Two blocks, each of mass M are connected by an extensionless string of length L. 1 Block is placed?

Question

Two blocks, each of mass M, are connected by at extensionless 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 frictionelss (massless) pulley.

a. Find the Lagrangian

b. Determine y(t) (assume y(t=0)=0 and y'(t=0)=0)

Answer 1

The Lagrangian is the KE-PE,
or M*v^2-2*M*g*d
where v is the speed of either block and d is the displacement of either block.
The partial differential equation will be skipped. I will provide the classic Newtonian solution for b.

b) The forces on the hanging block are
F=M*a=M*g-T
where T is the tension in the string
The positive direction is movement away from the pulley

The sliding block is
M*a=T

where sliding in the direction of the pulley is positive
Substitute T into the first equation and solve for a

M*a=M*g-M*a
a=g/2

Starting from rest, setting up as positive, and a height h for the block when it is released:
y(t)=h-.25*g*t^2

j

Answer 2

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