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Circular Motion Dynamics

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Circular Motion Dynamics Problem Solving Circular Motion Dynamics Problem 1: 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 about the center of mass of the system. If the stars are always a fixed distance s apart, what is the period of the orbit? Solution: Choose radial coordinates for each star with origin at center of mass. Let rˆ1 be a unit vector at Star 1 pointing radially away from the center of mass. Let rˆ2 be a unit vector at Star 2 pointing radially away from the center of mass. The force diagrams on the two stars are shown in the figure below. ! ! From Newton’s Second Law, F1 = m 1a 1 , for Star 1 in the radial direction is m m ˆ 1 2 2 r1 : "G 2 = "m1 r 1 ! . s We can solve this for r1 , m r = G 2 . 1 ! 2s 2 ! ! Newton’s Second Law, F2 = m 2a 2 , for Star 2 in the radial direction is m m ˆ 1 2 2 r2 : "G 2 = "m2 r 2 ! . s We can solve this for r2 , m r = G 1 . 2 ! 2s 2 Since s , the distance between the stars, is constant m2 m1 (m2 + m1 ) s = r + r = G + G = G 1 2 ! 2s 2! 2s 2 ! 2s 2 . Thus the angular velocity is 1 2 " (m2 + m1 ) # ! = $G 3 % & s ' and the period is then 1 2 2! # 4! 2s 3 $ T = = % & . % G m m & " ' ( 2 + 1 )( As will be seen later, this result is a variation of Kepler’s Third Law. Problem 2: Circular motion: banked turn A car of mass m is going around a circular turn of radius R , that is banked at an angle ! with respect to the ground. The coefficient of static friction between the tires and the road is µ . Let g be the magnitude of the gravitational acceleration. You may neglect kinetic friction (that is, the car’s tires do not slip). a) At what speed v should the car enter the banked turn if the road is very slippery 0 (i.e. µ ! 0 ) in order not to slide up or down the banked turn? b) Suppose µ tan! < 1. What is the maximum speed v with which the car can max enter the banked turn so that it does not slide up the banked turn? c) Suppose µ tan! < 1. What is the minimum speed v with which the car can min enter the banked turn so that it does not slide down the banked turn? d) Suppose the car enters the turn with a speed v such that v > v > v . Find an max 0 expression for the magnitude of the friction force. Solution: a) We will first consider the case where the friction f ! 0 is approximately zero. static Denote the speed of the car by v . Choose cylindrical coordinates as shown in the figure 0 below. Choose unit vectors rˆ pointing in the radial outward direction and kˆ pointing upwards. The force diagram on the car is shown in the figure below. ! ! Newton’s Second law, F = ma , becomes mv2 rˆ : ! N sin" = ! 0 (2.1) r kˆ : N cos! " m g = ma (2.2) z Because the car is traveling in a circle the acceleration in the z -direction is zero a z = 0 (2.3) and so the force equations (2.8) and (2.9) become v2 !N sin" = !m 0 (2.4) r N cos! = m g (2.5) Dividing these equations yields v2 tan! = 0 (2.6) r g We now solve for the minimum speed, v , necessary to maintain a circular path 0 1 2 vmin = r g tan! (2.7) ( ) b) We now consider the case when the car is traveling slow enough, i.e. with minimum speed v = vmin , such that it just starts to slip down the bank. The force diagram on the car is shown in the figure below. ! ! Newton’s Second law, F = ma , becomes mv 2 rˆ : " N sin ! + f cos ! = " (2.8) static r ˆ k : N cos ! + f static sin ! " m g = m a z (2.9) When v = vmin , the just-slipping condition is that the acceleration in the z -direction is zero and the static friction has its maximum value: a z = 0 (2.10) f = µ N (2.11) static and so the force equations (2.8) and (2.9) become v 2 ! N sin" + µN cos" = !m min (2.12) r N cos! + µN sin! = m g (2.13) Dividing these equations yields ! sin" + µ cos" v2 = ! min (2.14) cos" + µ sin" r g which can then be solved for the minimum speed, vmin , necessary to avoid sliding down the embanked turn. 1 # # sin! " µ cos! & & 2 v r g (2.15) min = % ( $ $% cos! + µ sin! '( ' This result should be checked for the limiting values of . In the limit 0 , µ µ ! v rg , the result we found in part a). In the limit tan , v 0 , which is min ! µ ! " min ! the static case of a block on an incline. c) Now let’s consider the case that the car is at the maximum speed such that it just starts to slip up the inclined plane. Then the direction of static friction points down the incline plane and the free body force diagram is shown in the figure below. The analysis is identical to the previous case except for changing the signs of the components of static friction. Thus Newton’s Second Law becomes mv 2 rˆ : " N sin ! " f cos ! = " (2.16) static r ˆ k : N cos! " fstatic sin! " m g = 0 (2.17) When v = vmax , the static friction has its maximum value given by Eq. (2.11), and so the force equations now become v 2 !N sin" ! µN cos" = !m max (2.18) r N cos! " µN sin! = m g (2.19) Dividing these equations yields ! sin" ! µ cos" v2 = ! max (2.20) cos" ! µ sin" r g which can then be solved for the maximum speed, vmax , necessary to avoid sliding down the embanked turn. 1 # # sin! + µ cos! & & 2 v max = r g (2.21) % $% cos! " µ sin! '( ( $ ' The figure below shows a plot of v2 / r g vs. µ when ! = 45! . The shaded area represents the set of points (µ,v2 / r g ) where the car remains in a circular path. Above that is the set of points in which the car will slid outward and below the set of points in which the car will slide inward. d) The analysis is the same as in part c) but the magnitude of the static friction is less than its maximum value. Hence we can multiply Eq. (2.16) by cos! and Eq. (2.17) . by sin! . 2 2 mv ! N sin" cos" ! fstatic cos " = ! cos" (2.22) r N cos! sin! " f sin2 ! " m g sin! = 0 (2.23) static Now add the two equations to yield 2 2 2 mv ! fstatic (cos " + sin ") ! m g sin" = ! cos" (2.24) r 2 2 Use the identity (cos ! + sin !) = 1 and solve Eq. (2.24) for the magnitude of the static friction force # v2 & f = m cos! " g sin! (2.25) static % r ( $ ' Problem 3 Sally swings a ball of mass m in a circle of radius R in a vertical plane by means of a massless string. The speed of the ball is constant and it makes one revolution every t 0 seconds. a) Find an expression for the radial component of the tension in the string T ( ) as a ! function of the angle ! the ball makes with the vertical 1 . Express your answer in terms of some combination of the parameters m, R , t and the gravitational constant g . 0 b) Is there a range of values of t for which this type of circular motion can not be 0 maintained? If so, what is that range? Solution: a) The free body diagram is shown in the figure below. ! ! Newton’s second Law F = ma in the radially inward direction becomes !T ! mg cos" = !mR# 2 . r Thus 1Note added after the fact: The ball moves in a circle, but Sally's hand cannot remain at the center of the circle if a constant speed is to be maintained. T = mR! 2 " mg cos# r Because the angular speed ! = 2" / t , the magnitude of the radial component of the 0 tension in the string is 4! 2 R T = m( " g cos#) r t 2 0 b) The magnitude of the radial component of the tension in the string T must r always be greater than zero. When / 2 3 / 2 , cos(!) " 0 hence T > 0 . For the ! " # " ! r range of angle 0 ! " < # / 2 and 3! / 2 " # < 2! , cos(!) > 0 and so the condition 4! 2 R T = m( " g cos#) > 0 r t 2 0 implies that R 2! > t0 . g cos" At ! = 0 , where mg makes the maximum contribution to the required radially inward directed force, cos(0) = 1, for all other values of ! in the range of angle 0 < ! < " / 2 and 3 / 2 2 , cos(!) < 1 Thus to maintain circular motion the period t must be less ! < " < ! 0 than a critical value (t ) 0 c R t0 < (t0 )c = 2! . g Problem 4 A ball of mass m is connected by a spring to shaft that is rotating with constant angular velocity ! . A student looking down on the apparatus sees the ball moving counterclockwise in a circular path of radius r .
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