1 (40) Gravitational Systems Two heavy spherical (radius 0.05R) objects are located at fixed positions along

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1 (40) Gravitational Systes Two heavy spherical (radius 0.05) objects are located at fixed positions along 2M 2M 0 an axis in space. The first ass is centered at r = 0 and has a ass of 2M. The second ass is centered at r = and has a ass of 2M. A very very sall ass,, is free to ove along the axis between the two fixed asses, << M. a (0) Define the zero of your gravitational potential energy of in the 2M-2M syste to be at r = ±. Accurately sketch the shape of the gravitational potential energy of for positions along the line that connects to two larger asses (between the asses). The gravitational potential between the two objects can be writte in ters of a diensionless variable, α such that r = α. We see that α goes fro 0 to and we can write the potential energy as U(α) = 2GM Plotting this, we get a function as follows: ( α + ) α. 2M 2M 0 U U(r) b (5) Consider the above region between the two heavy objects. At what value of r is the net gravitational force on equal to zero? The gravitational force is zero when the potential is a axiu or iniu. Thus, we need to set du/dα = 0. We have du dα = 2GM ( d α dα α + ) α 0 = 2GM α ( α 2 + ( α) 2 )

2 0 = ( α) 2 + α 2 0 = 2α c (5) or α = /2, so r = 0.5. The sall ass starts at rest fro the surface of the ass located at r =. What kinetic energy ust it have to be able to reach the surface of the object at r = 0? We need to find the energy to go fro the surface of the ass to the axiu of the potential at /2. U(/2) = 2GM (2 + 2) U(/2) = 8GM U(0.95) = 2GM ( ) U(0.95) = 42GM The change in the gravitational potential energy is equal to the needed kinetic energy, so K = 34GM 2 (40) A Coposite Syste A coposite syste is ade of two subsystes, one and two. We are told that. 2 E E 2 the entropy of the two subsystes are given by the functions S (E) = k B ln E 2 + c S 2 (E) = k B ln E + c 2 where E is the energy in a particular subsyste, k B is the Boltzaann constant and c and c 2 are constants. There is a total energy E which can be distibuted between the two subsystes. Answer the following questions in ters of E, k B, c and c 2. 2

3 a (0) If all of the energy E is placed in subsyste two, what is the teperature of subsyste two? We know that the teperature is so in subsyste two, we have so the teerature is T = ds de, = ds 2 T 2 de 2 = d T 2 de (k B ln E = d T 2 de ( 2 k B ln E = d T 2 de ( k B 2E T = 2E k B b (5) The two subsystes are placed in theral contact. When they finally coe into equilibriu, what fraction of the total energy, E, is in subsyste one? We need to equate the two teperatures to find theral equilibriu. Thus ds de = ds 2 de 2 ds = d k B ln E 2 de de ds = d 2k B ln E de de Thus, we have E 2 = E E and ds = 2k B de E ds 2 = k B de 2 2E 2 2k B = k B E 2E 2 2 = E 2(E E ) E = 4 5 E 3

4 c (5) Assuing that the two subsystes are in theral equilibriu, what is the heat capacity of the coposite syste? The heat capacity of our syste is C = de dt which eans that we need to express the energy of our syste as a function of teperature, or alternatively, we siply add the heat capcacities of the two systes together. Fro above, we have so E = 2k B T E 2 = 2 k BT 2 and we get that C = 2k B C 2 = 2 k B C = C + C 2 C = 5 2 k B. 3 (40) Pulling a Spring a (5) In ters of, g, k, L and F, what is the translational kinetic energy of the two-block syste in the final configuration? The change in translational kinetic energy is given as The net force acting on the syste is K tran = F net r c. F net = F 3 g F net = (F 3g) ẑ The change in the center of ass is found by coputing the intial and final positions of the center of ass. r c (init) = r c (init) = L 3 ẑ 4 (2)(0) + ()(L) ẑ 3

5 Two blocks are connected by a assless spring which is initially unstretched. The upper block has a ass while the lower block has a ass of 2. The spring has a constant of k and an external force F = F ẑ is applied to the upper block and gravity acts on the syste in the ẑ direction. Initially, both blocks are at rest. The lower block is centered at z = 0, the upper block is centered at z = L and the spring is not stretched. This is shown in the left-hand sketch in the figure. z 3L F L k Soe tie later, the upper ass is centered at z = 3L and the lower ass is centered at z = L and the blocks are oving. This is sketeched in the right-hand picture. 0 2 Initial Final r c (final) = r c (final) = 5L 3 ẑ (2)(L) + ()(3L) ẑ 3 r c = ( 5L 3 L 3 ) ẑ r c = 4L 3 ẑ Thus, ( 5L K tran = (F 3g) 3 ). b (25) You are told that F = 6g and k = 4g/L. What is the vibrational kinetic energy, K vib in the final configuration? In order to analyze this, we need our energy principle which states that E sys = W If we consider the syste to be everything except the external force, then the work is W = F r W = (6g)(2L) W = 2gL. 5

6 The syste energy includes ters for kinetic energy, gravitational potential energy, and spring potential energy. We will look at the two potential energies first. For gravity, we have and for the spring we have Thus, Finally, we know that 4 (40) A Diatoic Molecule U g = (3g) r c U g = (3g)( 4L 3 ) U g = 4gL U s = 2 k L2 U s = 4g 2 L U s = 2gL. E sys = W K + U g + U s = 2gL L2 K = 2gL 4gL 2gL K = 6gL. K = K tran + K vib K vib = K K tran K vib = 6gL 5gL K vib = gl. You are given a saple of a diatoic gas olecule, F 2, where you know that the ass of an F ato is 8 proton asses, ( p = kg = ev/c 2 ). In the lab you have an apparatus that easures two different eission spectra fro the gas. Series I corresponds to rotation and Series II to vibration. The energies of the observed photons in these series are: k Series I: ev, ev, ev, ev and ev. Series II: ev, ev, ev, ev, and ev. 6

7 a (0) Sketch the vibrational energy levels of this olecule, and show sufficient allowed transitions on your diagra to account for all of the lines that you observe and correctly label your transition energies. b (0) Sketch the rotational energy levels of this olecule, and show sufficient allowed transitions on your diagra to account for all of the line that you observe and correctly label your transition energies. In the figure below, the left-hand sketch is the vibrational energies, while the right-hand sketch is the rotational energies n=5 n=4 n=3 n=2 n= n= l=3 l=2 l= l=0 c (0) Given the data in the proble, deterine what the effective spring constant, k. (Hint: when two identical asses are connected to opposite ends of the spring, the syste behaves like one ass connected to a spring of constant 2k. The natural frequency of the syste is ω = 2k/, so the vibrational energy is ev = hω. We can convert this to Joules by noting that there are J/eV. We also know that h = Js so we have J Js = s = 2k 2k The ass of a flourine ato is about 8 proton asses, so we get = kg. This eans that we have the spring constants given by k = 2 (.3 03 s ) 2 ( kg) k = 2.6kg/s 2 7

8 d (0) Using the observed vibrational transitions, estiate the teperature of this gas. The largest transition energy is eV. We want to copare this to k B T. Assuing that the transition occurs, then we need k B T to be on the order of % of the vibrational energy, or ev. For 300 K, k B T is about 0.025eV, so we need a nuber that is about 5 K. 5 (40) A olling Disk A disk of ass and radius can roll without slipping down a slope which is at an angle θ relative to the surface of the Earth. This is shown in the figure to the right. The disk has a oent of inertia relative to an axis through its center pointing into the page of I = 2 2 and gravity, g acts toward the botto of the page. g I = (/2) 2 θ a (0) Identify a useful coordinate syste and express all forces acting on the disk in ters of coponents in your chosen coordinate syste. (HINT: is is strongly advisable to choose one of your coordinate axes in the direction of p.) The forces acting on our disk as it rolls down the slope are gravity, a noral f F g N θ z 2 y x force, and friction. The noral force in perpendicular to the slope, while friction is pointing up the slope. We also know that the change in oentu, p, ust be down the slope. Thus, we choose ˆx to be pointing down the slope, ŷ to be noral to the slope and up, and ẑ is out of the page towards us. These are shown in the figure. We now see that N = Nŷ f = f ˆx F g = g ( sin θˆx cos θŷ) 8

9 We also note that since p is in the ˆx direction, that the net force in the ŷ direction is zero. Thus, and N = g cos θ F net = ( g sin θ f) ˆx b (0) In ters of, g, θ and the frictional force f, what is the rate of change of speed of the center of the disk as it oves down the slope (dv/)? In order to do this, we will first note that the disk rolls without slipping as it goes down the slope. This eans that the speed of the center of the disk, v, is equal to the angular velocity about the center of the disk, ω, ultiplied by the disk s radius. v = ω () We now need to look at our physics principles to attack this proble. We start with the oentu principle which states that d p = F net dv ˆx = ( g sin θ f) ˆx dv = g sin θ f (2) c (0) In ters of, g, θ and the frictional force f, what is the rate of change of the angular velocity about the center of the disk as it rolls down the slope (dω/)? Now, using equation, we can relate v to ω, so we look for a physics principle that says soething about ω. For this, let us consider angular oentu about the center of the disk. Our principle says that d L = τ net For the left-hand side, L = I ω, and for the disk to roll down the slope, L ust be in the ẑ direction. Thus we have d L = I dω ( ẑ). 9

10 The torque about the center can be coputed by looking at the three forces acting on the disk. Gravity acts on the center of the disk, so the torque due to gravity is zero. The noral force and the radius to where the noral force acts are parallel, so the torque due to the noral force is also zero. That leaves friction. The vector fro the center of the disk to where friction acts is r = ŷ, so we get τ f = ( ŷ) ( f ˆx) τ f = f( ẑ) This gives us our angular oentu equation as I dω ( ẑ) = f( ẑ) dω = f I dω = 2f (3) d (0) In ters of, g and θ, what is the force due to friction, f, that acts on the disk? Cobining equation 3 with equation, we get dv Now, equating equations 2 and 4, we get that = 2f. (4) g sin θ f = 2f 3f = g sin θ f = g sin θ 3 0