Recent content by carrotstien

[cool un-allowed theory was here]

Agreed...I was thinking about the case that my professor told me about. Something that looked like light going from some source through A, and then destructively interfering at point B - and I remember him telling me that in such a case, light would never even leave point A. But I may be...

and by matter u mean the mirrors? i guess...i mean, theoretically you can use a gravity lens to a similar effect. But still, does the resultant propagating wave contain energy or not?

nevermind...i figured it out the moment of inertia with respect to the arbitrary point is just mr^2..without the added moment of interia about the center of mass - because the object isn't rotating! this forum isn't as active as i thought it'd be

Actually, DaleSpam...this isn't always true. What you speak of is true only in oppositely moving waves. However, when they move in the same direction, a flipped E-field results in a flipped B-field, so that 180-degree shifted waves traveling in the same direction would cancel each other out...

from wikipedia for "The most general equation for rotation of a rigid body in three dimensions about an arbitrary origin O with axes x, y, z is"... i thought that the sum of the torques on a system is just equal to d/dt(L)=d/dt(Iw)...apparently not - but i can't understand the notation...

nope..checked it out.. the torque due to gravity is the same if you treat it moving through the center, or sum the torques on every element of the structure all about the same point. \alpha for each element depends on the shape, as that would change it's position (such as the length of a...

after thinking about it...the only issue i can find is this perhaps the torque due to gravity isn't just the force (mg) applied at the center of mass, crossed with the \vec{r} but even if this is the case, and even if the torque due to gravity was some function of R...it doesn't seem...

point is...the \alpha should just be derived from a coordinate transformation (cart->polar)...the cartesian position function has absolutely nothing to do with the shape of the object..only the mass. \alpha should be the same

oopsie..right 'm' stays...the 'm' leaves when the torque is due to gravity which would be mg*r*sin(angle) ... for the case where the disc is being accelerated by gravity... \alpha = \frac{(mg)(r)cos(\theta) - 2mr\dot{r}\omega}{\frac{mR^2}{2}+mr^2} where \theta is the angle that \vec{r}...

Ok, so... \tau = \frac{dL}{dt} and L\equiv I\omega so \tau = \dot{I}\omega + I\dot{\omega} = \dot{I}\omega + I\alpha and i don't believe it'd be wrong to write (shouldn't be...right?) \alpha = \frac{\tau-\dot{I}\omega}{I} --- the situation I'm looking at is that of a disc (of radius 'R')...

yea, thanks, Nabeshin and all...i get it i just never knew that L was defined in terms of Iw the way you showed it. Now everything makes sense again.

yea, thanks, Nabeshin and all...i get it i just never knew that L was defined in terms of Iw the way you showed it. Now everything makes sense again.

I always thought that torque = I*d²θ/dt² so that, if there is any d²θ/dt² on an object with a moment of inertia (both with respect to the same point)..then there must be a torque applied. However, I've found a case where this isn't true. So, I'm assuming there is more to it then that...

well, I found the solution for the case where the string isn't wrapped around at all, and the point on the disc that the string is attached too is moving only perpendicularly to the string. (if it isn't moving perpendicularly, then the disc will start moving in such a way as to give the string...