Solution. will lead to a positive torque, while the bigger cat will give a negative torque. So,

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1 Physics 18 Fall 009 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 50 kg cat and a 0 kg bowl of tuna fish are at opposite ends of a 40 m long seesaw How far to left of pivot must a 40 kg cat stand to keep seesaw balanced? 134 Model: The see-saw is a rigid body The cats and bowl are particles Visualize: This is just a statics problem since none of components of system is moving The bowl and smaller cat will tend to twist seesaw one way, while heavier cat will twist it or way Thus, seesaw is subject to three torques If system is to balance, n sum of torques must be zero, τ i = τ net = 0 So, Solve: we just need The see-saw to add up is in rotational differentequilibrium torques, noting Calculate that bowl net torque and smaller about cat pivot point will lead to a positive torque, while bigger cat will give a negative torque So, τ net = 0 = ( FG ) ( 0 m) ( FG ) ( d) ( FG ) ( 0 m) 1 B τ net = τ bowl + τ small mgd = mg 1 ( cat τ 0 m) big cat = 0 mg B ( 0 m) Now, in general, τ = rf sin φ The force each component is just its mass times gravity ( m1 mb )( 0 m) ( 50 kg 0 kg)( 0 m) Since gravity is always pointing straight d = down, and since = seesaw is horizontal, = 15 m angle of each torque is 90 Thus, we have m 40 kg Assess: τ net The = smaller cat is τ bowl close + but τ small not cat all τ big cat way to end = by 0 bowl, which makes s combined mass = of m bowl rsmaller bowl g + cat m small and cat bowl r small of cat tuna g mis big greater cat r big than cat g = mass 0 of larger cat So, we just need to figure out r small cat, since all rest are known Solving for this gives r small cat = m big catr big cat m bowl r bowl 5() () = = 15 m m small cat 4 So, small cat is pretty close to bowl, but still a half a meter away 1

2 A 300 gram bird flying along at 60 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? 917 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 60 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 = 48 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(60) 10(30) 310 = 48 m/s

3 3 The Earth orbits sun at an average distance of about 150 million kilometers (it s a bit closer in winter, and a bit furr in summer), and (as you know) orbits it once a year Recalling that Netwon s gravitational constant is G N = N m /kg, determine mass of sun, assuming that Earth orbits sun in a perfect circle The gravitational force between Earth and Sun is F G = G m EM S r, where m E is mass of Earth, M S is mass of sun, and r is distance between m Now, since Earth is going around sun in a circle net force on Earth is just centripetal force, F cent = m E v /r If orbit is stable, n two forces are equal, G m EM S = m Ev r r Solving for solar mass gives M S = v r G Now, we know orbital period, T, which is one year Now, for an object moving around in a circle, n v = πr/t, and so M S = (π) r 3 GT Now, re are roughly π 10 7 seconds in a year (which you can easily check) So, we have everything that we need Plugging in numbers gives M S = (π) r 3 GT = (π) ( ) 3 ( )(π 10 7 ) kg The actual mass is kgs So we re very close! 3

4 4 A 500 gram, 80 cm-diameter, 15 cm long hollow can rolls without slipping across floor at 10 m/s If moment of inertia of a ring is I = MR, n what is can s kinetic energy? Since it s rotating, total kinetic energy of can is made up of two pieces, translational and rotational kinetic energies So, KE total = KE trans + KE rot Now, KE trans = 1 Mv, while KE rot = 1 Iω, where ω is angular velocity So, KE total = 1 Mv + 1 Iω Now, because can is rolling without slipping, v = Rω, or ω = v/r Plugging this into expression for kinetic energy gives KE total = 1 Mv + 1 I R v = 1 ( Mv 1 + I ) MR Now, since moment of inertia is I = MR, n KE total = 1 ( ) Mv 1 + MR = Mv MR This is total kinetic energy Plugging in numbers gives KE total = 05(1) = 1 Joule 4

5 Extra Credit Question!! The following is worth 10 extra credit points! When you turn on faucet a stream of water comes out as seen in figure The stream becomes narrower, tapering off, as it falls (a) Explain why this is (b) Bonus - 5 MORE extra credit points, for a total of 15! See if you can derive an equation for radius of stream as a function of height Hint: assume that stream is circular in cross-section (a) As water in stream falls it speeds up due to gravity Because it s speeding up, n diameter of stream changes This is a consequence of continuity equation, v 1 A 1 = v A, where v 1 and A 1 are velocity and area of stream at one position, and v and A are same quantities at a different point Solving for area at point gives A = v 1 v A If velocity at point is higher than that at point 1, n this means that area of point has to be smalller than that at point 1, to keep equation true (b) The continuity equation says that A = v 1 v A 1, where v 1 is speed at end of faucet, and A 1 is area of steam at faucet Now, since stream is circular A = πr, and A 1 = πr 1 So, r = v1 v r 1 Now, recall that, for constant acceleration, v = v0 gy If stream starts with a velocity v 1 and drops a distance h, n v = v1 + gh = v gh Plugging this into radius equation gives, after canceling off factor of v 1, v 1 r (h) = ( r gh v 1 ) 1/4 This is radius, as a function of fall distance, h Notice that it decreases as h goes up, as we should expect 5

Physics 18 Spring 2010 Midterm 2 Solutions

Physics 18 Spring 2010 Midterm 2 Solutions 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,