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Physics 325 Spring 2011 Problem Session 4 Solutions

# Physics 325 Spring 2011 Problem Session 4 Solutions - m f m...

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Physics 325 Spring 2011 Discussion 4 February 14, 2011 Problem 1 Suppose a rocket can eject fuel to the back (to accelerate) or to the front (to decelerate) at a constant rate dm/dt = - α . Assume there is no other force, so the equation of motion is m dv dt = uα, where u = u 0 when it is accelerating, and u = - u 0 when it is decelerating. The initial mass of the rocket is m i , and its mass when it is out of fuel is m f . Starting from rest, the rocket accelerates until its mass is m 1 . Afterwards, it decelerates until all of its fuel is used up. What is m 1 expressed in terms of m i and m f if the rocket stops moving when all of the fuel is gone? Problem 2 A force field has the form (in cylindrical coordinates): ~ F = - k r ˆ φ, where k is a constant. (a) Calculate H ~ F · d ~ around a unit circle centered at the origin. (b) Based on the previous result, is the force field conservative ? Why or why not? (c) Let W = k arctan y x . Calculate - ~ W . Solutions : Problem 1 When it is accelerating, we have m dv dt = u 0 α = - u 0 dm dt Z v 1 0 dv = - u 0 Z m 1 m i dm m v 1 = u 0 ln m i m 1 . 1

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When it is decelerating, its mass changes from m 1 to m f , while its velocity decreases from v 1 to 0 .
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Unformatted text preview: m f m 1 dm m-v 1 = u ln m f m 1 0 = u ln m i m 1 + u ln m f m 1 0 = u ln m i m f m 2 1 m i m f m 2 1 = 1 m 1 = √ m i m f Problem 2 (a) We have d ~ ‘ = 1 · dφ ˆ φ , and φ goes from to 2 π . I ~ F ± ± ± r =1 · d ~ ‘ =-Z 2 π k 1 ˆ φ · ˆ φdφ =-2 πk 6 = 0 . (b) A conservative force has zero line integral over any closed path. Thus, ~ F is not conservative, since there is a closed path for which the line integral of ~ F is not zero. (c) Recall that d ds arctan s = 1 1 + s 2 .-∂W ∂x =-k 1 + y 2 x 2 · ²-y x 2 ³ = k y r 2-∂W ∂y =-k 1 + y 2 x 2 · ² 1 x ³ = k x r 2-~ ∇ W = k r 2 ( y ˆ ı-x ˆ ) = k r (sin φ ˆ ı-cos φ ˆ ) =-k r ˆ φ. Although it seems that-~ ∇ W = ~ F , the potential W is undeﬁned at the origin. In fact, we cannot deﬁne a potential that is well-deﬁned everywhere , and this has to do with the force being non-conservative. 2...
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