Problem 1 Earth and a Geostationary Satellite (10 points) The earth is spinning about its axis with a period of 3 hours 56 minutes and 4 seconds. The equatorial radius of the earth is 6.38 10 6 m. The latitude of Cambridge, Mass is 4. a) Find the magnitude of the velocity and the centripetal acceleration with respect to the earth s axis of a person at MIT as they undergo circular motion about the earth s axis of rotation. A geostationary satellite goes around the earth once every 3 hours 56 minutes and 4 seconds, (a sidereal day, shorter than the noon-to-noon solar day of 4 hours) so that its position appears stationary with respect to a ground station. The mass of the earth is m e = 5.98 10 4 6 kg. The mean radius of the earth is R = 6.37 10 m. The universal constant of gravitation is G = 6.67 10 11 N m kg. Your goal is to find the radius of the orbit of a geostationary satellite. b) Describe what motion models this problem. c) What is the radius of the orbit of a geostationary satellite? Approximately how many earth radii is this distance? Solution: For recent values of the physical constants used in this problem, see http://pdg.lbl.gov/007/reviews/contents_sports.html - constantsetc. a) The rotational period of the earth is given by 1 1 ( ) ( ) T earth, 005 = (3 hr) 3600 s hr + (56 min) 60 s min + 4s = 86164 s. (1.1) Note that this period is less than 4hr. Twenty-four hours is one solar day (noon to noon), while the above period is one sidereal day; sidereal means with respect to the fixed stars, and you should be able to see why the two are different. A person at MIT undergoes circular motion about the axis of the earth. The radius of this circle is R = R sinθ e where θ is the angle between MIT and the axis of rotation. Since the latitude λ is measured form the equator, sinθ = sin(π / λ) = cosλ (the angle θ is sometimes called the colatitude ). Hence the radius of the orbit of a person at MIT is e ( ) R = R = R =. (1.) 6 6 esinθ ecos λ (6.38 10 m) cos 4.36 = 4.71 10 m Since the motion is uniform, during one period of rotation the person travels a distance
where v s = π R = vt (1.3) is the magnitude of the velocity and s = π R is the circumference. The magnitude of the velocity (the speed) is then given by v = π R T (1.4) and so for a person at MIT, the magnitude of the velocity is 6 ( ) π R ( π ) 4.71 10 m 1 v = = = 3.44 10 m s. (1.5) T 86164 s The magnitude a r of the centripetal acceleration is given by a r ( 3.44 10 m s 1 ) v = = = 6 R 4.71 10 m.51 10 m s. (1.6) b) 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. The acceleration of the satellite is directed towards the center of the circle, that is, along the radially inward direction. The figure below is close to a scale drawing.
c) Choose the origin at the center of the earth, and the unit vector ˆr along the radial direction. This choice of coordinates makes sense in this problem since the direction of acceleration is along the radial direction. Let r be the position vector of the satellite. The magnitude of r (we denote it as r s ) 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 = π / ω. The acceleration is directed inward, with magnitude rω ; in vector form, s a= rω ˆr. (1.7) s 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, F m r = ω rˆ grav s s. (1.8) The inward radial force on the satellite is the gravitational attraction of the earth, mm G rˆ = mω r r. ˆ (1.9) s e rs s s Equating the ˆr components, mm G s e rs s r s = mω. (1.10)
Solving for the radius of orbit of the satellite r s, Gme rs = ω 1/3. (1.11) The period T of the satellite s orbit in seconds is 86164 s (see Equation (1.1)) and so the angular speed is π π 5 1 ω = = = 7.91 10 s. (1.1) T 86164 s Using the values of ω, Gandm in Equation (1.11), we determine ; e 7 r s = 4. 10 m = 6.6 Re. (1.13) r s Problem (10 points): a) A bicycle tire rolls without slipping along the ground with its center of mass moving with speed v 0. You may neglect any rolling resistance. What is the speed with respect to the ground of each of the four marked points on the tire shown in the figure below? Explain each of your answers. b) A rider in a barrel of fun finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? Explain your reasoning.
Answers to (a): As an intermediate calculation (introducing an auxiliary parameter ), let the wheel radius be. The angular speed of the wheel about its center is then ω = v / R 0 0 wheel R wheel if the wheel is to roll without slipping. The speed of any point on the rim of the wheel with respect to the center is then v= ω0rwheel = v0. This should be obvious, and such a detailed calculation may seem pointless. However, for more complicated systems, or, if we needed to consider points other than those on the rim, we have to include this intermediate step. Speed of point a: This point is moving parallel to the velocity of the center of the wheel, so the speed is the sum v0 + v0 = v0, a perhaps apparent result. Speed of point b: This point has a horizontal velocity component and a vertical velocity component, also v0, and so the speed is v 0. Speed of point c: For this point, the velocity with respect to the center of the wheel is directed opposite to the velocity of the center of the wheel, and so the speed with respect to the ground is v0 v0 = 0. Or, we could say quite correctly that if the wheel is rolling without slipping, the contact point, being in contact with the ground, cannot be moving with respect to the ground. Speed of point d: This is the same speed as that of point b, v 0. The vertical component of velocity is of course down, but the magnitude is the same. Answer to (b): 1. Consider the vertical and horizontal forces on the rider separately. For the rider not to slip in the barrel, there must be no net vertical force acting on her. The two vertical forces acting on her are her weight down and a frictional force between her and the wall acting up. These forces must be equal in magnitude. Then there must also be v 0
a force acting on the rider that always points toward the center of the barrel to provide the rider with centripetal acceleration to keep her going in a circle. Where does that force come from? The normal component of the contact force exerted on her by the wall supplies that centripetal acceleration. You may wonder how the wall can exert a normal force on her if there is no other force acting to press her into the wall. Ordinarily we think of contact forces arising because some other force -- typically gravity -- is pushing an object against another object, and the contact force arises in response to that push. If you think carefully about the motion of the rider, her velocity is always directed tangent to the vertical wall of the barrel of fun, so the rider in a sense is constantly colliding with the part of the wall that is right behind her as it travels in a circle. So the wall behind her pushes out on her because otherwise her inertia would mean that she would go straight through the wall, tangent to her velocity at any moment. She exerts a contact force on the wall due to her inertia. As your reading remarked, one thing to be careful about is to realize that centripetal force is not an independent force that somehow arises because something is going in a circle. Rather, if an object is going in a circle, there must be a force acting on it from its environment -- a string tied to it or a contact force from an adjacent surface, for example -- which acts in the appropriate direction to be a centripetal force. So, in the case of the barrel of fun, there are three forces: gravity, the upward frictional force and the inward normal force, and the correct answer is diagram 1. Note that it would be possible to combine the friction force and the normal force into a single contact force. This contact force would have upward and inward components, hence directed up and to the left in the diagram. None of the five choices given matches this model (although a case could be made for other, but such this would need to be explained). Problem 3 Object Undergoing Circular Motion on the Inside of a Cone (10 points) A body of mass m slides without friction on the inside of a cone. The axis of the cone is vertical and gravity is directed downwards. The apex half-angle of the cone is θ as shown in the figure below. The path of the object happens to be a circle in a horizontal plane. The speed of the particle is v 0. 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. Include all relevant free-body diagrams in your answer.
Solution: The free body diagram of the forces acting on the object is shown in the figure below. We choose cylindrical coordinates with unit vectors ˆr pointing in the radial outward direction and ˆk pointing upwards. The two forces acting on the object are the normal force N of the wall on the object and the gravitational force mg (we are given that there is no friction). The two vector components of Newton s Second Law F= ma become: mv0 r ˆ : N cosθ =, r (3.1) k ˆ : Nsinθ mg = 0. (3.) These equations can be re-expressed as v0 Ncosθ = m, r (3.3) Nsinθ = mg. (3.4) We can divide these two equations, yielding Nsinθ mg N cosθ = v, (3.5) 0 m r rg tanθ =. (3.6) v 0 Equation (3.6) can be solved for the radius, v 0 tan r = θ. (3.7) g
Note that 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. Therefore the radially inward component of the normal force is also proportional to mass. Since this force is causing the object to accelerate inward, by Newton s Second Law, the mass will cancel and the acceleration is independent of the mass. Since the radial component of the acceleration is inversely proportional to the radius, ar = v0 / r, therefore the radius is also independent of the mass. The period for the orbit is given by T π r π v tanθ v g 0 = =. (3.8) 0 Problem 4: Circular Motion (10 points) A governor to control the rotational speed of a steam engine was invented by James Watt. Two spheres were attached to a rotating shaft by rigid arms that were free to rotate up and down about a pivot where they attached to the shaft, as in the diagram above. As the arms pivoted up and down they actuated a mechanism to control the throttle of the steam engine. Assume the rigid arms have length l and no mass. All of the mass is then concentrated in the two spheres at the end of the arms, each having mass m. a) Describe the acceleration of the spheres. Explaining and quantifying your knowledge of the acceleration will help you model the problem. Show all relevant free body diagrams. b) Show that there is a minimum angular velocity ω min below which the governor will not function as intended. c) Derive an expression for the radius r of the circular path followed by the spheres. Express your answer only in terms of as few of the quantities m, ω, l, and g (the acceleration due to gravity) as you can. (Do not use the angle φ in your answer).
Solutions: a) Each sphere moves in a horizontal circle of radius r = lsinφ with period T = π / ω, and hence an inward radial acceleration of magnitude a = ω r = ω lsinφ. (4.1) r Thus, there must be a net inward force of magnitude Fnet = m ar = mω lsinφ. This force is the vector sum of the gravitational force mg and the force F rod that the rod applies to the ball (since we call the period T, we don t want to use T for a tension force ). The gravitational force has no horizontal component, so the vertical component of F rod must balance the weight mg and the horizontal component of F rod must have magnitude Fnet m ar mω l = = sinφ. A free body diagram for the right-hand sphere is shown below (the diagram for the lefthand sphere is of course just a reflection and won t be shown here). b) Expressing the answer to part a) mathematically, the two components of Newton s Second Law are: kˆ : F cos φ mg = 0 rod rˆ : Frod sin = m lsin φ ω φ. (4.) Eliminating sinφ (why can we exclude the case sinφ = 0?) from the second expression in (4.) yields F = mω l and substitution into the first and rearranging yields rod
g cosφ =. (4.3) lω We must have cosφ < 1, or ω > g/ l, so ω min = g/ l. c) Using r = lsinφ = l 1 cos φ from part a) and can eliminate the angle φ with the result cos φ / ω = g l from part b) above, we ( ω ) r l 1 g / l l g / ω 4 = =. (4.4) Note that this is consistent with part b); for a positive argument of the square root, we must have ω > ω. min Extra: It s probably worth noting the differences between this problem and Problem 3. Except for notation, the free body diagrams are identical. For instance, in Problem 3 the speed was given and the angle was fixed. In Problem 4, as ω varies, the speed, radius and angle φ vary (which is the point of a governor). Problem 5: Whirling Objects, U-Control Model Airplane (10 points) A U-control airplane of mass M is attached by wires of length L and negligible mass to the pilot who controls the lift provided by the wing. (The wires control the plane s elevator.) The plane s engine keeps it moving at constant speed v. a) Briefly describe how you intend to model the motion of the object. What directions are you choosing for analyzing the components of your forces? Give your reasons. b) Find the total tension T in the wires when the plane is flown overhead in a circle so that the wires make an angle θ with the ground. Remember that the wings can provide lift only in the direction perpendicular to their area, that is, in a direction perpendicular to the wires. Think carefully before selecting the directions of the axes of your coordinate system. c) The plane will go out of control and crash if the tension is not maintained. Given a particular speed v of the plane, is there some angle θ crit which you would advise the pilot not to exceed? If possible, exhibit a speed v safe at which the plane would be safe at any angle. Solutions: (See the problem set for the original figure, which will not be reproduced here.)
a) The object (the airplane) is moving in a horizontal circle. The radius of the circle and the hence the period, for a given speed, will depend on the angle that the wires make with respect to the horizontal. Anticipating parts b) and c), the radius of the circle will be r = Lcosθ and hence the period of the circular motion will be π r/ v= πlcos θ / v. One choice of coordinates is that shown in the free body diagram on the left below, with ˆk directed vertically upward and ˆr directed radially outward. Two advantages to this choice are that the acceleration and the gravitational force are along the coordinate directions, and that this is the standard, already used in Problems 3 and 4 above. Another choice would be that shown on the right below. The advantage to this system is that the unknown forces, including the tension to be determined in part b), are along the coordinate axes, and need not be decomposed. However, the weight mg and the net force must be decomposed along the given axes. b) We ll do this using both of the coordinate systems, starting with the tilted system on the right below. Note that the î - and ĵ -directions are not the horizontal and vertical directions. Applying Newton s Second Law, with a = a as the magnitude of the centripetal acceleration, for the two coordinates, we have r ˆ i : T + mg sin θ = ma c osθ ˆ j : N mg cos θ = ma si n θ. (5.1)
Now we go back to the question asked: We want the find the tension but not the normal force. We know that a= v / r = v / Lcosθ, so the first expression in (5.1) leads almost immediately to ( / sin ) T = m v L g θ, (5.) a very tidy result. If we used the coordinate system on the left above, Newton s Second Law becomes kˆ : N cos θ T sin θ = mg r ˆ : Nsinθ + Tcosθ = ma. (5.3) The expressions in (5.3) are two equations in the two unknowns N and T and may be solved by standard methods. For our immediate purpose, however, eliminate N by multiplying the first by sinθ and the second by cosθ and then subtract the first from the second, leading to Equation (5.). Decide for yourself which coordinate system is better or easier, but keep in mind that either will work. c) Either way, we are now set to find the critical angle θ crit at which the tension as found in Equation (5.) would go to zero, 1 v θcrit = sin. (5.4) gl We see that if so we have v > gl, the critical angle is never reached, and all angles would be safe, v safe = gl. (5.5) Notice that if θ = 0 the rope lies in the plane of the airplane s circular orbit, sinθ = 0, and the tension T = mv / L> 0 for all velocities (assuming the normal force N, acting vertically upward, is sufficient to keep the plane flying). The other extreme value occurs when θ π /. This corresponds to the radius of the orbit r 0, cosθ 1 and the tension T mv / L mg. In order for the tension to stay positive, v> gl (which is ( ) consistent with θcrit π /). Extra: This problem did not ask for the normal force N (often called the lift ), but it s not hard to find N = mg v / L tan θ + cos θ. (5.6) (( ) )