Transcript Introduction to Electromagnetism

Forces and energies
fall CM lecture, week 2, 10.Oct.2002, Zita, TESC
• F  a, v, x (Ch.2 Probs.1,2)
• Quiz #1
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Potential energy (2.3)
Kinetic energy
Total mechanical energy
turning points (2.4)
changing mass (2.20)
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) = F0 + c t
2.2(a) F(x) = F0 + k x
Quiz #1
1. Given the time-varying vector A = a t2 i + b ect j + d cos wt
k, find the first and second time derivatives dA/dt and d2 A/dt2.
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 Fx = F0 cos wt.
3. Sketch the position, velocity, and acceleration versus time
for the problem above. Label your axes!
Review: Energies
F = m dv/dt = m v (dv/dx).
Trick: d(v2)/dx = 2v dv/d, so F = m/2 d(v2)/dx
Work done =  F dx = change in kinetic energy T, so
F = dT/dx and T =  F dx =  (m/2 d(v2)/dx) dx:
T = (1/2) m v2 = p2 /(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)
Practice calculating potential energy
#2.3: Find V = -  F dx for forces in 2.2.
(a) F(x) = F0 + k x
(b) F(x) = F0 e-kx
(c) F(x) = F0 cos kx
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: Fg = GmM/r2
Practice with kinetic and total energies
To solve for the motion x(t), integrate v = dx/dt where
T = 1/2 m v2 = 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
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 crosssectional area . Assume that the raindrop starts from rest and its
initial radius (R0) is so small that resistance is negligible. Show that

its speed increases linearly and radius
increases quadratically with
time. Let r0=density of water and r1 =density of humid air.
Hint: dm0/dt = k pr2 where
r0
r1
R0

d mair d
d
k
 r air  r air
 r1v
dt A
dt
dt