Recent content by Jenny Physics

Typically when we have a wave equation like $$u_{tt}=[g-\ddot{Y}](xu_x)_x$$ we solve it by imposing a boundary condition at the support that would be in this case ##u(L,t)=Y(t)##. So ##Y(t)## appears in the equation (as the double time derivative) and in the boundary condition?

Yes it is ill-defined. I believe the intention of small vertical displacements means ##Y(t)## is small as well (as in a small oscillation of the supporting point say).

So ##g(t)=g-\ddot{Y}##? (thinking of non inertial force). Or is it ##g(t)=g+\ddot{Y}##?

Since at rest the chain elements all lie vertically, shouldnt it be instead if a vertical section length ##a## is tilted at angle ##d\theta## with the vertical then its vertical extent reduces to ##a\cos(d\theta)## and so its vertical size reduces by ##a-a\cos (d\theta)=a(1-\cos...

I don't understand the ##a(1-\cos d\theta)## shouldn't it be ##a\cos d\theta\approx a[1-(d\theta)^{2}/2]##? As far as the vertical equation, I believe you are saying because both elements at x and ##x+dx## experiment the same vertical acceleration of the moving support it should be...

Ceiling: Yes it should be a point support, the drawing is a bit misleading. Displacement meaning as far as the transverse displacement of the chain. The chain as a whole does move vertically but its elements relative displacement is negligible vertically (thats how I understand it). Thinking...

In the figure assume the "ceiling" moves with motion ##Y(t)##, i.e. it is a point support. Applying Newton's law in the vertical direction ##T(y).\hat{y}=\rho y[g+\frac{d^{2}Y}{dt^{2}}]## If ##\theta## is the angle between ##T## and ##\hat{y}## that means ##|T|\cos\theta=\rho...

It doesn't depend on the length of the lower rod only because of the symmetry. But what if the pendulum were not attached right at the middle of the lower rod? How could I derive the equations without using torques?

You are right, this was a misunderstanding. I edited the question.

I am not sure which other forces I should consider besides those 3. I cannot consider tensions due to the massless rod on the masses since those will not add up to zero.

##\phi+\theta+\alpha=\pi/2## that would mean that ##\alpha=\pi/2-\phi-\theta## and hence ##AB=a\cos\alpha=a\sin(\phi+\theta)##?

The length of ##OB=L\sin\phi##. The length of AB requires some trigonometry involving ##\theta## which is not obvious to me.

It should then be ##r_{left}=[(L-a\cos\theta)\sin\phi,(L-a\cos\theta)\cos\phi]##? That will not work when ##\theta=\pi/2##

Just trying to find the angular momentum using the definition.

It is a pendulum swinging back and forth and so both ##\phi,\theta## change with time.