(a) The potential energy in Cartesian coordinates is V (y(x(s))) = fr²² = fr² gpy(x(s)) ds where ds = |dx (s)² + dy (x (s))² ⇒ ds = dy (x (s))² 1 + dx(s) dx (s)² SO x2 V(y(x)) = gp ´y(x) √y'(x)² + 1 dx (b) Since the length of the chain is constant 1 = = S²² · L² √y wx)² + 1 dx ds= the potential energy, with the constraint included, is V(y(x), y'(x), α)=gp L (a + y(x)) √y'(x)² + 1 dx √x1 where the constant / was ignored. The potential energy can be rewritten as V(y(x), y'(x), a)=gp • fr³² L(x(x), y'(x), a) dx Hanging chain Consider the shape of a hanging rope as shown. (a) Express the potential energy of the rope as the function y(x) inte- grated between x₁ and x2. Here x; are the horizontal Cartesian coordi- nates of the suspension points, and y(x) is the vertical Cartesian coordi nate of the chain at x. (b) Find the length of the chain as an function of y(x) integrated between x1 and x2. Since the hanging chain minimizes the potential, interpret the potential energy as an 'action' to be minimise, add the constraint expressing the fixed length of the chain using a Lagrangian multiplier a, and find the corresponding 'Lagrangian'. d- I I I -Xo x=0 d-xo λ
(a) The potential energy in Cartesian coordinates is V (y(x(s))) = fr²² = fr² gpy(x(s)) ds where ds = |dx (s)² + dy (x (s))² ⇒ ds = dy (x (s))² 1 + dx(s) dx (s)² SO x2 V(y(x)) = gp ´y(x) √y'(x)² + 1 dx (b) Since the length of the chain is constant 1 = = S²² · L² √y wx)² + 1 dx ds= the potential energy, with the constraint included, is V(y(x), y'(x), α)=gp L (a + y(x)) √y'(x)² + 1 dx √x1 where the constant / was ignored. The potential energy can be rewritten as V(y(x), y'(x), a)=gp • fr³² L(x(x), y'(x), a) dx Hanging chain Consider the shape of a hanging rope as shown. (a) Express the potential energy of the rope as the function y(x) inte- grated between x₁ and x2. Here x; are the horizontal Cartesian coordi- nates of the suspension points, and y(x) is the vertical Cartesian coordi nate of the chain at x. (b) Find the length of the chain as an function of y(x) integrated between x1 and x2. Since the hanging chain minimizes the potential, interpret the potential energy as an 'action' to be minimise, add the constraint expressing the fixed length of the chain using a Lagrangian multiplier a, and find the corresponding 'Lagrangian'. d- I I I -Xo x=0 d-xo λ
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How is the highlighted section the lagrangian? what about the kinetic energy?
Transcribed Image Text:(a) The potential energy in Cartesian coordinates is
V (y(x(s))) = fr²²
= fr²
gpy(x(s)) ds
where
ds = |dx (s)² + dy (x (s))² ⇒ ds =
dy (x (s))²
1 +
dx(s)
dx (s)²
SO
x2
V(y(x)) = gp ´y(x) √y'(x)² + 1 dx
(b) Since the length of the chain is constant
1 =
= S²²
· L² √y wx)² + 1 dx
ds=
the potential energy, with the constraint included, is
V(y(x), y'(x), α)=gp
L
(a + y(x)) √y'(x)² + 1 dx
√x1
where the constant / was ignored. The potential energy can be rewritten as
V(y(x), y'(x), a)=gp • fr³² L(x(x), y'(x), a) dx
Transcribed Image Text:Hanging chain
Consider the shape of a hanging rope as shown.
(a) Express the potential energy of the rope as the function y(x) inte-
grated between x₁ and x2. Here x; are the horizontal Cartesian coordi-
nates of the suspension points, and y(x) is the vertical Cartesian coordi
nate of the chain at x.
(b) Find the length of the chain as an function of y(x) integrated
between x1 and x2. Since the hanging chain minimizes the potential,
interpret the potential energy as an 'action' to be minimise, add the
constraint expressing the fixed length of the chain using a Lagrangian
multiplier a, and find the corresponding 'Lagrangian'.
d-
I
I
I
-Xo
x=0
d-xo
λ
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