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Lecture

# L04_notes.pdf

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University of Toronto St. George

Physics

PHY354H1

Erich Poppitz

Winter

Description

PHY254 Lecture 4
▯ Readings: Morin 3.5 & 1.4(numerical stu▯), supplementary notes. Con-
tinue reading compwiki website.
▯ In this lecture we review polar coordinates for solving problems and
discuss solving F=ma numerically
F=ma as an ODE when F=F(x)
Last time:
dv d x
m = F(x) or in another common form m = F(x) (1)
dt dt2
dv dv dx dv
a = dt = dx dt = dx v (2)
dv
mv dx = F(x) (3)
Now we can use separation of variables:
Z v 0 0 Z x 0 0
mvdv = F(x)dx ) m v dv = F(x )dx )
v0 x0
▯v Z x Z x
1 mv 0▯ = F(x )dx ) 1mv ▯ mv = 2 F(x )dx 0 (4)
2 v0 x0 2 2 0 x0
You might recognize this as a standard Work/Kinetic energy equation.
Example: Coming up soon, but its slightly hidden...
Motion in plane, polar coordinates
▯ Polar coordinates are convenient for motion in a plane, and are used
a lot in central ▯eld motion and in cases where motion is restricted to
circles.
▯ Polar coordinates are curvilinear and known as a \curvilinear coordi-
nate system".
▯ Velocity and acceleration look di▯erent in polar coordinates.
1 ▯ We start with the following de▯nitions: the coordinate transformation
(x;y) ! (r;▯):
(x;y) = r(cos▯;sin▯) (5)
and the transformation of unit vectors (i;j) ! (r ^;▯):
^ = cos▯i + sin▯j; ▯ = ▯sin▯i + cos▯j ^ (6)
_ 2 2
▯ We will use the dot notation: f = df=dt, f = d f=dt .
▯ Unlike Cartesian unit vectors, the radial and azimuthal (this just means
in the direction of theta increasing) unit vectors have non-zero time
derivatives:
^ = ▯sin▯▯i + cos▯▯j = ▯▯ _^ (7)
2 and
▯ = ▯▯r^ (8)
▯ Now the position vector
r = xi + yj = rr^ (9)
The velocity
v = _ = r^+ rr^ = r_^+ r▯▯^ (10)
Thus, the radial component of the velocity is r _ and the azimuthal
_
component of the velocity is r▯. Now the acceleration:
a = = r ^+ r__ + _▯▯ + r▯▯ + r▯▯_
= (r ▯ r▯ )r^+ (2r_▯ + r▯)▯
▯ Now, Newton’s second law is F = m~ a. Expressing F = F rr^+ F ▯▯, we
▯nd
_2 _
F r m(r ▯ r▯ ); F ▯ m(2r _▯ + r▯) (11)
▯ Example: Find the radius of orbit r of a planet in uniform circu-
lar motion around the sun under the in
uence of gravity with F = r
▯GMm=r , if the period is 1 y.
▯ Answer: In this case, r _ = 0, ▯ = ! = 2▯=1y, = 0, F▯= 0. So
▯ ▯1=3
GMm 2 GM 8
▯ r 2 = ▯mr! ) r = ! 2 ▯ 1:5 ▯ 10 km (12)
▯ Example: Morin pg 75 #23: Consider a particle that peels an angular
force only, of the form F▯= mr _▯. Show that r_ = Alnr + B, where
A and B are constants of integration, determined by initial conditions.
▯ Answer: Using the force equations in this case we get the following
two equations:
2
0 = m(r▯ r▯ ) & mr _▯ = m(r▯ + 2r_▯) (13)
Where do we go from here? If you start ▯dgeting with both equations
you might stumble upon the answer eventually, but is there a more
3 systematic way? Probably. Here’s my attempt to provide one. Choose
one of the equations, identify the type of ODE (like we did in last
lecture) and integrate it. In this example, things may seem stranger
because of all the dots over variables, b_ is still a velocity and can
be treated as such. I will start with the second equation. Simplifying
it a bit gives:
_
▯r_▯ = r▯ (14)
Separating the variables gives:
_ ▯
▯ = _ (15)
r ▯
Both sides are a derivative over a variable. If I make the substitution:
! = ▯ then I can write the equation:
▯ _ = !_ or if you prefer1 dr = ▯ 1 d! (16)
r ! r dt ! dt
Multiplying both sides by dt and integrating both sides gives:
Z Z
1dr = ▯ 1d! ) lnr = ▯ln! + C (17)
r !
Notice that rather th

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