Where are we? phys16/lectures/sl02.pdf Where are we? Forces of the form F(v) Example: F(v) =...

Click here to load reader

  • date post

    24-Feb-2021
  • Category

    Documents

  • view

    0
  • download

    0

Embed Size (px)

Transcript of Where are we? phys16/lectures/sl02.pdf Where are we? Forces of the form F(v) Example: F(v) =...

  • Where are we?

    Forces of the form F (v)

    Example: F (v) = −m Γ v Another example: F (v) = −mβ v2 Forces of the form F (x)

    Review of the harmonic oscillator

    Linearity and Time Translation Invariance

    Back to F (v) = −m Γ v

  • Forces of the form F (v)

    Formal solution of F = ma

    F (v) = ma

    F (v) = m dv

    dt

    dt = dv m

    F (v)

    t− t0 = ∫ t t0

    dt′ = ∫ v(t) v(t0)

    dv′ m

    F (v′) solve this for v(t) as a function of t

  • Forces of the form F (v)

    Formal solution of F = ma

    F (v) = ma

    F (v) = m dv

    dt

    dt = dv m

    F (v)

    t− t0 = ∫ t t0

    dt′ = ∫ v(t) v(t0)

    dv′ m

    F (v′) solve this for v(t) as a function of t

  • Forces of the form F (v)

    Formal solution of F = ma

    F (v) = ma

    F (v) = m dv

    dt

    dt = dv m

    F (v)

    t− t0 = ∫ t t0

    dt′ = ∫ v(t) v(t0)

    dv′ m

    F (v′) solve this for v(t) as a function of t

  • Forces of the form F (v)

    Formal solution of F = ma

    F (v) = ma

    F (v) = m dv

    dt

    dt = dv m

    F (v)

    t− t0 = ∫ t t0

    dt′ = ∫ v(t) v(t0)

    dv′ m

    F (v′) solve this for v(t) as a function of t

  • Forces of the form F (v)

    Formal solution of F = ma

    F (v) = ma

    F (v) = m dv

    dt

    dt = dv m

    F (v)

    t− t0 = ∫ t t0

    dt′ = ∫ v(t) v(t0)

    dv′ m

    F (v′) solve this for v(t) as a function of t

  • Forces of the form F (v)

    Formal solution of F = ma

    F (v) = ma

    F (v) = m dv

    dt

    dt = dv m

    F (v)

    t− t0 = ∫ t t0

    dt′ = ∫ v(t) v(t0)

    dv′ m

    F (v′) solve this for v(t) as a function of t

  • Forces of the form F (v)

    Formal solution of F = ma

    F (v) = ma

    F (v) = m dv

    dt

    dt = dv m

    F (v)

    t− t0 = ∫ t t0

    dt′ = ∫ v(t) v(t0)

    dv′ m

    F (v′) solve this for v(t) as a function of t

  • Example: F (v) = −β v = −m Γ v Frictional force appropriate for small velocities in viscous fluids.

  • ma = m dv

    dt = F (v) = −β v = −m Γ v

    dt = − dv Γ v

    Γ dt = −dv v

    ∫ t t0

    Γ dt′ = − ∫ v(t) v(t0)

    dv′ 1

    v′

    Γ (t− t0) = − ln v(t) + ln v(t0) eΓ (t−t0) = v(t0)/v(t)

    v(t) = v(t0) e −Γ (t−t0)

  • ma = m dv

    dt = F (v) = −β v = −m Γ v

    dt = − dv Γ v

    Γ dt = −dv v

    ∫ t t0

    Γ dt′ = − ∫ v(t) v(t0)

    dv′ 1

    v′

    Γ (t− t0) = − ln v(t) + ln v(t0) eΓ (t−t0) = v(t0)/v(t)

    v(t) = v(t0) e −Γ (t−t0)

  • ma = m dv

    dt = F (v) = −β v = −m Γ v

    dt = − dv Γ v

    Γ dt = −dv v

    ∫ t t0

    Γ dt′ = − ∫ v(t) v(t0)

    dv′ 1

    v′

    Γ (t− t0) = − ln v(t) + ln v(t0) eΓ (t−t0) = v(t0)/v(t)

    v(t) = v(t0) e −Γ (t−t0)

  • ma = m dv

    dt = F (v) = −β v = −m Γ v

    dt = − dv Γ v

    Γ dt = −dv v

    ∫ t t0

    Γ dt′ = − ∫ v(t) v(t0)

    dv′ 1

    v′

    Γ (t− t0) = − ln v(t) + ln v(t0) eΓ (t−t0) = v(t0)/v(t)

    v(t) = v(t0) e −Γ (t−t0)

  • ma = m dv

    dt = F (v) = −β v = −m Γ v

    dt = − dv Γ v

    Γ dt = −dv v

    ∫ t t0

    Γ dt′ = − ∫ v(t) v(t0)

    dv′ 1

    v′

    Γ (t− t0) = − ln v(t) + ln v(t0) eΓ (t−t0) = v(t0)/v(t)

    v(t) = v(t0) e −Γ (t−t0)

  • ma = m dv

    dt = F (v) = −β v = −m Γ v

    dt = − dv Γ v

    Γ dt = −dv v

    ∫ t t0

    Γ dt′ = − ∫ v(t) v(t0)

    dv′ 1

    v′

    Γ (t− t0) = − ln v(t) + ln v(t0) eΓ (t−t0) = v(t0)/v(t)

    v(t) = v(t0) e −Γ (t−t0)

  • ma = m dv

    dt = F (v) = −β v = −m Γ v

    dt = − dv Γ v

    Γ dt = −dv v

    ∫ t t0

    Γ dt′ = − ∫ v(t) v(t0)

    dv′ 1

    v′

    Γ (t− t0) = − ln v(t) + ln v(t0) eΓ (t−t0) = v(t0)/v(t)

    v(t) = v(t0) e −Γ (t−t0)

  • 1 2 3 4 5

    0.2

    0.4

    0.6

    0.8

    1

    ↑ v(t) v(t0)

    Γ (t−t0) →

  • x(t) = x(t0) + ∫ t t0

    dt′ ẋ(t′)

    x(t) = x(t0) + ∫ t t0

    dt′ v(t′)

    = x(t0) + ∫ t t0

    dt′ v(t0) e−Γ (t ′−t0)

    u = e−Γ (t ′−t0) du = −dt′ Γ e−Γ (t′−t0)

    x(t) = x(t0)− v(t0) Γ

    ∫ e−Γ (t−t0)

    1 du

    x(t) = x(t0) + v(t0)

    Γ

    ( 1− e−Γ (t−t0))

  • x(t) = x(t0) + ∫ t t0

    dt′ ẋ(t′)

    x(t) = x(t0) + ∫ t t0

    dt′ v(t′)

    = x(t0) + ∫ t t0

    dt′ v(t0) e−Γ (t ′−t0)

    u = e−Γ (t ′−t0) du = −dt′ Γ e−Γ (t′−t0)

    x(t) = x(t0)− v(t0) Γ

    ∫ e−Γ (t−t0)

    1 du

    x(t) = x(t0) + v(t0)

    Γ

    ( 1− e−Γ (t−t0))

  • 1 2 3 4 5

    0.2

    0.4

    0.6

    0.8

    1

    ↑ x(t)−x(t0)

    v(t0)/Γ

    Γ (t−t0) →

  • More general but still very simple case

    F (v) = F0 −m Γ v

  • More general but still very simple case

    F (v) = F0 −m Γ v

    v(t) = v(t0) e −Γ (t−t0) +

    F0 m Γ

    ( 1− e−Γ (t−t0)

    )

  • More general but still very simple case

    F (v) = F0 −m Γ v

    v(t) = v(t0) e −Γ (t−t0) +

    F0 m Γ

    ( 1− e−Γ (t−t0)

    )

    v(∞) = F0 m Γ

    independent of v0

    Arisotelian physics if Γ is large - v ∝ F0

  • 1 2 3 4 5

    0.5

    1

    1.5

    2

    2.5

    3

    ↑ v(t)

    F0/mΓ

    Γ (t−t0) →

    v0 > F0 mΓ

    v0 < F0 mΓ

  • Another example: F (v) = −mβ v2 Frictional force for rapid movement through a thin medium like a gas — force arises because the body is bumping into molecules and knocking them out of the way. The force from each collision is ∝ v, and the # per unit time is also ∝ v

  • ma = m dv

    dt = F (v) = −mβ v2

    β dt = −dv v2

    ∫ t t0

    β dt′ = − ∫ v(t) v(t0)

    dv′

    v′ 2

    1

    v(t0) + β (t− t0) = 1

    v(t) − 1

    v(t0) − 1

    v(t0)

    v(t) = v(t0)

    1 + β v(t0) (t− t0)

  • ma = m dv

    dt = F (v) = −mβ v2

    β dt = −dv v2

    ∫ t t0

    β dt′ = − ∫ v(t) v(t0)

    dv′

    v′ 2

    1

    v(t0) + β (t− t0) = 1

    v(t) − 1

    v(t0) − 1

    v(t0)

    v(t) = v(t0)

    1 + β v(t0) (t− t0)

  • ma = m dv

    dt = F (v) = −mβ v2

    β dt = −dv v2

    ∫ t t0

    β dt′ = − ∫ v(t) v(t0)

    dv′

    v′ 2

    1

    v(t0) + β (t− t0) = 1

    v(t) − 1

    v(t0) − 1

    v(t0)

    v(t) = v(t0)

    1 + β v(t0) (t− t0)

  • ma = m dv

    dt = F (v) = −mβ v2

    β dt = −dv v2

    ∫ t t0

    β dt′ = − ∫ v(t) v(t0)

    dv′

    v′ 2

    β (t− t0) = 1 v(t)

    − 1 v(t0)

    v(t) = v(t0)

    1 + β v(t0) (t− t0)

  • ma = m dv

    dt = F (v) = −mβ v2

    β dt = −dv v2

    ∫ t t0

    β dt′ = − ∫ v(t) v(t0)

    dv′

    v′ 2

    β (t− t0) = 1 v(t)

    − 1 v(t0)

    1

    v(t0) + β (t− t0) = 1

    v(t)

  • ma = m dv

    dt = F (v) = −mβ v2

    β dt = −dv v2

    ∫ t t0

    β dt′ = − ∫ v(t) v(t0)

    dv′

    v′ 2

    1 + v(t0)β (t− t0) v(t0)

    = 1

    v(t)

    1

    v(t0) + β (t− t0) = 1

    v(t)

  • ma = m dv

    dt = F (v) = −mβ v2

    β dt = −dv v2

    ∫ t t0

    β dt′ = − ∫ v(t) v(t0)

    dv′

    v′ 2

    1 + v(t0)β (t− t0) v(t0)

    = 1

    v(t)

    v(t) = v(t0)

    1 + β v(t0) (t− t0)

  • F (v) = −β v = −m Γ v ⇒ v(t) = v(t0) e−Γ (t−t0)

    F (v) = −mβ v2

    ⇒ v(t) = v(t0) 1 + β v(t0) (t− t0)

  • F x(t)

    F0 ⇒ x(t0) + v(t0) (t− t0) + F0 2m

    (t− t0)2

    −m Γ v ⇒ x(t0) + v(t0) Γ

    ( 1− e−Γ (t−t0)

    )

    −mβ v2 ⇒ x(t0) + 1 β

    log ( 1 + β v(t0) (t− t0)

    )

  • Forces of the form F (x)

    Lots of examples (if you don’t look too closely) - gravity - the electric force. Many situations in which there is v dependence, but it is small and can be ignored.

  • m dv

    dt = F (x)

    m dv

    dt v = F (x)

    dx

    dt 1

    2 m

    d

    dt v2 = F (x)

    dx

    dt 1

    2 m

    ∫ t t0

    dt′ d

    dt′ v(t′)2 =

    ∫ t t0

    dt′ F (x(t′)) dx