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classical. Special Cases of the 2 nd Law: “Recipes”. 1d discussion for now. Easily generalized to 3d. Newton’s 2 nd Law equation for a particle has some equivalent forms ( F = total external force ) which might be useful in different cases: F = ma = m(dv/dt) = m(d 2 x/dt 2 ) (1)

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classical

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  1. classical

  2. Special Cases of the 2nd Law: “Recipes” • 1d discussion for now. Easily generalized to 3d. • Newton’s 2nd Law equation for a particle has some equivalent forms (F =total external force) which might be useful in different cases: F = ma = m(dv/dt) = m(d2x/dt2) (1) Also, from the chain rule: dv/dt = (dv/dx)(dx/dt) = v(dv/dx)  F = mv(dv/dx) (2) • In general, F = F(x,v,t) • Goal: Given F, & initial conditions, find v(t), x(t)  F = ma can be a horrendous differential equation! • It is easier in cases where F is a function of x, v, t separately.

  3. 2nd Law F = ma = m(dv/dt) = m(d2x/dt2) = mv(dv/dx) • General comments about these equivalent forms: F = m(d2x/dt2):2 time integrations to get x(t) F = m(dv/dt):1 time integration to get v(t) F = mv(dv/dx):Useful if F = F(v) or F=F(x)

  4. F = ma = m(dv/dt) = m(d2x/dt2) = mv(dv/dx) • Consider in 4 special cases: 1. F = constant 2. F = F(t), a function of time only 3. F = F(v), a function of velocity only 4. F = F(x), a function of position only • We’ve already seen examples of 1., 2., & 3. Will look at 4. in detail soon. Its helpful to (briefly) summarize general procedures in the 4 cases.

  5. Case 1 F = Constant  a = (d2x/dt2) = F/m = constant • Obviously, get the familiar, 1d kinematics equations for constant acceleration (Physics I!). • Given, initial conditions: t = 0, x = x0, v = v0 • Get: v = v0 + at x = x0 + v0t + ½at2 v2 = (v0)2 + 2a(x-x0)

  6. Case 2 F = F(t) Time dependent forces  m(dv/dt) = F(t) (1) • Initial conditions: t = 0, x = x0, v = v0 • Integrate directly. General solution(integrals: limits on t are 0  t, limits on v are v0 v, limits on x are x0 x): (1)  dv = (F(t)/m) dt ∫dv =∫(F(t)/m)dt  v(t) = v0 + ∫(F(t)/m)dt = (dx/dt) (2) (2) dx = v0dt + [∫(F(t)/m)dt]dt  x(t) = x0 + v0t + ∫[∫(F(t)/m) dt]dt (3)

  7. Case 3 F = F(v) Velocity dependent forces (like retarding forces). • Method 1: Direct Integration  m(dv/dt) = F(v) (4) • Initial conditions: t = 0, x = x0, v = v0 • General solution(integrals: limits on t are 0  t, limits on v are v0 v, limits on x are x0 x): (4)  dt = [m/F(v)]dv ∫dt = t = t(v) = m∫[1/F(v)] dv (5)(Gives t(v)) If possible, algebraically invertt(v) in (5) to get: v(t) = (dx/dt) (6) Then, integrate (6) to get x(t): (6)  dx = v(t)dt ∫dx = ∫v(t)dt  x(t) = x0 + ∫v(t)dt (7)

  8. Case 3 F = F(v) Velocity dependent forces (like retarding forces). • Method 2: m(dv/dt) = F(v) (4) But, m(dv/dt) = mv(dv/dx).  (4) becomes: mv(dv/dx) = F(v) (8) • Initial conditions: t = 0, x = x0, v = v0 • General solution(integrals: limits on t are 0  t, limits on v are v0 v, limits on x are x0 x): Get v(t) from (4) as before (previous page). Then, Integrate (8) to get x(v): (8)  dx = [(mv)/F(v)]dv ∫dx = m∫[v/F(v)] dv x(v) = x0 + m∫[v/F(v)]dv (9) Finally, substitute v(t) into x(v): x[v(t)]  x(t) (10)

  9. Case 4 F = F(x): Position dependent forces.  m(dv/dt) = F(x) = m(d2x/dt2) (11) • Initial conditions: t = 0, x = x0, v = v0 • (11) is a 2nd order differential equation to solve for x(t). Then compute v(t) = (dx/dt). Use standard methods of differential equations. We’ll discuss this in special cases as the course proceeds. • Alternatively (& preferably!) use energy methods. Introduce a potential energy functionU(x) such that F  - (dU/dx). Analyze the motion using this function, as will be discussed in detail in the energy discussion in the next section.

  10. F = F(x): Outline using energy. More details next section. • (11)  m(dv/dt) = F(x) • Initial conditions: t = 0, x = x0, v = v0 • Integrate: (limits on v:v0 v, limits on x:x0 x): mv dv = F(x)dx  m∫v dv = ∫F(x)dx  (½)mv2 – (½)m(v0)2 = ∫F(x)dx (12) • Introduce a potential energy functionU(x) such that F  - (dU/dx). Also define kinetic energyT  (½) mv2 (12) T - T0 = -∫(dU/dx)dx or T - T0 = U(x0) - U(x) and T + U(x) = T0 + U(x0) =CONSTANT Come back to this in the next section.

  11. In General • Newton’s 2nd Law: F = (dp/dt), p = mv • If we know F = F(t), integrate this to get (limits on t are 0  t): Δp  p - p0 = ∫F(t) dt • Δp Impulse produced by F Δp ∫F(t) dt  “Impulse-Momentum Theorem”

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