# Physics 321 Hour 22 Central Force Problems

```Physics 321
Hour 22
Central Force Problems
Straight-line Motion in Spherical or Cylindrical
Coordinates Coordinates
y
r
r
x
a
t
We want to treat r like a Cartesian coordinate, but if we do, if
appears that there’s a force on the mass even when there is
no force!
Understanding the “Fictitious” Force
Think of r like x :
πΉ
0
a
The force acts in the +r direction.
r
Understanding the “Fictitious” Force
π₯=π
π¦ = π£π‘
→ π=
π2 + π£ 2 π‘ 2
2π£ 2 π‘ π£ 2 π‘
π=
=
2π
π
π£ 2 π£ 2π‘ π£ 2π‘ π£ 2
π£ 2π‘2
π=
− 2
=
1− 2
π
π π
π
π
π£ 2 π2 − π£ 2π‘ 2
π£ 2 π2
π£ 2 π2
=
=
= 3
2
2
π
π
π π
π
Understanding the “Fictitious” Force
ππ£ 2 π2
πΉ = ππ =
π3
π = ππ£π
π2
πΉ=
ππ 3
We can define a corresponding potential
energy:
π=−
πΉππ = −
π2
π2
ππ = +
3
ππ
2ππ 2
Before we even consider real forces, we have
to “add” a repulsive centrifugal force or
centrifugal potential to the problem:
π2
πΉ=
ππ 3
π2
π=+
2ππ 2
Centrifugal Potential
πππππ‘
β2
=+
2μπ 2
ππππ = πππππ‘ + ππ
ππ = −πΊ
ππ
π
Central Force
• Force is in the radial direction in spherical
coordinates
• The curl is always zero – even if the force isn’t
inverse square
• Source at the origin
πΌ
πΌπ
πΉ = π π = π+1
π
π
• Source at π1
πΌ(π2 − π1 )
πΉ21 =
|π2 − π1 |π+1
Central Force
• Source at π2
πΉ12
• Potential energy
πΌ(π1 − π2 )
=
|π2 − π1 |π+1
πΌ π−1
π π =
,
π−1
π2 − π1
π≠1
Inverse-square Force
• Source at π1
πΉ21
• Potential energy
πΌ(π2 − π1 )
=
|π2 − π1 |3
πΌ
π π =
π2 − π1
Example
moon_orbit.nb
```