Forest UCM Energy Line1D

From New IAC Wiki
Jump to navigation Jump to search

The equation of motion for a system restricted to 1-D is readily solved from conservation of energy when the force is conservative.

T+U(x)= cosntant E
T=EU(x)
12m˙x2=EU(x)
˙x=±2(EU(x))m
±m2(EU(x))dx=dt=tti=t

The ambiguity in the sign of the above relation, due to the square root operation, is easily resolved in one dimension by inspection and more difficult to resolve in 3-D.

The velocity can change direction (signs) during the motion. In such cases it is best to separte the inegral into a part for one direction of the velocity and a second integral for the case of a negative velocity.


Free fall

Consider a rock dropped at t=0 from a tower of height h.

The potential energy stored in the rock at any instant is given by

U(x)=mgx

Note
The potential is highest at x=0 and becomes negative as x increases

The initial total energy is

Etot=T+U=00=0
t=±m2(EU(x))dx
=±m2(0(mgx))dx
=±12gxdx=±(2gx)12dx
=±(2g)122x=2xg


or

x=12gt2

spring example (problem 2.8)

Consider the problem of a mass attached to a spring in 1-D.

F=kx

The potential is given by

U(x)=F(x)dx=12kx2
t=±m2(EU(x))dx=dt
=m2xx0(EU(x))12dx
=m2xx0(E12kx2)12dx
=m2Exx0(1(xk2E)2)12dx

let

sinθ=xk2E and ω=km
cosθdθ=dxk2E

then

t=m2Exx0(1sin2θ)12dx
=m2Exx0dxcosθ
=m2Exx0cosθdθcosθ2Ek
=mkxx0dθ
=1ωθθ0dθ
θ=ωt+θ0
sinθ=sin(ωt+θ0)
2Emsinθ=2Emsin(ωt+θ0)
x=2Emsin(ωt+θ0)
x=Asin(ωt+θ0)
A=2Em = amplitude of oscillating motion
U(x)=12kx2=12kA2sin2(ωt+θ0)
E=T+U(x)=12kA2


Forest_UCM_Energy#Energy_for_Linear_1-D_systems