Recent content by so_gr_lo

$$ ds^2 = -(1-\frac{2M}{r'})(\frac{\frac{r'}{2M} -1+2M}{\frac{r'}{2M}-1})(dr')^2+(1-\frac{2M}{r'})^-1 (dr')^2 $$

data I'm trying to fit

thats the result for a cylinder about its symmetry axis, not sure if that’s the axis I’m supposed to use

Using the equation above I get Xcm = 0.022 m. I set the origin be at the left of the vertical rod parallel to its centre of mass as in the diagram. But I’m not sure if the equation is correct for 3d. for the moments of inertia I am using I = Icm + md^2 = (mr^2)/2 + md^2 where d is the...

Okay, but if I substitute that into the equations I get m1(d2x2/dt2) - d2x1/dt2) = sx and similar for m2, how does this help with combining them?

So x = x2-x1-l

I used between x = x2-x1

Is d2x/dt2 = d2x2/dt2 - d2x1/dt2 ?

Yes the masses are at the ends

Expresign extension in term of x1/2 gives x2-x1 = x Which could be substituted into each of the motion equatiosn but I'm not sure how that helps

tried writing the x position as x = Acos(wt) (ignoring the phase) so that d2x / dt2 = -w2x Substituting that into the individual motion equations would get the required result for the individual masses, but I am not sure how to combine the equations to get the reduced mass

The chain rule allows you to deal with composite functions, but since I don’t actually have the y and x components written in t explicitly maybe it’s not necessary. I think the flow lines need to equal F at each point since they are the tangent vector. In that case dy/dx = y’(t)/x’(t) then dy/dx...

I gave 2 different statements because I’m not sure if I am supposed to use the chain rule or not. The problem is that I don’t know how to turn the vector into a scalar so that I can write it as a differential.

Or that F = dx/dt + dy/dt

dy/dx = dy/dt x dt/dx = F2/F1 ?