Recent content by radagast_

The asterisk usually refers to the optimal solution (it can also indicate a complex conjugate, but this is not the case :) ). "min" refers to the minimum of..., ie "Give me the minimal value of this function". "arg" tells you "You calculated the minimum, right? so don't give me the minimal...

Thanks.

Hello. I am watching this video: https://www.youtube.com/watch?v=xIQtnQ9XPbE and he says there: I see that the wave turns into a "normal" sinus in the end, but it's not constant. I think, that even on an unmatched transmission line, the waves should go back and forth, but eventually...

Thanks! I did it with energies and it worked perfectly (through one line :) ).

Hello! http://img151.imageshack.us/img151/6571/cques1vd5.gif Homework Statement A particle with mass m is thrown in lateral speed V_0 inside a hollow half-ball with radius R. At the beginning of it's motion the ball has an angle of \theta_0 from the perpendicular. The gravitational force will...

Thanks tiny-tim and Oerg!

ahh. so F=2mv. For some reason I got confused thinking the momentum difference is integral of F and not F itself... ~(_8^[I] D'oh! Thus 2I\omega=F(h-r) => 2*\frac{mR^2}{2}*\frac{v}{R}=2mv(h-r) Thus h=\frac{3R}{2} correct?

well, I have F=ma, and 2mv=integral{f*dt}. from both I can't get anything... because the acceleration is 0. what am I missing?

Oreg, I don't understand why I need to separate the nail force, instead of separating the r vector to the center of mass, in which I'll get N=(r-h)*F. Conservation of angular momentum is what I started with. I still remain with a missing ratio - F/Fmu ... I don't know where to get more information.

I don't understand - the cylinder is spinning - so there's a torque which generates \omega, right? how can you say there isn't (im not talking about force, but torque). So you're saying I should do something like this: 2I\omega=F(h-r). I need also to use \omega=\frac{v}{R}. but then what can I...

That's my problem: I get from that this equation: the nail's torque needs to equal both directions' friction torque: (h-R)F=2F\mu*R, by the angular momentum conservation. from this I get \frac{F}{F\mu}=\frac{2R}{h-R} and I remain with a ratio of the forces I don't know how to get.

Do you mean I need to break the F generated from the nail to perpendicular and tangential, and relate to the tangential one? and then I need to do (Ft is the tangential F) F_t*R=2F\mu*R? About your second line - do you mean the linear impulse or angular?

Homework Statement http://img503.imageshack.us/img503/3535/11uv8.gif a cylinder with a given radius R rolls to the right without slipping (\omega= \frac{v}{R}). It hits a nail fixed to the pillar. find h, the height of the nail above the floor, in which the cylinder will roll back without...

Well, the object isn't moving perpendicular to the force: the normal is at an angle (not 90) relative to the x axis, and the object's moving on the x axis. the momentum equation is: (a=alpha, b=beta, ub=ballspeed after, ut=tablespeed after) m*v0*cos(a) = m*ub*cos(b-a) + M*ut*cos(a)

ok. The forces on the block are mg to -y^, and on the moment of the impact also normal perpendicular to the slope, right?