==== Solution to coupled differential equations (September 18 - Lecture)  ====

After setting up Newton's Law for the Lorentz Force on a particle with 
arbitrary velocity in a B-field directed along the z-axis, we found
the coupled differential equations (below). We got as far as 
the y equation of motion but just wrote down the x solution.
The complete solution is below. 


We started with the differential equations:\\
<math>
(1)\ddot{y}=-\omega\dot{x} \;\;\; (2)\ddot{x}=\omega\dot{y} \;\;\; </math> where <math> \omega=\frac{qB}{m}
</math>\\

Integrating with respect to time once gives:\\
<math>
(3)\dot{y}=-\omega x +C_1 \;\;\; (4)\dot{x}=\omega y + C_2
</math>\\

Substitute eq. 4 into eq. 1: which yields\\
<math>
\ddot{y}=-\omega^2 y - C_2 
</math>\\

Solution is of the form (check by direct substitution):\\
<math>
(5)y=-Acos(\omega t+\phi)+y_0 \;\;\; y_0 = -C_2/\omega
</math>\\

Take the time derivative of eq. 5 and substitute into eq. 3. This shows you why the amplitude, A,  and phase, <math>\phi</math>, must be the same in both solutions and why you get the sin function.\\
The time derivative of eq. 5 is:\\
<math>
(6)\dot{y}=A \omega sin(\omega t+\phi) \\
</math>\\

Substitute into 3:\\
<math>
A \omega sin(\omega t+\phi) = -\omega x +C_1
</math>\\

Now just solve for x:\\
<math>
x= -Asin(\omega t+\phi)+x_0\\
x_0=C_1/\omega
</math>

===== 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?