1. Forces and energies fall CM lecture, week 2, 10.Oct.2002, Zita, TESC F  a, v, x (Ch.2 Probs.1,2) Quiz #1 Potential energy (2.3) Kinetic energy Total mechanical energy turning points (2.4) changing mass (2.20)

2. Review: Force  a, v, x Given a force F= ma = dp/dt, find the resultant velocity v and x. For time-dependent forces: a(t) = F(t)/m, v(t) =  a(t) dt, x(t) =  v(t) dt, For space-dependent forces: F(x) = ma = m dv/dt where dv/dt = dv/dx * dx/dt = v dv/dx and  v dv = 1/m  F dx. 2.1(a) F(t) = F 0 + c t 2.2(a) F(x) = F 0 + k x

3. Quiz #1 1. Given the time-varying vector A = a t 2 i + b e ct j + d cos  t k, find the first and second time derivatives dA/dt and d 2 A/dt 2. 2. Find the velocity (v) and the position (x) as functions of time (t) for a particle of mass m which starts from rest at x=0 and t=0, subject to a force F x = F 0 cos  t. 3. Sketch the position, velocity, and acceleration versus time for the problem above. Label your axes!

4. Review: Energies F = m dv/dt = m v (dv/dx). Trick: d(v 2 )/dx = 2v dv/d, so F = m/2 d(v 2 )/dx Work done =  F dx = change in kinetic energy T, so F = dT/dx and T =  F dx =  (m/2 d(v 2 )/dx) dx: T = (1/2) m v 2 = p 2 /(2m) Work done = loss of potential energy V, so F = -dV/dx and V = -  F dx Work =  F dx = -  dV = -V(x) + V(0) = T(x) - T(0)

5. Practice calculating potential energy #2.3: Find V = -  F dx for forces in 2.2. (a) F(x) = F 0 + k x (b) F(x) = F 0 e -kx (c) F(x) = F 0 cos kx

6. Conservation of mechanical energy Recall that -V(x) + V(0) = T(x) - T(0) Total mechanical energy E = T(x) + V(x) = T(0) + V(0) is conserved in the absence of friction or other dissipative forces. Example: Escape velocity and black hole: F g = GmM/r 2

7. Practice with kinetic and total energies To solve for the motion x(t), integrate v = dx/dt where T = 1/2 m v 2 = E - V Note: x is real only if V < E  turning points where V=E. Solve for v and find locations (x) of turning points for F = -kx

8. What if mass is not constant? F = dp/dt = d(mv)/dt = m dv/dt + v dm/dt 2.20: A falling (spherical) raindrop grows as moist air condenses on it. Assume that its mass increases at a rate proportional to its cross- sectional area. Assume that the raindrop starts from rest and its initial radius (R 0 ) is so small that resistance is negligible. Show that its speed increases linearly and radius increases quadratically with time. Let  0 =density of water and  1 =density of humid air. Hint: dm 0 /dt = k  r 2 where R0R0 00 11