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Physics 170
Summary of Results from Lecture
The position vector ~r(t) can be resolved into its Cartesian
components: ~r(t) = x(t)^i + y(t)^j + z(t)k^.
Kinematical Variables
Rates of Change
Velocity ~v(t) = d~r(t)=dt
Acceleration ~a(t) = d~v(t)=dt = d2~r(t)=dt2
Kinematics with Constant Acceleration
( ) = ~r(0) + ~v(0)t + 21~at2.
The motion is specied by the constant acceleration ~a during the interval and the
initial state of motion (~r(0); ~v(0)) at the beginning of the interval.
This describes (approximately) the \free fall" motion of an object on the earth where
~a = g^
j and where ^
j is a unit vector perpendicular to the surface.
~
r t
Useful results for Free Fall/Projectile Motion
Trajectory: y(x) = x tan() gx2=(2vo2 cos()2) (released from origin with speed vo
and projection angle .
Time of ight: T = 2vo sin()=g
Maximum height: h = (vo sin())2=2g
Horizontal Range: R = ((vo2=g)) sin(2)
Kinematics in 2D Polar Coordinates
In two dimensions it is often useful to specify the position in polar coordinates, giving
the distance of the particle from the observer and the direction of the line of sight.
( ) = r(t)^r(t)
(1)
where r^(t) is a unit vector directed outward along the line of sight. If ^(t) is a unit
vector perpendicular to r^(t) directed counterclockwise (in the direction of increasing
) then
~
r t
( ) = r(t)^r(t)
d~
r (t)
~v (t) =
= r_ r^ + r_^
dt
d~v (t)
= (r r_2 )^r + (2_r_ + r)^
~a(t) =
~r t
dt
= m~a, i.e. in an inertial frame of reference changes
of motion (measured by ~a) are responses to external motive forces described by F~ .
The magnitude of the gravitational force on an object of mass m near the
surface of the earth is the weight W = mg where g 9:8 m=sec2.
Newton's Second Law of Motion
Weight
~
F
A string under tension mediates a force: the
magnitude of the force from a section of string is the tension T and the direction
of this force is tangent to the string pointing towards the section. (The string can
pull but can't push.) In general the tension can vary as a function of position in the
string. An ideal string is regarded as massless and unstretchable.
The normal force N is a force perpendicular to the plane of contact
with a rigid object. The frictional force is the force parallel to the plane of contact.
We distinguish the case of static friction when there is no slipping and kinetic friction
when there is relative motion
static Ff sN
kinetic Ff = k N
Forces from Strings Under Tension
Contact Forces
The force from a Hooke's Law spring Fel = ks where s is the spring
extension (positive s denotes extension and negative s is compression, the negative
sign in the force law says that the elastic force opposes the extension/compression).
The equation of motion of a mass m connected to a Hooke's Law spring attached to
a xed support is
Elastic Forces
=
mx
kx
(2)
q
with solutions x(t) = A sin(!t)+ B cos(!t) and ! = k=m. A and B are determined
by the initial state of the motion: A = x_ (0)=! and B = x(0).
For a system of particles with masses mi and positions ~ri, the \center
of mass" is the mass-weighted position
Center of Mass:
~ cm
R
The equationPofN motion for R~ cm
Letting M = i=1 mi we have
PN mi~ri
= Pi N m
i
=1
i=1
(3)
is often simpler than for the masses themselves.
 = F~ext
~ cm
MR
(4)
For an isolated system, initially at rest this leads to the center of mass principle R~ cm
is constant.
The linear momentum of a particle with mass m moving at
velocity ~v is p~ = m~v. The linear momentum of an collection of particles is the
Linear Momentum:
(vector) sum of their linear momenta: P~ = Pi p~i. The total linear momentum changes
in response to an external force P~_ = F~ext .
The impulse J~ = R F~ (t)dt. The impulse is the change
of the linear momentum: P~ = J~.
The dynamics of a system of mass M that
releases mass at a rate dmr =dt (with relative speed ur ) and accumulates mass at a rate
dma =dt (with relative speed ua ), in the presence of an external force F is described
by
Impulse Momentum Theorem:
Momentum Flow and Mass Transport:
M
dM
dt
=
F
r
+ ur dm
dt
ua
dma
dt
(5)
The
M dV =dt = ur dmr =dt specializes this result to the case of an
isolated rocket that accelerates by releasing mass at a steady rate.
For motion in one dimension
Zb
1
1
2
2
Kab = 2 mvb 2 mva = W (a ! b) = a F (x) dx
(6)
For motion in higher dimensions the work is computed by projecting the force into
direction of the displacement and integrating over the trajectory of the particle (this
requires evaluation of a line integral), thus
rocket equation
Work Energy Theorems:
Kab
Zb
1
1
2
2
= 2 mvb 2 mva = W (a ! b) = a F~ (r) d~r
(7)
The work done by a conservative force is identied with a change
of the potential energy. The change in potential energy is the negate of the work done
by a conservative force
Conservative Forces:
U (a ! b) = U (b)
( )=
U a
Zb
a
~ d~
F
r
(8)
U is the same for any path that connects the same two endpoints.
Three Potential Energy Functions:
Potential energy of a Hooke's law spring with compression/extension s : U (s) = ks2=2.
Potential energy of a mass m a distance y above the surface of the earth: U (y) = mgy
with g 9:8 m=s2.
Gravitational potential energy of two point masses m1 and m2 with separation s :
U (s) = Gm1 m2 =s (note the sign) with G 6:67 10 11 N m2 =kg2 .
The total mechanical energy E = K + U . In terms of the
total mechanical energy, the work energy theorem states
Conservation of Energy
Eb
= Ea + Wnc(a ! b)
(9)
when Wnc < 0 the total mechanical energy decreases.
In a two body
collision, in one dimension the nal state
is completely constrained by the conservation of linear momentum and conservation
of energy. For an elastic collision of two masses m1 and m2 with initial velocities v1i
and v2i the nal velocities are determined by mass ratios only (they don't depend on
any details of the force laws if the kinetic energy is unchanged).
Two Body Collisions
elastic
v1f
=
v2f
=
m1
m1
m2
+ m2
v1i
+ m 2+m2m
1
2
v2i
2m1 v + m2 m1 v
1i
2i
m1 + m2
m1 + m2
(10)
For an inelastic collision, linear momentum is conserved but the kinetic energy changes.
The change of the kinetic energy is Q where
1 m v2 + 1 m v2 = 1 m v2 + 1 m v2 + Q
2 1 1i 2 2 2i 2 1 1f 2 2 2f
(11)
Positive Q means that energy is lost; both positive Q (inelastic) and negative Q
(superelastic) are possible. A
has the largest possible Q
consistent with the conservation of momentum. For a two body collision, this requires
a situation where the two objects \stick together" in the nal state.
Here the outcome of the collision is only
partially constrained by the conservation laws:
completely inelastic collision
Collisions in More than One Dimension
m1~v1i
+ m2~v2i = m1~v1f + m2~v2f
1 m v2 + 1 m v2 = 1 m v2 + 1 m v2 elastic collision
2 1 1i 2 2 2i 2 1 1f 2 2 2f
(12)
(13)
In two(three) dimensions, the conservation laws provide three(four) independent constraints for the four(six) unknown components of the nal velocities ~v1f and ~v2f . Thus
the outcome of the collision is not constrained by the conservation laws. An analysis
of the two body collision in more than one dimension can be accomplished by additional
of e.g. the scattering angle, which provides one(two) additional
constraints on 2D(3D) collisions.
measurement