Physics 321
Hour 7
Energy Conservation – Potential Energy in
One Dimension
Bottom Line
• Energy is conserved.
• Kinetic energy is a definite concept.
• If we can determine the kinetic energy at all
points in space by knowing it at one point
in space, we can invent a potential energy
so that energy can be conserved.
• Kinetic energy is related to work.
• Potential energy must also be related to
work.
Kinetic Energy
1
𝑇 = 𝑚𝑣 2
2
1
𝑃 = 𝑇 = 𝑚2𝑣 𝑣 = 𝐹𝑣
2
Does that make sense?
Is it sometimes true?
Is it always true?
Work-Energy Theorem
1
𝑝∙𝑝
𝑇 = 𝑚𝑣 ∙ 𝑣 =
2
2𝑚
𝑝∙𝑝 𝑝∙𝑝 𝑝∙𝑝
𝑃=𝑇=
+
=
=𝐹∙𝑣
2𝑚
2𝑚
𝑚
This is a useful relation – but we’ll go one step
further:
𝐹 ∙ ∆𝑟
∆𝑇 =
∆𝑡 → 𝑑𝑇 = 𝐹 ∙ 𝑑 𝑟
∆𝑡
2
∆𝑇|21 =
𝐹 ∙ 𝑑 𝑟 = 𝑊1→2
1
• Positive work increases kinetic energy, negative
work decreases kinetic energy
Gravity
Depending on initial velocity,
an object can move freely
under the influence of gravity
in many different paths from
1 to 2.
y
1
2
In each case:
2
𝑊=
2
𝐹 ∙ 𝑑𝑟 =
1
2
𝐹𝑦 𝑑𝑦 = −𝑚𝑔
1
= −𝑚𝑔 𝑦2 − 𝑦1 = −𝑚𝑔∆𝑦
And on any closed path 𝑊 = 𝐹 ∙ 𝑑 𝑟=0
𝑑𝑦
1
x
Gravity
y
1
This means that ∆𝑇 from
1 to 2 is independent of
path. If we know 𝑇1 we
also know 𝑇2 .
2
x
Potential Energy
If ∆T is dependent only
on the end points of a
path (not on the path or
on time), we can define
a potential energy.
Otherwise, we can not.
𝑊=
y
1
𝐹 ∙ 𝑑 𝑟 = −𝑚𝑔∆𝑦 = −∆𝑈
2
x
Stokes’ Theorem I
P1
y
𝐹 ∙ 𝑑𝑟 =
𝑃1
𝐹 ∙ 𝑑𝑟 +
𝑃2
𝐹 ∙ 𝑑𝑟
𝑃3
P3
P2
Note the path integrals
on the mid-line cancel.
𝐹 ∙ 𝑑𝑟 =
𝑃1
x
𝐹 ∙ 𝑑𝑟
𝑖
𝑃𝑖
Pi
P1
Stokes’ Theorem II
y
𝐹 ∙ 𝑑𝑟 =
𝐹 ∙ 𝑑𝑟
𝑃1
=
𝑖
=
1
∆𝐴𝑖
𝐹 ∙ 𝑑 𝑟 ∆𝐴𝑖
𝑃𝑖
1
lim
∆𝐴→0 ∆𝐴
𝐹 ∙ 𝑑𝑟 =
𝑃1
𝑃𝑖
𝑖
𝐹 ∙ 𝑑𝑟 𝑑𝐴
𝑃
𝑐𝑢𝑟𝑙 𝐹 ∙ 𝑑 𝐴
Pi
P1
x
Potential Energy
If the curl of 𝐹 is zero and 𝐹 has no explicit time
dependence, we can define a potential energy.
Otherwise, we can not.
In general, if 𝑈 = − 𝐹 ∙ 𝑑 𝑟, 𝐹 = −𝛻𝑈.
Since 𝛻 × (𝛻𝑈) ≡ 0 the curl of 𝐹 must be zero.
Differential Form of the Curl
𝑐𝑢𝑟𝑙 𝐹 𝑧 =
𝐹𝑦 𝑥 + ∆𝑥, 𝑦 − 𝐹𝑦 𝑥, 𝑦 ∆𝑦
−
∆𝑥∆𝑦
𝐹𝑥 𝑥, 𝑦 + ∆𝑦 − 𝐹𝑥 𝑥, 𝑦 ∆𝑥
∆𝑥∆𝑦
𝜕𝐹𝑦 𝜕𝐹𝑥
=
−
𝜕𝑥
𝜕𝑦
𝑥
𝛻 × 𝐴 = 𝜕/𝜕𝑥
𝐴𝑥
𝑦
𝜕/𝜕𝑦
𝐴𝑦
𝑧
𝜕/𝜕𝑧
𝐴𝑧
y+Δy
y
x
x+Δx
Spherical Coordinates
It’s important to go back and forth between
spherical and Cartesian coordinates. Know these:
𝑥 = 𝑟 sin 𝜗 cos 𝜑
𝑦 = 𝑟 sin 𝜗 sin 𝜑
𝑧 = 𝑟 cos 𝜗
𝑟2 = 𝑥2 + 𝑦2 + 𝑧2
𝑦
tan 𝜑 =
𝑥
𝑥2 + 𝑦2
tan 𝜗 =
𝑧
Cylindrical Coordinates
It’s also important to go back and forth between
cylindrical and Cartesian coordinates. Know
these:
𝑥 = 𝑟 cos 𝜗
𝑦 = 𝑟 sin 𝜗
𝑧=𝑧
𝑟2 = 𝑥2 + 𝑦2
𝑦
tan 𝜑 =
𝑥
Vector Calculus in Mathematica
vec_calc_ex.nb