AME451_HW_Solution_2

AME451_HW_Solution_2 - AME 451 Homework Solution 02 a. qi -...

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AME 451 Homework Solution 02 a. 12 io o dV qq q dt −− = Since VH A = , 1 o H q R = , 2 o Hh q r = Then i HHh qA H Rr = ± b. Assume HH > , therefore for tank 1, 3 o q is outflow; for tank 2, 3 o q is inflow. For tank 1: 11 113 oo dA H dt −−= 1 1 1 o H q R = , 3 3 o q R = Then, 112 1 13 HHH H RR = ± For tank 2: 22 223 dA H qq q dt −+= , 2 2 2 o H q R = Then, 212 2 23 H H −+ = ± For tank 3: 33 245 ooo dA H qqq dt , 3 4 4 o H q R = , 3 5 5 o H q R = Then, 2 H AH RRR ± c. Assume the radius of top surface of liquid is r From 2 hr r Hr r = , 2 rr rh r H =+ dV dt −= , o h q R = , h r h H r r r h H r r r r r r h V 2 2 2 2 1 2 3 2 2 1 2 2 2 2 ) ( 3 1 ) ( 3 1 π + + = + + =
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h r h h H r r r h h H r r R h q i ± ± ± 2 2 2 1 2 2 2 2 1 2 ) ( π + + = Another way to calculate the volume is as following: 0 () h VA h d h = , and 22 12 2 ( ) rr Ah r h r H ππ == + Then 3 2 2 2 00 1 ( ) ( ) 3 hh h d h h r d h hr h r h HH H −− + = + + ∫∫ d. Let 6 1 = R , 3 2 = R , 4 1 = L , 3 2 = L Then, 0 )) ( ) ( ( )) ( ) ( ( ) ( 2 1 2 3 1 1 1 1 = + + + t i t i R dt t i t i d L e t i R 0 )) ( ) ( ( )) ( ) ( ( 2 3 2 1 3 1 = + dt t i t i d L E dt t i t i d L 0 )) ( ) ( ( )) ( ) ( ( 3 2 2 1 2 2 = + + E dt t i t i d L t i t i R Therefore, e t i t i t i t i = + ) ( 3 ) ( 9 ) ( 4 ) ( 4 2 1 3 1 ± ± E t i t i t i = ) ( 3 ) ( 4 ) ( 7 2 1 3 ± ± ± E t i t i t i t i = + ) ( 3 ) ( 3 ) ( 3 ) ( 3 1 2 3 2 ± ± e. Assume when 0 ) ( , 0 = = t x θ , the springs are at the equilibrium point. ) sin ) ( ( 1 L t x k f s = sin 1 2 a k f s = Mg f g = 2 mg f g = 1 cos b c f c ± = For mass M: Since = ± ± I T 2 3 1 ML I = cos cos ) ( cos 2 cos cos 2 2 1 b f d t F L f a f L f T c g s s + + = ±
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This note was uploaded on 07/20/2009 for the course AME 451 taught by Professor Flashner during the Fall '07 term at USC.

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AME451_HW_Solution_2 - AME 451 Homework Solution 02 a. qi -...

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