Hi Tiny-Tim,
I do use the cross section for the mean free path, d.
I guess you are not in the United States since we are the only country that still uses a.m. and p.m. when the rest of the world works under 24-hour clock. Don't know why this country doesn't change, but what I mean by 1:00...
Homework Statement
Electrical breakdown (e.g, lightning) is caused by an avalanche process. If a free electron gains enough energy between collisions to ionize the neutral molecules when they collide with them, then those two electrons will gain enough energy between collisions to ionize the...
Anyone? Is the question confusing or do you not know what I am asking? My main concern is if there does exist a Coriolis Force and why would there be one.
Homework Statement
Figure is attached.
In an amusement park ride called the hammer a large beam rotates in a ver- tical plane about a central pivot (see figure). Cages are attached to the end of the beam; these rotate independently. The riders are strapped to the walls of the cages. The...
Wesley,
There was a mistake in my derivation. What you have is right. I got the same result. But I don't understand what you said in your last post about the derivatives.
What I did was:
\ddot{\theta} = \tan{\theta} (\dot{\theta})^2 - \frac{g}{a} \tan{\theta}
I then used the hint and...
It came with a lot of thinking until I found my mistake with the help of Mathematica.
So: T = \frac{1}{2} m a^2 \frac{d\theta}{dt}^2 (\cos{\theta})^2
U = \frac{1}{2} m g a (\sin{\theta})^2
Plugging this into the lagrangian and doing the derivation I get the final equation of motion...
Hi Wesley,
I did the very exact thing that you said. The part that I am stuck is on simplifying it nicely to find this equation of motion and the substitution.
@ gabba
Wesley is right. This is only for a two dimensional surface. There is no z involved.
Homework Statement
A particle is free to slide along a smooth cycloidal trough whose surface is given by the parametric equations:
x = \frac{a}{4}(2 \theta + \sin{2 \theta})
y = \frac{a}{4}(1 - \cos{\theta})
where 0 <= \theta <= \pi and a is a constant.
(sorry, TeX is not working for...
Thank you fzero for the help & dr_sasha I will still present this question in discussion since you requested, even though this problem is now officially solved for me.
It is not that difficult. I think the main leap in this problem is conceptually figuring out what is going on and...
True. But we should end up to the same result. I will try doing it your way as well. But if we know that F_x = F sin \theta and x = r sin \theta
Then
m \ddot{x} = F_x = F sin \theta.
with F = -G M m_p /r^2
and M = 4/3 \pi r^3 \rho
we get:
m \ddot{x} = - G 4/3 \pi \rho r sin...
I mean the way I did it just assumed r and x would change which will also is related to the angle you mentioned. I guess they are equivalent, just different way of writing it.
I was thinking of that route as well. But would you not have to also know how \theta will change as the particle follows its path? In your equation, it seems that both F will change due to r and \theta will change as well.
Also, should it not be cosine instead of sine? I mean in part (a) we had...