8.01SC Classical Mechanics: Chapter 9.3

9.3 Worked Examples Circular Motion Example 9.1 Geosynchronous Orbit A geostationary satellite goes around the earth once every 23 hours 56 minutes and 4 seconds, (a sidereal day, shorter than the noon-to-noon solar day of 24 hours) so that its position appears stationary with respect to a ground station. The mass of the earth is me = 5.98 × 1024 kg . The mean radius of the earth is Re = 6.37 × 106 m . The universal constant of gravitation is G = 6.67 × 10−11 N ⋅ m 2 ⋅ kg −2 . What is the radius of the orbit of a geostationary satellite? Approximately how many earth radii is this distance? Solution: The satellite’s motion can be modeled as uniform circular motion. The gravitational force between the earth and the satellite keeps the satellite moving in a circle (In Figure 9.4, the orbit is close to a scale drawing of the orbit). The acceleration of the satellite is directed towards the center of the circle, that is, along the radially inward direction. Figure 9.4 Geostationary satellite orbit (close to a scale drawing of orbit). Choose the origin at the center of the earth, and the unit vector r̂ along the radial direction. This choice of coordinates makes sense in this problem since the direction of acceleration is along the radial direction. 9-1   Let r be the position vector of the satellite. The magnitude of r (we denote it as rs ) is the distance of the satellite from the center of the earth, and hence the radius of its circular orbit. Let ω be the angular velocity of the satellite, and the period is T = 2π / ω . The acceleration is directed inward, with magnitude rsω 2 ; in vector form,  a = −rsω 2r̂ . (9.3.1) Apply Newton’s Second Law to the satellite for the radial component. The only force in this direction is the gravitational force due to the Earth,  Fgrav = −msω 2 rs r̂ . (9.3.2) The inward radial force on the satellite is the gravitational attraction of the earth, −G ms me r̂ = −msω 2 rs r̂ . rs 2 (9.3.3) ms me = msω 2 rs . 2 rs (9.3.4) Equating the r̂ components, G Solving for the radius of orbit of the satellite rs , 1/ 3 ⎛Gm ⎞ rs = ⎜ 2 e ⎟ . ⎝ ω ⎠ (9.3.5) The period T of the satellite’s orbit in seconds is 86164 s and so the angular speed is ω= 2π 2π = = 7.2921× 10−5 s −1 . T 86164 s (9.3.6) Using the values of ω , G and me in Equation (9.3.5), we determine rs , rs = 4.22 ×107 m = 6.62 Re . (9.3.7) Example 9.2 Double Star System Consider a double star system under the influence of gravitational force between the stars. Star 1 has mass m1 and star 2 has mass m2 . Assume that each star undergoes uniform circular motion such that the stars are always a fixed distance s apart (rotating counterclockwise in Figure 9.5). What is the period of the orbit? 9-2 Figure 9.5 Two stars undergoing circular orbits about each other Solution: Because the distance between the two stars doesn’t change as they orbit about each other, there is a central point where the lines connecting the two objects intersect as the objects move, as can be seen in the figure above. (We will see later on in the course that central point is the center of mass of the system.) Choose radial coordinates for each star with origin at that central point. Let r̂1 be a unit vector at Star 1 pointing radially  away from the center of mass. The position of object 1 is then r1 = r1 rˆ1 , where r1 is the distance from the central point. Let r̂2 be a unit vector at Star 2 pointing radially away  from the center of mass. The position of object 2 is then r2 = r2 r̂2 , where r2 is the distance from the central point. Because the distance between the two stars is fixed we have that s = r1 + r2 . The coordinate system is shown in Figure 9.6 9-3 Figure 9.6 Coordinate system for double star orbits The gravitational force on object 1 is then  Gm1m2 F2,1 = − r̂1 . s2 The gravitational force on object 2 is then  Gm1m2 F1,2 = − r̂2 . s2 The force diagrams on the two stars are shown in Figure 9.7. Figure 9.7 Force diagrams on objects 1 and 2 Let ω denote the magnitude of the angular velocity of each star about the central point.   Then Newton’s Second Law, F1 = m 1 a1 , for Star 1 in the radial direction r̂1 is −G m1 m2 s 2 = −m1 r1 ω 2 . We can solve this for r1 , r1 = G m2 . ω 2s2   Newton’s Second Law, F2 = m 2 a 2 , for Star 2 in the radial direction rˆ2 is 9-4 −G m1 m2 s2 = −m2 r2 ω 2 . We can solve this for r2 , r2 = G m1 . ω 2s2 Because s , the distance between the stars, is constant s = r1 + r2 = G m2 m (m + m ) + G 21 2 = G 2 2 2 1 2 2 . ω s ω s ω s Thus the magnitude of the angular velocity is ⎛ (m + m )⎞ ω = ⎜G 2 3 1 ⎟ s ⎝ ⎠ 12 , and the period is then 2π ⎛ 4π 2 s3 ⎞ T= = ω ⎜⎝ G(m2 + m1 ) ⎟⎠ 12 . (9.3.8) Note that both masses appear in the above expression for the period unlike the expression for Kepler’s Law for circular orbits. Eq. (9.2.7). The reason is that in the argument leading up to Eq. (9.2.7), we assumed that m1 << m2 , this was equivalent to assuming that the central point was located at the center of the Earth. If we used Eq. (9.3.8) instead we would find that the orbital period for the circular motion of the Earth and moon about each other is T= 4π 2 (3.82 × 108 m)3 = 2.33× 106 s , (6.67 × 10−11 N ⋅ m 2 ⋅ kg −2 )(5.98 × 1024 kg +7.36 × 1022 kg) which is 1.43 × 104 s = 0.17 d shorter than our previous calculation. Example 9.3 Rotating Objects Two objects 1 and 2 of mass m1 and m2 are whirling around a shaft with a constant angular velocity ω . The first object is a distance d from the central axis, and the second object is a distance 2d from the axis (Figure 9.8). You may ignore the mass of the strings and neglect the effect of gravity. (a) What is the tension in the string between the inner 9-5 object and the outer object? (b) What is the tension in the string between the shaft and the inner object? Figure 9.8 Objects attached to a rotating shaft Solution: We begin by drawing separate force diagrams, Figure 9.9a for object 1 and Figure 9.9b for object 2. (a) (b) Figure 9.9 (a) and 9.9 (b) Free-body force diagrams for objects 1 and 2   Newton’s Second Law, F1 = m 1 a1 , for the inner object in the radial direction is r̂ : T2 − T1 = −m1 d ω 2 .   Newton’s Second Law, F2 = m 2 a 2 , for the outer object in the radial direction is r̂ : − T2 = −m2 2d ω 2 . The tension in the string between the inner object and the outer object is therefore T2 = m2 2d ω 2 . 9-6 Using this result for T2 in the force equation for the inner object yields m2 2d ω 2 − T1 = −m1 d ω 2 , which can be solved for the tension in the string between the shaft and the inner object T1 = d ω 2 (m1 + 2m2 ) . Example 9.4 Tension in a Rope A uniform rope of mass m and length L is attached to shaft that is rotating at constant angular velocity ω . Find the tension in the rope as a function of distance from the shaft. You may ignore the effect of gravitation. Solution: Divide the rope into small pieces of length Δr , each of mass Δm = (m / L)Δr . Consider the piece located a distance r from the shaft (Figure 9.10). Figure 9.10 Small slice of rotating rope The radial component of the force on that piece is the difference between the tensions evaluated at the sides of the piece, Fr = T (r + Δr) − T (r) , (Figure 9.11). Figure 9.11 Free-body force diagram on small slice of rope The piece is accelerating inward with a radial component ar = −r ω 2 . Thus Newton’s Second Law becomes Fr = −Δmω 2 r (9.3.9) T (r + Δr) − T (r) = −(m / L)Δr rω 2 . Denote the difference in the tension by ΔT = T (r + Δr) − T (r) . After dividing through by Δr , Eq. (9.3.9) becomes 9-7 ΔT = −(m / L) rω 2 . Δr (9.3.10) In the limit as Δr → 0 , Eq. (9.3.10) becomes a differential equation, dT = −(m / L)ω 2 r . dr (9.3.11) From this, we see immediately that the tension decreases with increasing radius. We shall solve this equation by integration T (r) − T (L) = r′=r r dT 2 dr′ = −(m ω / L) ∫ dr′ ∫L r′ dr′ r′= L = −(mω 2 / 2L)(r 2 − L2 ) (9.3.12) = (mω / 2L)(L − r ). 2 2 2 We use the fact that the tension, in the ideal case, will vanish at the end of the rope, r = L . Thus, (9.3.13) T (r) = (mω 2 / 2L)( L2 − r 2 ). This last expression shows the expected functional form, in that the tension is largest closest to the shaft, and vanishes at the end of the rope. Example 9.5 Object Sliding in a Circular Orbit on the Inside of a Cone Consider an object of mass m that slides without friction on the inside of a cone moving in a circular orbit with constant speed v0 . The cone makes an angle θ with respect to a vertical axis. The axis of the cone is vertical and gravity is directed downwards. The apex half-angle of the cone is θ as shown in Figure 9.12. Find the radius of the circular path and the time it takes to complete one circular orbit in terms of the given quantities and g . Figure 9.12 Object in a circular orbit on inside of a cone 9-8 Solution: Choose cylindrical coordinates as shown in the above figure. Choose unit vectors r̂ pointing in the radial outward direction and k̂ pointing upwards. The force diagram on the object is shown in Figure 9.13. Figure 9.13 Free-body force diagram on object The two forces acting on the object are the normal force of the wall on the object and the gravitational force. Then Newton’s Second Law in the r̂ -direction becomes −mv 2 − N cosθ = r and in the k̂ -direction becomes N sin θ − m g = 0 . These equations can be re-expressed as v2 r N sin θ = m g . N cosθ = m We can divide these two equations, N sin θ mg = N cosθ mv 2 / r yielding tan θ = rg . v2 This can be solved for the radius, v2 r = tan θ . g The centripetal force in this problem is the vector component of the contact force that is pointing radially inwards, Fcent = N cosθ = m g cot θ , 9-9 where N sin θ = m g has been used to eliminate N in terms of m , g and θ . The radius is independent of the mass because the component of the normal force in the vertical direction must balance the gravitational force, and so the normal force is proportional to the mass. Example 9.6 Coin on a Rotating Turntable A coin of mass m (which you may treat as a point object) lies on a turntable, exactly at the rim, a distance R from the center. The turntable turns at constant angular speed ω and the coin rides without slipping. Suppose the coefficient of static friction between the turntable and the coin is given by µ . Let g be the gravitational constant. What is the maximum angular speed ω max such that the coin does not slip? Figure 9.14 Coin on Rotating Turntable Solution: The coin undergoes circular motion at constant speed so it is accelerating inward. The force inward is static friction and at the just slipping point it has reached its maximum value. We can use Newton’s Second Law to find the maximum angular speed ω max . We choose a polar coordinate system and the free-body force diagram is shown in the figure below. Figure 9.15 Free-body force diagram on coin The contact force is given by    C = N + fs = N k̂ − fsr̂ . The gravitational force is given by  Fgrav = −mgk̂ . (9.3.14) (9.3.15) Newton’s Second Law in the radial direction is given by 9-10 − fs = −m Rω 2 . (9.3.16) Newton’s Second Law, Fz = maz , in the z-direction, noting that the disc is static hence az = 0 , is given by Thus the normal force is N − mg = 0 . (9.3.17) N = mg . (9.3.18) As ω increases, the static friction increases in magnitude until at ω = ωmax and static friction reaches its maximum value (noting Eq. (9.3.18)). ( fs ) max = µ N = µ mg . (9.3.19) At this value the disc slips. Thus substituting this value for the maximum static friction into Eq. (9.3.16) yields 2 µ mg = mRωmax . (9.3.20) We can now solve Eq. (9.3.20) for maximum angular speed ω max such that the coin does not slip µg ωmax = . (9.3.21) R 9-11