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--- classes:2009:fall:phys4001.100:home [2009/09/19 09:38] – hennessy
+++ classes:2009:fall:phys4001.100:home [2009/10/18 17:55] (current) – x500_brehm046
@@ Line 1: / Line 1: @@
-====== Ask a Question ======
 ==== Solution to coupled differential equations (September 18 - Lecture)  ====
@@ Line 48: / Line 47: @@
 ===== Dan K =====
 The wiki works!
 -Dan K
+Okay, so I have a little question regarding 2-11. I've finished the problem but I'm not sure if one of my steps is valid (and some minus signs I need to iron out, but that's no big deal). I'm pretty new to Latex so I'll see how this goes (gotta learn some time anyway).
+I ended up with this differential question:\\
+<math>
+\frac{dv}{dt} = -g - kv^2\\
+</math>
+I found it useful to take the reciprocal of both sides to:\\
+<math>
+\frac{dt}{dv} = \frac{1}{-g - kv^2}\\
+</math>
+The question is, can I DO this? With a few more steps I can get a solution, but it seems to be a rather cheesy way to manipulate the differentials and I'm not sure if it holds water logically. It also seems to assume not only that the inverse exists, but also imposes something on its derivative which may not be correct.
+===== Hemisphere Problem From Lecture =====
+We need to determine when the normal force goes to zero
+since we are asked when the block leaves the surface after sliding down a smooth (frictionless) surface. The block starts at the top of the hemisphere on which it sits.
+The normal force is always radial and the component of gravity in the radial direction is <math>mgcos\theta</math>. The sum of these two forces produces the centripetal acceleration.
+Therefore we have:\\
+(1)<math>\frac{mv^2}{a}=mgcos\theta-N</math>\\
+There is no friction. Gravity is conservative. And the normal force does no work since it acts perpendicular to the displacement of the block. Therefore we can use energy conservation to eliminate <math>mv^2</math> from our force equation. Energy conservation gives:\\
+(2)<math>E_{initial}=mga=\frac{mv^2}{2}+mgacos\theta</math>\\
+Solving (1) for <math>mv^2</math> and substituting into (2) and solving for N yields (with a little algebra):\\
+<math>N=mg(3cos\theta-2)=mg(3\frac{z}{a}-2)</math>\\
+Which is 0 when <math>z=\frac{2}{3}a</math>\\
+=====General Question=====
+Is there any homework for Oct. 20th. I don't see it posted that is why I am asking.
+I'm a different student curious about this too.  Also, when will solutions for the HW 5 be posted?

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