For every planet in a solar system, R3/T2 = GM/(4π2) Thus, ⎛R ⎞ 3/ 2 RSat /TSat = REarth /TEarth or TSat = ⎜ Saturn ⎟ TEarth 3 2 3 2 ⎜ ⎟ ⎝ R Earth ⎠ Calculation yields TSat = (1.43 x 109/1.49x 108)3/2· (1 year) = 29.7 year