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# crittenden_HW_example - 211S10 Written Homework#7(1 A stick...

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Unformatted text preview: 211S10 Written Homework #7 (1) A stick of length L has non-uniform density r @ x D = r x + m x 2 , where x = 0 is at one end of the stick. Find the center of mass. Unit Test data to check your analysis: r L m r sys 1 2 3 22 15 3 1 2 9 13 2 3 1 17 8 2 2 3 13 9 By definition, (1 r sys = 1 m sys ‡ r @ m D „ m In this case the mass of the stick is different at every point, so we have to do some work even to find m sys . Since the total mass is the sum of all the little bits of mass at each distance from the end, (2 m sys = ‡ „ m = ‡ r „ x and we’re given the density as a function of position so (3 m sys = ‡ L I r x + m x 2 M „ x = 1 2 r x 2 + 1 3 m x 3 F L = 1 2 r L 2 + 1 3 m L 3 Now we only need to evaluate (4 ‡ r @ m D „ m and we do the same thing as for the mass, we first replace the little bit of mass with the little bit of position times the density at that position and then we replace the general abstract ‘position at little bit of mass m ’ with the specific thing it is here, namely, the position x , (5 ‡ r @ m D r „ x = ‡ x r „ x But r is not constant, so we have to replace that with the given function and the integration limits are the end points of the stick, so we need to evaluate (6 ‡ L x I r x + m x 2 M „ x = 1 3 r L 3 + 1 4 m L 4 Thus (7 x sys = 1 3 r L 3 + 1 4 m L 4 1 2 r L 2 + 1 3 m L 3 I see little point in doing any algebra on this. Asking mathematica for the answer ans = ‡ L x I r x + m x 2 M „ x ì ‡ L I r x + m x 2 M „ x 1 4 L 4 m + 1 3 L 3 r 1 3 L 3 m + 1 2 L 2 r which is clearly the same thing. (2) A block of mass m 2 is at rest attached to an initially uncompressed spring, spring constant k , and sits on a rough surface, friction coefficient m k . Another block, mass m 1 , is slid along the surface and strikes the first block with speed at impact v i ....
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crittenden_HW_example - 211S10 Written Homework#7(1 A stick...

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