1. classical

2. 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) 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. 2 nd Law F = ma = m(dv/dt) = m(d 2 x/dt 2 ) = mv(dv/dx) General comments about these equivalent forms: F = m(d 2 x/dt 2 ): 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(d 2 x/dt 2 ) = 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 = (d 2 x/dt 2 ) = F/m = constant Obviously, get the familiar, 1d kinematics equations for constant acceleration (Physics I!). Given, initial conditions: t = 0, x = x 0, v = v 0 Get: v = v 0 + at x = x 0 + v 0 t + ½at 2 v 2 = (v 0 ) 2 + 2a(x-x 0 )

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

7. F = F(v) Velocity dependent forces (like retarding forces). Method 1: Direct Integration  m(dv/dt) = F(v) (4) Initial conditions: t = 0, x = x 0, v = v 0 General solution (integrals: limits on t are 0  t, limits on v are v 0  v, limits on x are x 0  x): (4)  dt = [m/F(v)]dv  ∫ dt = t = t(v) = m ∫ [1/F(v)] dv (5) (Gives t(v)) If possible, algebraically invert t(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) = x 0 + ∫v(t)dt (7) Case 3

8. 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 = x 0, v = v 0 General solution (integrals: limits on t are 0  t, limits on v are v 0  v, limits on x are x 0  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) = x 0 + m∫[v/F(v)]dv (9) Finally, substitute v(t) into x(v):  x[v(t)]  x(t) (10) Case 3

9. F = F(x): Position dependent forces.  m(dv/dt) = F(x) = m(d 2 x/dt 2 ) (11) Initial conditions: t = 0, x = x 0, v = v 0 (11) is a 2 nd 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 function U(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. Case 4

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

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