Physics 225 - Dr. Shapiro
Final Exam, Spring 1998
Physics 225 - Dr. Shapiro
Problem #1) The block shown attached to the spring in the diagram below has a mass of 0.15 kg. The surface is frictionless. The block is pulled so that the spring is extended 10 cm from its equilibrium length. At t = 0 the block is released and undergoes simple harmonic motion. You measure the period of the motion, and find that it is 0.25s. a) What is the linear frequency of the motion? b) What is the angular frequency of the motion? c) What is spring constant? d) What is maximum acceleration of the block?
a) f = 1/T = 4 Hz b) omega = (2Pi)f = 25.1 rad/s
c) omega = sqrt(k/m) so k = m(omega)2 = 0.15 kg·(25.1 s-1)2 = 94.5 N/m
d) a = -(omega)2xmcos((omega)t + phi), and since the maximum value of cos = 1, the maximum value of the acceleration will be (omega)2xm = (25.1 s-1)2·0.1m = 63.0 m/s2
Problem #2) The wooden beam shown in the diagram is 3.0 m long and has a mass of 50 kg. It rests on two narrow supports. The support on the left is located 0.1 m in from the left end of the beam, while the support on the right is 0.25 m in from the right end of the beam. a) What is the total upward force exerted on the beam by the supports? b) What is the upward force exerted on the beam by the left support? c) What is the upward force exerted on the beam by the right support? d) What is the largest mass that can be hung from the right hand end of the beam, without the beam tipping?
a) The downward force is just the weight of the beam, mg so the upward force, Fl + Fr from the supports must equal this for the beam to be in equilibrium, so Fl + Fr = 50 kg·9.8 m/s2 = 490 N.
b) The weight of the beam acts at the center of the beam. Since the beam is in equilibrium, the torques must balance. Taking torques about the center of the beam, we have Fr(1.5 m - 0.25 m) = Fl(1.5 m - 0.1 m) or Fr·1.25 m = Fl·1.4 m, so Fr = Fl·1.12, thus 2.12Fl = 490 N and Fl = 231.1 N
c) Fr = 490 N - 231.1 N = 258.9 N
d) To find the largest mass that can be hung from the right hand end of the beam without the beam tipping, we take torques about the right hand support position. In this case the torque due to the added mass is just balanced by the torque from the weight of the beam. Thus, (490 N)(1.5 m - 0.25 m) = Mg·0.25 m, or 612.5 Nm/(2.45 m2/s2) = 250 kg
Problem #3) A child is playing with a large rubber playground ball. The ball has a radius of 0.4 m, and can be considered to be a thin shell of mass M so its rotational inertial is I = (2/3)MR2. She starts the ball rolling (without slipping) uphill on a section of the playground that makes a 5° angle with the horizontal. Assuming that the ball starts its roll with vCM = 1.0 m/s, how far will it go before it stops and starts rolling back downhill?
The ball will roll uphill until all of its initial kinetic energy has been converted into gravitational potential energy. It's initial kinetic energy consists of two parts; its translational kinetic energy (1/2)MvCM2 and its rotational kinetic energy (1/2)I(omega)2, where omega is the angular velocity about the CM. However, since the ball rolls without slipping the angular velocity about the contact point and about the CM are the same. Thus, R·omega = vCM, so (1/2)I(omega)2 = (1/2)(2/3)MR2(vCM/R)2 = (1/3)MvCM2. The total kinetic energy to start with is then M(1/2 +1/3)1.0 m2/s2, and this equals Mgh, where h is the vertical distance that the ball rises. So h = 0.8333/9.8 m = 0.0850 m. Then D·sin(5°) = 0.085 m and D = 0.975 m
Problem 4) Geostationary sattelites appear to remain stationary because they are in equatorial orbits and have the same rotational velocity as the earth. How high above the earth does a sattelite have to be in order to remain in a geostationary orbit? (You can assume that the earth is a perfect sphere, and that the sattelite’s orbit is circular. Also, G = 6.67 · 10-11 m3/kg·s2, Rearth = 6,370 km, and Mearth = 5.98 · 1024 kg.)
For the satellite to appear to be stationary above the equator, its angular velocity must be exactly the same as that of the earth, which is (2Pi)/86,400 rad/s. The gravitational force must provide the centripetal acceleration to keep satellite in orbit, so GmM/(R+h)2 = m(omega)2(R+h), where h is the height above the earth's surface. Thus, (R+h)3 = GM/(omega)2 and R+h = ((6.67·10-11m3/kg·s2)(5.98·1024 kg)/5.29·10-9 s-2)1/3 = 4.22·107 m. Since R = 0.637·107 m. h = 3.59·107 m = 3.59·104 km
Problem 5) The railroad boxcar on the right is moving to the right at 0.5 m/s, while the boxcar on the left is moving to the right at 0.75 m/s. When the faster boxcar hits the slower one, the automatic couplers cause them to join together. The boxcar on the left has a mass of 20,000 kg, while the one on the right has a mass of 30,000 kg. a) After the cars join together, what is their velocity? b) Compare the kinetic energy of the cars before and after the collision. What happened to the kinetic energy that was "lost"? c) Suppose that it takes about 0.05 s for the cars to join together. What is the average force on one of the cars during this process?
a) From conservation of momentum we have m1v1 + m2v2 = (m1 + m2)vf, so 20,000 kg·0.75 m/s + 30,000 kg·0.5 m/s = (50,000 kg)vf so vf = (30,000 kg·m/s)/50,000 kg = 0.6 m/s
b) The K.E. before the collision is (1/2)m1v12 + (1/2)m2v22 = 5,625 J + 3,750 J = 9,375 J. After the collision, the K.E. is (1/2)(m1 + m2)vf2 = 9,000 J. So energy has been "lost" in the collision (it is a completely inelastic collision). Actually, this energy was converted into acoustic, thermal, and vibrational energy during the "collision".
c) The impulse-momentum theorem can be used to find the average force on one of the cars during the collision. J = Fave(delta-t) = pf - pI so if we look at the lighter car we find that Fave = 20,000 kg(0.6 m/s - 0.75 m/s)/0.05s = -60,000 N
Problem 6) Bundles of newspapers are being unloaded from a truck by sliding them down a ramp one at a time. Each newspaper bundle has a mass of 25 kg. The truck bed is 1.5 m above the ground. The ramp makes a 45° angle with the ground, and the coefficient of kinetic friction between the ramp and a newspaper bundle is 0.10. At the top of the ramp a worker gives each bundle a push so that it starts down the ramp with a speed of 0.35 m/s. a) How fast is the bundle going when it gets to the bottom of the ramp? b) How much work has been done on the bundle by the gravitational force? c) How much work has been done on the bundle by the frictional force?
a) This problem can be solved either by direct application of Newton's second law or by energy methods. Using the Newton's law approach the downhill force is mg·sin(45°), while the uphill force due to friction is mg·cos(45°)·0.10. Thus the net downhill force is 25 kg·0.707·9.8 m/s2·0.9 and the net downhill acceleration is 0.707·9.8 m/s2·0.9 = 6.24 m/s2. The time required to slide downhill can be found from 1.5 m/sin(45°) = v0t + (1/2)at2, which gives 2.12 m = 0.35(m/s)t + 3.12(m/s2)t2. This yields t = 0.770 s, and then v = v0 + at = 0.35 m/s + 6.24(m/s2)0.770 s = 5.15 m/s
Using energy methods, the final K.E. is (1/2)mv02 + mgh - Ffd, where Ffd is the work done by the frictional force. (1/2)mv02 = 1.531 J, mgh = 367.5 J, and since d = 2.12 m and Ff = 0.1·9.8(m/s2)cos(45°)25 kg = 17.3 N the work done by friction is 36.7 J. So the final K.E. is 332.3 J = (1/2)mvf2. This yields 5.16 m/s for vf.
b) mgh = 367.5 J
c) Ffd = 36.7 J
Problem 7) You have two objects, a hoop and a solid cylinder. Both have exactly the same mass and radius. If you roll both the hoop and the cylinder down the same incline, which gets to the bottom first? Explain your answer.
The solid cylinder will get to the bottom first. The solid cylinder has the smaller rotational inertia. Thus, more of its energy is translational, and the object with the highest CM velocity (and thus the highest translational energy) will reach the bottom first.
Problem 8) A grandfather clock (basically a pendulum) is placed in an elevator. When the elevator is stationary, the period of the clock pendulum is measured to be T.
If the elevator is moving upward at constant velocity, will the period of the clock pendulum be (a) larger than T, (b) smaller than T, (c) the same as T? Explain your answer.
The correct answer is (c) the same as T. As long as the pendulum is not accelerating, the force on the pendulum bob will still be mg downward. Thus, the period will remain = 2Pi·sqrt(L/g).
Now if the elevator is accelerating upward is the period of the clock pendulum (a) larger than T, (b) smaller than T, or (c) the same as T? Again explain your answer.
In this case the net downward force on the pendulum bob is m(g-a), where a is the acceleration upward. Thus the period becomes 2Pi·sqrt(L/(g-a)), which is larger than T (answer (a)). This means that the clock slows down while it accelerates upward.
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