Mindscrape
Sep1-07, 05:49 PM
Two blocks, each of mass M, are connected by an extensionless, uniform string of length l. One block is placed on a smooth horizontal surface, and the other block hangs over the side, the string passing over a frictionless pulley. Describe the motion of the system when the string has a mass m.
By Hamilton's principle
L = T - U
the kinetic energies will be
T = 1/2 m \dot{x}^2 + 1/2 m \dot{y}^2
and if the potential is defined to be zero at the horizontal, the potential will be
U = -Mgy + U_{string}
This is the part I need a quick help on. The x block has a zero potential because it stays along the horizontal where the zero potential is defined, and the hanging block will have a potential of -Mgy, and I know that the mass of the string contributing to the potential will increase until finally it reaches as the string moves down. So I was thinking that
U_{string} = -\frac{m}{t}*g*y
That gives the mass per unit time for a given length y, which would also be
U_{string} = -m g \dot{y}
But units don't work out correctly unless I divide U_string by t, which would create a discontinuity and not make any sense. I don't know why I am having so much trouble with such a simple prospect.
By Hamilton's principle
L = T - U
the kinetic energies will be
T = 1/2 m \dot{x}^2 + 1/2 m \dot{y}^2
and if the potential is defined to be zero at the horizontal, the potential will be
U = -Mgy + U_{string}
This is the part I need a quick help on. The x block has a zero potential because it stays along the horizontal where the zero potential is defined, and the hanging block will have a potential of -Mgy, and I know that the mass of the string contributing to the potential will increase until finally it reaches as the string moves down. So I was thinking that
U_{string} = -\frac{m}{t}*g*y
That gives the mass per unit time for a given length y, which would also be
U_{string} = -m g \dot{y}
But units don't work out correctly unless I divide U_string by t, which would create a discontinuity and not make any sense. I don't know why I am having so much trouble with such a simple prospect.