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Homework 32 - homework 32 – BAUTISTA ALDO – Due 4:00 am...

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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 back-and-forth 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( ω t + φ ) (1) with the angular frequency ω equal to the angular velocity of the rotating crankshaft. For the problem at hand, ω = 4890 rpm = 512 . 08 rad / s . (2) The amplitude A of the piston’s motion is fixed by the engine’s design. To find out the amplitude for the piston at hand, consider the extreme ponts of the piston’s 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 piston’s velocity. Differ- entiating Eq. (1) WRT time, we find v ( t ) = d x dt =- A × ω × sin( ω t + φ ) , (6) which ranges between + A × ω and- A × ω . In magnitude, the maximal speed of the piston is | v | max = A × ω (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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Homework 32 - homework 32 – BAUTISTA ALDO – Due 4:00 am...

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