Problem Set #13 Solutions


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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.0L: Physics I January 3, 06 Prof. Alan Guth Problem Set #3 Solutions Due by :00 am on Friday, January in the bins at the intersection of Buildings 8 3rd floor) and 6 4th floor). Please put your name and recitation number clearly on the top page of your problem set. SOURCES: University Physics, Volume, 3th Edition, by Hugh D. Young and Roger A. Freedman AddisonWesley, 0). This is the required textbook for the course. Essentials of Introductory Classical Mechanics, 6th Edition, by Wit Busza, Susan Cartwright, and Alan H. Guth, now available to EVERYBODY at 8.0L/studyguide/index.html. Also known as the BCG Study Guide. READING: Chapter 0, Sections DYNAMICS OF ROTATIONAL MOTION: Work and Power in Rotational Motion; Angular Momentum; Conservation of Angular Momentum; Gyroscopes and Precession) and Chapter 3, Sections : GRAVITATION: Newton s Law of Gravitation; Weight; Gravitational Potential Energy; The Motion of Satellites; Kepler s Laws and the Motion of Planets; Spherical Mass Distributions; Apparent Weight and the Earth s Rotation; Black Holes). OPTIONAL ADDITIONAL READING: Busza, Cartwright, and Guth BCG), Chapter 9, pp Optional means that you should read this only if you find it useful.) NOTE: Your written solutions must include a brief commentary in addition to any equations or graphs used to arrive at your answer. For example, this commentary should explain your basic strategy for solving the problem and also highlight the concept before applying an equation. We also want you to first solve a problem algebraically before you put in the particular numbers relevant to the problem. We ll say that again: solve the answer in terms of variables before you start sticking in numbers! This makes it easier for you when you check your work and also for the grader to follow your logic. 3: Square pendulum Young and Freedman 3th edition) problem 4.79, page 468. same as Young and Freedman th edition) problem 3.7, page 45.) In doing this problem, you may want to refer to Section 4.6, The Physical Pendulum, pp of the 3th edition Section 3.6, pp in th edition). A square object of mass m is constructed of four identical uniform thin sticks, each of length L, attached together. This object is hung on a hook at its upper corner. If it is rotated slightly to the left and then released, at what frequency will it swing back and forth?
2 8.0L Problem Set #3, p. This is a physical pendulum, the frequency of oscillation is given by f = mgd π I The center of mass of the square object is simply its geometrical center, ) d = L cos 45 = L. ) Now we need to calculate the moment of inertia of the square I about an axis through the pivot, perpendicular to the plane of the square. By parallelaxis theorem, I = I cm + md = I cm + ml 4. 3) To calculate I cm, think of the square as 4 sticks, each stick contributes a moment of inertia of m 4 ) L + m 4 I = 4 Substitute into ) gives L ). Again by parallelaxis theorem, m 4 ) L + m 4 ) ) L + ml = 5 6 ml 5) f = π mg L 5 = 6 ml π : Parachutist and turntable Based on Young and Freedman 3th edition) exercise 0.45, page 337. same as Young and Freedman th edition) exercise 0.43, page 347.) 4) g L. 6) A large wooden turntable in the shape of a flat uniform disk has a radius R and a total mass M. The turntable is initially rotating at angular frequency ω 0 about a vertical axis through its center. Suddenly, a parachutist of mass m makes a soft landing on the turntable at a point near the outer edge. a) Find the angular speed of the turntable after the parachutist lands. Assume that you can treat the parachutist as a particle.)
3 8.0L Problem Set #3, p. 3 Since there is no net external torque around the rotation axis, angular momentum of the system of parachutist and turntable is conserved. With L 0 = I 0 ω 0 and L = I ω, L 0 = L implies Since I 0 = MR and I = I 0 + mr we have ω = I 0 I ω 0 7) I 0 I = + m M. 8) Therefore ω = kg 0 kg 3.00 rad/s =.38 rad/s. 9) b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal? The kinetic energy of the system before the parachutist lands K 0 = I 0ω0 = ) MR ω0 0) after the landing, = 4 0 kg.00 m) 3.00 rad/s) ) = 080 J ) K = I ω = I = kg 0 kg ) I0 ω 0 = I 0 K 0 = I I + m M K 0 3) 080 J = 498 J 4) The kinetic energy decreases due to the negative work done by friction between the parachutist and turntable) on the parachutist and the turntable.
4 8.0L Problem Set #3, p : Gyroscope Young and Freedman 3th edition) exercise 0.55, page 337. same as Young and Freedman th edition) exercise 0.5, page 348.) A gyroscope is precessing about a vertical axis. Describe what happens to the precession angular speed if the following changes in the variables are made, with all other variables remaining the same: a) the angular speed of the spinning flywheel is doubled; Apply equation 0.33, Ω = dφ dt = τ z = W r L z Iω 5) where W is the weight. ω ω Ω Ω 6) b) the total weight is doubled; W W Ω Ω 7) Here we assume that the added weight does not change I or r) c) the moment of inertia about the axis of the spinning flywheel is doubled; I I Ω Ω 8) assuming this does not change W or r) d) the distance from the pivot to the center of gravity is doubled. r r Ω Ω 9) e) What happens if all four of the variables in parts a) through d) are doubled? If all 4 variables are doubled, Ω Ω unchanged.
5 8.0L Problem Set #3, p : Inelastic collision A slender uniform metal rod with mass M and length l is pivoted at its upper end and hangs vertically. A small blob with mass m traveling horizontally with velocity v 0 strikes the lower end of the rod and sticks to it. a) Find the angular momentum of the blob with respect to the pivot prior to the collision. Angular momentum of the blob with respect to the pivot p, L blob = mv 0 lˆk 0) b) Find the angular momentum of the blob+rod system with respect to the pivot immediately after the collision. Using angular momentum conservation immediately after the collision. The angular momentum of the blob + rod, L satisfy L = 0 + L blob = mv 0 lˆk ) c) Calculate the moment of inertia of the blob+rod system around the pivot immediately after the collision. About the pivot, net net moment of inertia is I = ml + 3 Ml = m + 3 ) M l ) d) Find the angular velocity of the blob+rod system after the collision.
6 8.0L Problem Set #3, p. 6 From L = I ω, the angular velocity of the blob + rod system right after the collision is ω = L I ˆk = m v 0 m + M l ˆk 3) 3 e) Take θ max to be the maximum angle to the vertical made by the blob+rod system following the collision. What is difference in the gravitational potential energy of the blob+rod system between θ max and θ = 0? The difference in gravitational potential energy is U = Uθ = θ max ) Uθ = 0) 4) = Mg l cos θ max) + mgl cos θ max ) 5) ) = M + m gl cos θ max ) 6) f) Find θ max in terms of the given quantities: M, l, m, v 0 and as always) g. Use energy conservation, from right after collision to the maximum angle θ max, U = K 7) ) M + m gl cos θ max ) = Iω = L I = mv 0l) ) 8) m + M 3 l cos θ max = m + M 3 θ max = arccos m ) m + M ) v 0 gl m + M 3 m ) m + M ) v 0 gl ) 9) 30) 35: Stopping a bicycle An idealized bicycle+rider consists of a block representing the rider, and two wheels of radius R with centers separated by distance L. Suppose that the center of mass of the system total mass M) is midway between the wheels at a height R above the ground. The rider squeezes the front brakes, locking the front wheel. What is the maximum deceleration such that the system slows down without the rear wheel leaving the ground and the rider going over the handlebars)?
7 8.0L Problem Set #3, p. 7 When the rider squeezes the front brake, the front wheel is locked and starts slipping. The bike decelerates with a magnitude of a due to the friction between the front wheel and the ground. From Newton s nd Law, f = Ma. Since f = µn f, as a increases, N f increases the weight Mg shifts to the front. Apart from the deceleration a, friction f also contributes to a torque about the c.m. that tends to lift the rear wheel; this torque is cancelled by the torque from N f and N r when a is small. But when a is big enough, N f Mg and N b 0. This is when the rear wheel leaves the ground. In addition, when the rear wheel is just about to leave the ground, we have a y = 0, α = 0 about c.m.), so N b = 0 L N b + N f Mg = 0 = f = Mg 3) τ = L Nf gr) = 0 4R so a = f M = g L 3) 4R This is the maximum deceleration such that the bike slows down without the rear wheel leaving the ground bike tilting around the front wheel). 36: Backspin on cue ball In a game of billiards, a skillful player hits a cue ball with just enough backspin so that the ball stops after traveling a short distance away from where is was hit. Suppose the initial velocity of the ball was v 0 after being hit). Assume the cue ball is a solid sphere with mass m and radius r. How much backspin that is, what negative initial angular velocity ω 0 ) would be required for this to occur? Solve this problem two ways: a) Consider the force of friction and use equations describing the linear kinematics and rotational kinematics.
8 8.0L Problem Set #3, p. 8 Newton s nd Law: Torque about the c.m.: Kinematics: f = µ k N = µ k mg = a x = f m = µ kg 33) τ = fr = µ k mgr 34) α = τ I = µ kmgr /5)mr 35) = 5µ kg r 36) v = v 0 + a x t, ω = ω 0 + αt 37) Thus when v = 0, t = v 0 µ k. For the ball to stop, ω also has to vanish at the same g time. That is, 0 = ω 0 + αt. ω 0 = αt = 5µ kg r v 0 µ k g = 5v 0 r b) Find a reference point about which the net torque is zero and use the concept of conservation of angular momentum. In this approach you will probably want to use the theorem that says when a rigid body translates and simultaneously rotates about an axis of symmetry, then the angular momentum about any fixed point S can be written as L = r cm P tot + I cm ω, where r cm is a vector from the reference point S to the center of mass, P tot is the total momentum of the object, I cm is the moment of inertia of the object about the rotation axis through the center of mass, and ω is the angular velocity. You obviously) should get the same answer in both methods. Which method was easier? 38)
9 8.0L Problem Set #3, p. 9 Choose the reference point to be Q along the direction of v 0 on the table. Notice that around Q, f contributes 0 torque whereas the contribution from N and mg cancel, therefore the net torque around Q is 0. Angular momentum about this point is conserved. The initial angular momentum is L = I cm ω } {{ } 0 + m r v 0 = I } {{ } cm ω 0ˆk + mv0 r)ˆk 39) i) ii) where i) is the angular momentum about the c.m. and ii) is the angular momentum due to the motion of the center of mass. The final angular momentum L = 0, since the motion stops completely. Since angular momentum about Q is conserved, we have L = L and therefore L = 0, which implies ω 0 = mv 0r I cm = mv 0r = 5v 0 5 mr r 37: Geosynchronous satellites Young and Freedman 3th edition) problem 3.5, page 43. same as Young and Freedman th edition) problem.55, page 45.) Many satellites are moving in a circle in the earth s equatorial plane. They are at such a height above the earth s surface that they always remain above the same point. a) Find the altitude of these satellites above the earth s surface. Such an orbit is said to be geosynchronous.) 40) For satellites to remain above the same point on the earth means they have to have the same orbital period as earth Now using equation 3. T = d = ) s = s 4) T = πr3/ GmE = r = is the radius of the satellite s orbit. r = T Gm E π) ) s) N m kg ) kg) 4π ) 3 4) 43) = m 44) The altitude of the satellites h satisfy r = h + R E, thus h = r R E = m m = m 45) b) Explain, with a sketch, why the radio signals from these satellites cannot directly reach receivers on earth that are north of 8.3 N latitude.
10 8.0L Problem Set #3, p. 0 Draw the tangent line to the earth from the same satellite as in the picture. Radio signals from the satellite are blocked by the earth from reaching higher latitudes beyond the tangent point. ) ) θ = cos RE = cos 6 m r =.4 rad 46) m = ) 38: Gravity near surface of the Earth Young and Freedman 3th edition) problem 3.57, page 43. same as Young and Freedman th edition) problem.6, page 46.) There are two equations from which a change in the gravitational potential energy U of the system of a mass m and the earth can be calculated. One is U = mgy Eq. 7. of Y&F 3th edition). The other is U = Gm E m/r Eq. 3.9). As shown in Section 3.3, the first equation is correct only if the gravitational force is a constant over the change in height y. The second is always correct. Actually, the gravitational force is never exactly constant over any change in height, but if the variation is small, we can ignore it. Consider the difference in U between a mass at the earth s surface and a distance h above it using both equations, and find the value of h for which Eq. 7.) is in error by %. Express this value of h as a fraction of the earth s radius, and also obtain a numerical value for it. [In the th edition, the references are Eq. 7., Eq..9, and Section.3.] The fraction error is expressed as α = Gm E m R E mgh Gm Em R E +h = gh Gm E From Equation 3.4, g = Gm E, so RE R E R E +h α = R E + h R E ) 48) = g Gm E R E R E + h) 49) = h R E 50)
11 8.0L Problem Set #3, p. Therefore if the fraction error is %, α = % h R E = 0.0 = h = 0.0R E = m 5) 39: Falling hammer Young and Freedman 3th edition) problem 3.65, page 433. same as Young and Freedman th edition) problem.67, page 46.) A hammer with mass m is dropped from rest from a height h above the earth s surface. This height is not necessarily small compared with the radius of the earth. If you ignore air resistance, derive an expression for the speed v of the hammer when it reaches the surface of the earth. Your expression should involve h, R E, and m E, the mass of the earth. Apply conservation of energy, and note that we need to use 3.9) U = Gm Em for r gravitational potential energy since h is not necessarily small compared to R E K + U = K + U, 0 + U = Gm Em, h + R E Gm ) Em = h + R E mv + U = Gm Em 5) R E Gm ) Em R E ) v = Gm e R E R E + h Gm E h v = R E R E + h) 53) 54) 55) 30: Spacecraft escape Young and Freedman 3th edition) problem 3.67, page 433. same as Young and Freedman th edition) problem.69, page 46.) The astronomical data needed for this problem can be found in Appendix F of the textbook, in either the th or 3th editions. A spacecraft is to be launched from the surface of the earth so that it will escape from the solar system altogether. a) Find the speed relative to the center of the earth with which the spacecraft must be launched. Take into consideration the gravitational effects of both the earth and the sun, and include the effects of the earth s orbital speed, but ignore air resistance.
12 8.0L Problem Set #3, p. Apply energy conservation to find the escape velocity from the solar system, assume that the gravitational potential energy between the spacecraft and the earth/the sun dominate. R E is the radius of the earth; R ES is the distance between the earth and the sun. mv Gmm E R E Gmm S R S = 0 56) The escape velocity v in the reference frame of a distance star) me v = G + m ) S r E R S = N m kg 4 kg m +.99 ) 030 kg.50 0 m 57) 58) = m/s 59) Now assume the spacecraft is launched in the direction of the earth orbital motion around the sun, the speed relative to the center of the earth is v = v πr ES T = m/s π.50 0 m ) s 60) = m/s 6) b) The rotation of the earth can help this spacecraft achieve escape speed. Find the speed that the spacecraft must have relative to the earth s surface if the spacecraft is launched from Florida at the point shown in the figure. The rotation and orbital motions of the earth are in the same direction. The launch facilities in Florida are 8.5 north of the equator. The rotation and orbital motion of the earth are in the same direction. The speed of a point on the surface of the earth at an angle θ from the equator is v = πr E cos φ T E, where T E = ) s = s. Therefore launched from Florida, the spacecraft must have a speed v relative to the earth s surface. v = v πr e cos 8.5 T E 6) = m/s π m) cos ) s = m/s 64) c) The European Space Agency ESA) uses launch facilities in French Guiana immediately north of Brazil), 5.5 north of the equator. What speed relative to the earth s surface would a spacecraft need to escape the solar system if launched from French Guiana?
13 8.0L Problem Set #3, p. 3 Similarly, launched from French Guiana, the space craft must have an escape speed of v 3 relative to the earth s surface, v 3 = v πr e cos 5.5 T E 65) = m/s π m) cos ) s = m/s 67) SuperChallenge Problem : Elliptical orbit Adapted from Young and Freedman 3th edition) problem 3.77, page 434. same as Young and Freedman th edition) problem.77, page 47.) For up to 0 points of extra credit, you may do the following totally optional problem. For a description of how extra credit problems will be processed, see Problem Set 9. Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is at a distance r p from the earth s center; at the high point, or apogee, it is a distance r a from the earth s center. a) What is the period of the spacecraft s orbit? Since a = r a + r p, we have T = πa3/ GmE = π Gm E ra + r p ) 3 68) b) Using conservation of angular momentum, find the ratio of the spacecraft s speed at perigee to its speed at apogee. Using conservation of angular momentum about the earth, mr a v a = mr p v p 69) v p = r a 70) v a r p c) Using conservation of energy, find the speed at perigee and the speed at apogee. Apply conservation of energy to apogee and perigee K a + U a = K p + U p 7) mv a Gm Em = r a mv p Gm Em 7) r p
14 use result from b) v p v p v a = Gm E r p r a ) rp r a ) ) 8.0L Problem Set #3, p. 4 = Gm Er a r p ) r a r p 73) = Gm Er a r p ) r a r p 74) r a vp r a r p = Gm E r a r p ra rp Gm E r a v p = r p r a + r p ) v a = r pv p Gm E r p = r a r a r a + r p ) = Gm Er a r p r a + r p ) 75) 76) 77) d) It is necessary to have the spacecraft escape from the earth completely. If the spacecraft s rockets are fired briefly at perigee, producing a sudden change in the rocket s speed, by how much would the speed have to be increased to achieve this? What if the rockets were fired at apogee? Which point in the orbit is more efficient to use? For the spacecraft to escape, E has to be at least 0, K + U = 0. At apogee: mṽ p Gm Em Gm E = 0 = ṽ p = 78) r p r p the speed has to be increased by ) Gm E ra v p = ṽ p v p = r p r a + r p 79) At perigee: mṽ a Gm Em GmE = 0 = ṽ a = 80) r a r a the speed has to be increased by GmE v a = ṽ a v a = r a rp r a + r p ) 8) It is not easy to see which of these two expressions is larger, but if one inserts numbers one finds immediately that v is always smaller at perigee. This implies
15 8.0L Problem Set #3, p. 5 that less fuel would be needed to fire the rockets at perigree, since the fuel needs are determined entirely by the change in velocity. Note that the spacecraft flies essentialy through a vacuum, so it does not interact with the Earth, and its speed relative to the Earth is of no relevance. The same change in velocity will always require the same amount of fuel, regardless of the initial velocity relative to the Earth.) SuperChallenge Problem : Navigation from Earth to Mars Young and Freedman 3th edition) problem 3.87, page 435. same as Young and Freedman th edition) problem.87, page 48.) For up to 0 points of extra credit, you may do the following totally optional problem. For a description of how extra credit problems will be processed, see Problem Set 9. Figure P3.87 The most efficient way to send a spacecraft from the earth to another planet is by using a Hohmann transfer orbit Fig. P3.87). If the orbits of the departure and destination planets are circular, the Hohmann transfer orbit is an elliptical orbit whose perihelion and aphelion are tangent to the orbits of the two planets. The rockets are fired briefly at the departure planet to put the spacecraft into the transfer orbit; the spacecraft then coasts until it reaches the destination planet. The rockets are then fired again to put the spacecraft into the same orbit about the sun as the destination planet. a) For a flight from earth to Mars, in what direction must the rockets be fired at the earth and at Mars: in the direction of motion, or opposite the direction of motion? What about for a flight from Mars to the earth? The orbital radius of the Earth is r E =.50 0 m, and the orbital radius of Mars is r M =.8 0 m. One can see from the figure that the major axis of the Hohmann transfer orbit is just the sum of these two, so the semimajor axis for the elliptical Hohmann transfer orbit is given by a = r M + r E =.89 0 m. 8)
16 8.0L Problem Set #3, p. 6 If the spacecraft does not fire its engines, it will of course remain in its circular orbit at radius r E. To reach the larger radii of the Hohmann transfer orbit, it must increase its speed and energy), and so the rockets must be fired in the opposite direction of the motion. Once the spacecraft reaches the orbit of Mars, if it did not fire its rockets, it would continue on the elliptical orbit and fall back to smaller radii, as it approached once more the radius of the Earth s orbit. To maintain the larger radius of Mars orbit, it must again increase its speed and energy), which means that the rockets must again be fired in the direction opposite that of the motion. For the returning flight from Mars to Earth, the precedure is reversed, and the rockets must be fired in the direction of the motion so as to decreased the speed and thus decreased the mechanical energy. b) How long does a oneway trip from the the earth to Mars take, between the firings of the rockets? The time of the oneway trip would be one half the period of the transfer orbit. T = πa3/ Gms t = T = πa3/ Gms = π.89 0 m) N m kg kg 83) 84) = s = 59 d 85) c) To reach Mars from the earth, the launch must be timed so that Mars will be at the right spot when the spacecraft reaches Mars s orbit around the sun. At launch, what must the angle between a sun Mars line and a sun earth line be? Use data from Appendix F. During this time t, Mars will pass through an angle of θ = π t = π 59 d 86) T M 687 d =.37 rad = 36 87) Since the spacecraft passes through an angle of 80 about the Sun, at launch, the angle between the SunMars line and the SunEarth line must be = 44. The earth must be 44 behind Mars in the direction of the orbital motion.
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