- #1
Jenny Physics
- 111
- 4
- Homework Statement
- Derive the equation of motion of a chain whose supporting point moves vertically with motion Y(t).
Assume displacement of the chain in the vertical y direction is much smaller than the chain length.
Thus displacement of the chain happens only in the x direction.
- Relevant Equations
- u is the transverse displacement, g is gravity, L is the chain length and ##\rho## is the mass density per unit length of the chain. ##\hat{x},\hat{y}## are the unit vectors along the x and y directions.
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 y[g+\frac{d^{2}Y}{dt^{2}}]##
In the horizontal direction
##[T(y)-T(y-\Delta y)].\hat{x}=\rho\Delta y\frac{\partial^{2}u}{\partial t^{2}}##
Taking the limit ##\Delta y\rightarrow 0##
##\frac{\partial}{\partial y}[T(y).\hat{x}]\equiv \frac{\partial}{\partial y}[|T(y)|\sin\theta]\equiv \frac{\partial}{\partial y}[\rho y[g+\frac{d^{2}Y}{dt^{2}}]\tan\theta]=\rho y\frac{\partial^{2}u}{\partial t^{2}}##
since ##\tan\theta=\frac{\partial u}{\partial y}## the equation of motion is (assuming ##\rho=constant##)
##(g+\frac{d^{2}Y}{dt^{2}})\frac{\partial}{\partial y}[ y\frac{\partial u}{\partial y}]=y\frac{\partial^{2}u}{\partial t^{2}}##
Is this right? In particular should I include the acceleration of the supporting point moving vertically (##d^{2}Y/dt^{2}##)?
Usually the motion of the supporting point is used when we set the boundary condition at the top of the string, but here I am including it in the equation itself, so I suspect it is wrong?
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