- #1
Potatochip911
- 317
- 3
Homework Statement
A rope with uniform density ##\lambda=\frac{m}{L}## is placed on a frictionless table with an initial length ##y_0## hanging through the hole. Derive a differential equation for the position of the bottom of the rope and then using this solve for the time required for the rope to leave the table.
Homework Equations
##p=mv##
##F=\frac{dp}{dt}##
##p=\dot{m}v+m\dot{v}##
The Attempt at a Solution
I'm currently stuck with determining which differential equation I've derived is the correct one (although they only differ slightly).
The first method is just by analyzing the motion in the y direction. The acceleration will be given by ##ma=\lambda gy## where ##m## is the mass of the entire rope and ##y## (a function of time) is the length of the rope that's fallen through the hole. We can rewrite ##m## as ##\lambda L## from which we obtain: ##a=\frac{gy}{L}## or written more conveniently $$\ddot{y}-\frac{g}{L}y=0$$
The second method is the way my professor suggested to go about solving it. We consider the forces in both the ##x## and ##y## directions. To do this I considered the mass on the table ##m_T## and the mass falling off the table ##m_F## separately. Defining the ##x## and ##y## axis so the movement on the x and y-axis will be positive, in the x direction we have $$F_x=\frac{dp}{dt}=\dot{m_T}v+m_T\dot{v}\hspace{5mm} \mbox{and} \hspace{5mm} F_y=\dot{m_F}v+m_F\dot{v}$$ Then using the relation ##\dot{m_T}=-\dot{m_F}## we can express the first equation as ##F_x=-\dot{m_T}v+m_T\dot{v}##. Now the same force ##\lambda gy## will act on both of the sections therefore $$F_x+F_y=2\lambda gy=\dot{v}(m_F+m_T)$$ Now using ##m_F=y\lambda## and ##m_T=\lambda(L-y)## we get ##2\lambda gy=\dot{v}\lambda L## which after cancelling terms and changing to a more convenient form: $$\ddot{y}-\frac{2g}{L}y=0$$
I feel like something is definitely off about the second method particularly with how the force is acting on the masses but I just can't see what.