Physics 18 Spring 2010 Midterm 2 Solutions

Transcription

1 Physics 18 Spring 010 Midterm s For midterm, you may use one sheet of notes with whatever you want to put on it, front and back. Please sit every or seat, and please don t cheat! If something isn t clear, please ask. You may use calculators. All problems are weighted equally. PLEASE BOX YOUR FINAL ANSWERS! You have full length of class. If you attach any additional scratch work, n make sure that your name is on every sheet of your work. Good luck! 1. A hollow cylinder (with moment of inertia, I = MR, where M is mass, and R is radius) of radius 10 cm and mass 5.0 kg rolls without slipping at a constant speed of 8.0 cm/sec. (a) What is its angular momentum? (b) What is its rotational kinetic energy? (c) What is its total kinetic energy? (a) The angular momentum is L = Iω, where ω is angular velocity. Since cylinder is rolling without slipping we know that v = ωr, so ω = v/r. Thus, L = Iω = Iv R = MvR = = 0.04 kg m /s. (b) The rotational kinetic energy is KE rot = 1 Iω. Again, plugging in ω = v/r and I = MR gives KE rot = 1 Iω = 1 I v R = 1 Mv = 1 5 (0.08) = J. (c) The total kinetic energy is made up of rotation piece that we found in part (b), and also translational piece, KE trans = 1 Mv. But, we saw in part (b) that rotational kinetic energy was precisely equal to 1 Mv, meaning that translational and rotational kinetic energies are equal. So, total kinetic energy is just twice value found in part (b), KE total = J. 1

2 . A 300 gram bird flying along at 6.0 m/s sees a 10 g insect heading straight toward it with a speed of 30 m/s. The bird opens its mouth wide and enjoys a nice lunch. What is bird s speed immediately after swallowing? Model: We will define our system to be bird + bug. This is case of an inelastic collision because bird and bug move toger after collision. Horizontal momentum is conserved because re are no external forces acting on system during collision in impulse approximation. Visualize: Let s draw bird and insect. Solve: The conservation of momentum equation pf x = pix is ( m1+ m Because ) vf x = m 1( v insect ix) + m( v gets i ) 1 x eaten ( 300 by g + 10 g bird, ) v ( )( ) ( )( ) f x = this 300 is g an 6.0 inelastic m/s + collision. 10 g 30 We m/s can determine final speed of bird/insect system by applying law of conservation of vf x = 4.8 m/s Assess: We momentum. left masses The in grams, initial rar momentum than convert is just to kilograms, initial because momentum mass of units bird, cancel minus out from both sides of equation. initial Note momentum that ( v of ix) is negative. insect (minus, since insect is initially heading to left). So, p i = p i bird p i insect = m 1 v i1 m v i. Now, after bird eats insect, y are a single system of total mass m t = m 1 +m, and is moving at a speed v f. So, final momentum is p f = (m 1 + m ) v f. Conservation of momentum says that p i = p f. Solving for v f gives Plugging in numbers gives v f = m 1v i1 m v i m 1 + m. v f = m 1v i1 m v i m 1 + m = 300(6.0) 10(30) 310 = 4.8 m/s.

3 3. While our Sun doesn t have enough mass to ever become a black hole, suppose that it could. (a) What would be Schwarzschild radius of Sun? (b) Using Kepler s law, determine orbital period in years of Earth around this new black hole. Hint: for this problem you might find following information useful: G N = N m /kg, M Sun = kg, M Earth = kg, c = m/s, and orbital radius of Earth is meters. (a) The Schwarzschild radius of Sun is given by Plugging in values gives R S = G NM Sun c. R S = G NM Sun c = ( ) = 970 meters, or about 3 kilometers. (b) Kepler s law relates period of orbit to its radius, r, by T = 4π G N M Sun r 3. Since mass of Sun didn t change upon becoming a black hole, orbital period is exactly same as it always was, namely one year. We can check this just by plugging in numbers 4π T = 4π r G N M 3 = ( ) 3 Sun = seconds, which is a year! 3

4 4. Many people have imagined that if y were to float top of a flexible snorkel tube out of water, y would be able to brea through it while walking underwater. However, y generally do not take into account just how much water pressure opposes expansion of chest and inflation of lungs. Suppose you can just brea while lying on floor with a 400 N (90 lb) weight on your chest. (a) What is total pressure on your chest from this weight (ignoring atmospheric pressure), assuming your chest has a frontal area of m? (b) How far below surface of water could your chest be for you to still be able to brea? (Note: you don t need to include atmospheric pressure since we re asking about additional pressure due to water.) (a) If you can only support 400 N of force over an area of m, n pressure associated with this force is P = F/A = 400/0.090 = 4444 Pa. (b) The total pressure at a depth h is just P (h) = P 0 + ρgh, where P 0 = 1 atm is atmospheric pressure, and ρ is density of water. So, pressure due only to water (above atmospheric pressure) is just P water = ρgh. So, for a given pressure, depth beneath water is h = P ρg. So, depth corresponding to pressure found in part (a) is h = P ρg = = 45 cm! So, you could only go down 45 centimeters before you couldn t brea in and out! 4

5 Extra Credit Question!! The following is worth 10 extra credit points! Newton s law of gravity tells us that gravitational acceleration in orbit is not zero. The shuttle and astronauts are falling toger around Earth, and so astronauts appear to be weightless. Explain why shuttle astronauts suffer from adverse biological effects such as muscle atrophy even though y are not actually weightless. Hint: consider what leads to muscle atrophy, and wher that effect is present in orbit. The weight that we feel is due to normal force. The astronauts in orbit are in constant free-fall, with no normal force acting on m, and so y don t feel ir weight. So, without compensating normal force to fight against, muscles begin to weaken, not needing to work so hard or do as much anymore. Over time this can lead to muscle atrophy, among or problems. 5