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Unformatted text preview: homework 32 BAUTISTA, ALDO Due: Apr 19 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. Consider a car engine running at constant speed. That is, the crankshaft of the en gine rotates at constant angular velocity while each piston moves backandforth in its cylin der according to the rules of simple harmonic motion. 4890 rpm 10 . 1 cm Suppose the two extremal positions x max and x min of a piston are 10 . 1 cm from each other. When the crankshaft of the engine rotates at 4890 rpm (revolutions per minute), what is the maximal speed  v  max of the piston? Correct answer: 25 . 86 m / s (tolerance 1 %). Explanation: Although a piston in a car engine is not by itself a harmonic oscillator, the crank con necting it to the rotating crankshaft imposes the simple harmonic motion on the piston x ( t ) = x middle + A cos( t + ) (1) with the angular frequency equal to the angular velocity of the rotating crankshaft. For the problem at hand, = 4890 rpm = 512 . 08 rad / s . (2) The amplitude A of the pistons motion is fixed by the engines design. To find out the amplitude for the piston at hand, consider the extreme ponts of the pistons motion: Ac cording to Eq. (1), x max = x middle + A, x min = x middle A, (3) hence the distance between the extremes x max x min = 2 A. (4) For the problem at hand, x max x min = 10 . 1 cm , hence A = 10 . 1 cm 2 = 5 . 05 cm . (5) Now consider the pistons velocity. Differ entiating Eq. (1) WRT time, we find v ( t ) = d x dt = A sin( t + ) , (6) which ranges between + A and A . In magnitude, the maximal speed of the piston is  v  max = A (7) = (5 . 05 cm) (512 . 08 rad / s) = 25 . 86 m / s . Question 2 Part 1 of 1. 10 points. Two identical stars, a fixed distance D apart, revolve in a circle about their mutual center of mass, as shown below. Each star has mass M and speed v . G is the universal gravitational constant. D v v M M Which of the following is a correct relation ship among these quantities? 1. v 2 = G M D 2 2. v 2 = 2 G M D homework 32 BAUTISTA, ALDO Due: Apr 19 2006, 4:00 am 2 3. v 2 = 4 G M 2 D 4. v 2 = 2 G M 2 D 5. v 2 = G M 2 D correct 6. v 2 = M G D 7. v 2 = G M D 8. v 2 = 4 G M D Explanation: From circular orbital movement, the cen tripetal acceleration is a = v 2 D 2 ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
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