Introductory Physics Week 10 @K301 2015/06/19 Week 10 @K301 Introductory Physics
Summary of week 9
We studied
velocity and acceleration in polar coordinate system
In polar coordinate system, both ~er and ~eθ are vector functionsof time t.
Week 10 @K301 Introductory Physics
velocity and acceleration formula
Cartesian coordinate system
~r = x~ex + y~ey
~v = x~ex + y~ey
~a = x~ex + y~ey
Polar coordinate system
~r = r~er
~v = r~er + rθ~eθ
~a = (r − rθ2)~er + (rθ + 2rθ)~eθ
The velocity formula in polar coordinate system is easy tounderstand - the vector sum of an outward radial velocity r~erand a transverse velocity rθ~eθ.
The acceleration formula in polar coordinate system is difficultto interpret - especially the term 2rθ. This term is called”Coriolis term”.
Week 10 @K301 Introductory Physics
The simple pendulum
A particle P is suspended from a fixed point O by a lightinextensible string of length l. P is under uniform gravity, andno registance force acts on it. The string is taut. Find thesubsequent motion.
θ
P
mg
T
eθ
er
O
ez
ex
Week 10 @K301 Introductory Physics
The simple pendulum
The equation
θ +(gl
)sin θ = 0
is a non-linear differential equation, which is difficult to solve.
When θ � 1, we can approximate sin θ ∼ θ.
(sin θ = θ − 1
3!θ3 +
1
5!θ5 + · · · )
We can linearize the differential equation as
θ +(gl
)θ = 0,
which is a Simple Harmonic Motion.
Week 10 @K301 Introductory Physics
The simple pendulum
θ +(gl
)θ = 0,
θ = C cos(ωt+ δ)
(ω =
√g
l
)The period of motion is
T =2π
ω= 2π
√l
g,
which is independent of mass of the weight or the amplitudeof the pendulum.
Week 10 @K301 Introductory Physics
non-inertial frame
We have learned that the motion of a particle in an inertialframe is determined by the differential equation
m~a = m~r = ~F ,
where m and ~r are the mass and the position of the particle,~F is the force acting on the particle.This equation of motion is valid for all inertial frames.
What happens if we observe the motion of the particle in anon-inertial frame?
Week 10 @K301 Introductory Physics
example: Acceleration in a straight line
We consider the case in which we observe the motion of aparticle from a train which is accelerating with a constantacceleration.We have a fixed inertial frame (Frame 1), and a moving (butnot rotating) non-inertial frame (Frame 2)
Week 10 @K301 Introductory Physics
example: Accleration in a straight line
The relations between quantities observed in Frame 1 and 2
~r = ~r′ + ~D
~v = ~r = ~ ′r + ~D = ~v′ + ~V
~a = ~v = ~ ′v + ~V = ~a′ + ~A
Week 10 @K301 Introductory Physics
example: Accleration in a straight line
Frame 1 is an inertial frame. When force ~F is acting on theparticle, the equation of motion in Frame 1 is
m~a = ~F
In Frame 2,
m(~a′ + ~A) = F
m~a′ = F −m~A
This means, when we observe the motion of the particle fromnon-inertial Frame 2, it seems an additional force −m~A isacting on the particle in additional to the real force ~F . This isan example of fictitious forces.
When a train accelerates, passengers on the train feels like theyare being pushed back. This is because of the fictitious force.
Week 10 @K301 Introductory Physics
other fictitious forces
There are other fictitious forces, which include centrifugalforce and Coriolis force. Both fictitious forces appears whenwe observe the motion from rotating (thus non-inertial) frame.
When a car makes a turn, passengers on the car feels like theyare being pushed outward. This is because of the centrifugalforce.
Week 10 @K301 Introductory Physics
Equations of motion for a particle under a central
force
Let’s consider the motion of a particle in a central force~F = F (r)~er. The equations of motion are
m(r − rθ2) = F (r), m(rθ + 2rθ) = 0
∵ in polar coodinate system,
~r = r~er
~v = r~er + r~er = r~er + rθ~eθ
~a = r~er + r~er + rθ~eθ + rθ~eθ + rθθ~eθ
= r~er + rθ~eθ + rθ~eθ + rθ~eθ + rθ2(−er)= (r − rθ2)~er + (2rθ + rθ2)~eθ
Week 10 @K301 Introductory Physics
centrifugal force and centripetal force
The equation of motion for r direction is
m(r − rθ2) = F (r)
This equation of motion can be written as
mr = F (r) +mrθ2
The term mrθ2 is the centrifugal force, which is a fictitiousforce.
Don’t confuse with centripetal force, which is the force thatmakes a particle’s path curved. In circular motion, thedirection of the centripetal force is toward the center of thecircle. In the above case, F (r) is the centripetal force.
Week 10 @K301 Introductory Physics
Work in 3-dim motion
If the particle’s kinetic energies are K1 and K2 at times t1 andt2,
K2 −K1 =
∫ t2
t1
dK
dtdt =
∫ t2
t1
~F · ~vdt
Definition
The scalar quantity
W =
∫ t2
t1
~F · ~vdt
is called the work done by the force ~F .
Week 10 @K301 Introductory Physics
Work in 3-dim motion
In rectilinear motion: if F is a force field F (x),
W =
∫ t2
t1
F (x)vdt =
∫ t2
t1
F (x)dx
dtdt =
∫ x2
x1
F (x)dx
In 3-dim motion: if ~F is a force field ~F (~r),
W =
∫ t2
t1
~F (~r) · ~vdt =
∫ t2
t1
~F · d~rdtdt =
∫ ~r2
~r1
~F · d~r
The integral on the right side of equation is a line integral.
Week 10 @K301 Introductory Physics
Line integral
Line integral of a scalar function is an expression of the form∫ b
a Pfdl = lim
n→∞
n∑i=1
f(~ri)∆li
a
b∆l
i
ri
This differ from ordinarydefinite integrals in that the range of integration is not aninterval of the x-axis, but a path in three-dimentional space.
Week 10 @K301 Introductory Physics
Surface integral and Volume integral
Surface integral of a scalar function is an expression of theform ∫
SfdA = lim
n→∞
n∑i=1
f(~ri)∆Ai
Volume integral of a scalar function is an expression of theform ∫
Vfdτ = lim
n→∞
n∑i=1
f(~ri)∆τi
Week 10 @K301 Introductory Physics
Line integral
The length of circumference is
L =
∫Pdl =
∫ 2π
0
Rdθ = 2πR
Rθ
R
dl = Rdθ
Week 10 @K301 Introductory Physics
Surface integral
The area of circle is
S =
∫SdA =
∫ R
0
∫ 2π
0
rdrdθ = 2π1
2R2 = πR2
Week 10 @K301 Introductory Physics
Work differs on the path
In general, the work done by force ~F (~r) differs depending onthe path the particle traveled.
Example : ~F = x~ex + xy~ey
O (0, 0) A (1, 0)
C (1, 1)B (0, 1)
x
y
Week 10 @K301 Introductory Physics
Work differs on the path
Example : ~F = x~ex + xy~ey O (0, 0) A (1, 0)
C (1, 1)B (0, 1)
x
y
O→A→CW =
∫ 1
0xdx+
∫ 1
0ydy = 1
2+ 1
2= 1
O→B→CW =
∫ 1
00dy +
∫ 1
0xdx = 0 + 1
2= 1
2
Week 10 @K301 Introductory Physics
Conservative force
Forces which can be written as ~F (~r) = − gradV (~r), whereV (~r) is a scalar function of position (= a scalar field), arecalled conservative forces.
definition of gradient of a scaler field
When V (~r) is a scalar field,
gradV (~r) ≡ ∂V
∂x~ex +
∂V
∂y~ey +
∂V
∂z~ez
∂f
∂x,∂f
∂y,∂f
∂zis called a partial derivative of function f .
Week 10 @K301 Introductory Physics
partial derivative
Take derivative with respect to one of the variables, keepingthe others constant.
definition of partial derivative
∂f(x1, ..., xn)
∂xi
≡ lim∆xi→0
f(x1, ..., xi + ∆xi, ..., xn)− f(x1, ..., xn)
∆xi
Cf.df(x)
dx≡ lim
∆x→0
f(x+ ∆x)− f(x)
∆x
Week 10 @K301 Introductory Physics
gradient of scalar field
Examples:
V (~r) = xy2z3
gradV (~r) = y2z3~ex + 2xyz3~ey + 3xy2z2~ez
V (~r) =1
2kr2 =
1
2k(x2 + y2 + z2)
gradV (~r) = kx~ex + ky~ey + kz~ez
= k~r = kr~er
Week 10 @K301 Introductory Physics
meaning of gradient
When V (x) is a function of (only) x,
V (x+ δx)− V (x) =dV
dxδx
When V (~r) is a function of x, y and z, and ~et is a unit vector,
V (~r + δt · ~et)− V (~r)
= V (x+ δtx, y + δty, z + δyz)− V (x, y, z)
=∂V
∂xδtx +
∂V
∂yδty +
∂V
∂zδtz
= (gradV · ~et)δt
Week 10 @K301 Introductory Physics
meaning of gradient
V (~r + δt · ~et)− V (~r)
δt= gradV · ~et
This means that
gradV points the direction of the greatest rate ofincrease of V
the magnitude of gradV is the rate of the increase
Week 10 @K301 Introductory Physics
Work done by a conservative force
When ~F (~r) = − gradV (~r),
W =
∫ ~r2
~r1
~F (~r) · d~r =
∫ ~r2
~r1
(− gradV (~r)) · d~r
= −∫ ~r2
~r1
(∂V
∂x~ex +
∂V
∂y~ey +
∂V
∂z~ez
)· (dx~ex + dy~ey + dz~ez)
= −∫ ~r2
~r1
(∂V
∂xdx+
∂V
∂ydy +
∂V
∂zdz
)= −
∫ ~r2
~r1
dV = V (~r1)− V (~r2)
Week 10 @K301 Introductory Physics
Work done by a conservative force
As we see in the previous slide, the work done by theconservative force
W =
∫ ~r2
~r1
~F (~r) · d~r
is equal toV (~r1)− V (~r2).
This means the work done by a conservative force isindependent on the particle’s path (determined only bythe initial position ~r1 and the final position ~r2).
Week 10 @K301 Introductory Physics
Potential energy in 3-dim motion
When ~F (~r) = − gradV (~r), V (~r) is called the potential
energy function of the force ~F .
From the previous slide,
K2 −K1 = V (~r1)− V (~r2)
K2 + V (~r2) = K1 + V (~r1)
E = K + V is called the mechanical energy.
Week 10 @K301 Introductory Physics
Conservation of mechanical energy
Conservation of mechanical energy
When a particle moves in a conservative force field, themechanical energy (the sum of its kinetic and potentialenergy) remains constant in the motion.
Week 10 @K301 Introductory Physics
Central force
Central force is a force whose magnitude only depends onthe distance r of the particle from the origin and is directedalong the line joining them.
~F = h(r)~er
A central force is conservative with potential energy
V (r) = −H(r) = −∫h(r)dr
Week 10 @K301 Introductory Physics
Example: spring force
One end of a spring is fixed at the origin, and the other isattached to a particle. The force ~F acting on the particle is
~F (~r) = −k~r = −kr~erand its potential energy function is
V (~r) = −∫krdr =
1
2kr2 + C
By taking r = 0 as the reference point,
V (~r) =1
2kr2
Week 10 @K301 Introductory Physics
Example: spring force
Let’s confirm the potential energy V for the restoring force
~F = −k~r is given by1
2kr2.
V (~r) =1
2kr2 =
1
2k(x2 + y2 + z2)
~F = − gradV (~r)
= −(∂V
∂x~ex +
∂V
∂y~ey +
∂V
∂z~ez
)= −kx~ex − ky~ey − kz~ez = −k(x~ex + y~ey + z~ez)
= −k~rOK.
Week 10 @K301 Introductory Physics
Example: Gravitational Force
Gravitational force exerted by a fixed point at the origin,
~F = −GMm
r2~er
is a central force (so it is conservative).Its potential V is
V (r) = −∫ (−GMm
r2
)dr = −GMm
r+ C
By taking r =∞ as the reference point,
V (r) = −GMm
r
Week 10 @K301 Introductory Physics
Example: Gravitational Force
Let’s confirm the gradient gives the gravitational force.
∂V
∂x=
∂
∂x
(−GMm
r
)=
∂
∂x
(− GMm
(x2 + y2 + z2)1/2
)=
(−1
2
)(− GMm
(x2 + y2 + z2)3/2
)2x
=GMm
(x2 + y2 + z2)3/2x
=GMm
r3x
Week 10 @K301 Introductory Physics
Example: Gravitational Force
Thus,
− gradV = −(∂V
∂x~ex +
∂V
∂y~ey +
∂V
∂z~ez
)= −GMm
r3(x~ex + y~ey + z~ez)
= −GMm
r3~r
= −GMm
r3r~er
= −GMm
r2~er
OK.
Week 10 @K301 Introductory Physics
Escape speed
A particle of mass m is projected from the surface of a planetwith speed v0. Regard the planet as a fixed symmetric sphereof mass M and radius R. The conservation of mechanicalenergy applies.
1
2mv2 − GMm
r=
1
2mv2
0 −GMm
R
v2 = v20 + 2GM
(1
r− 1
R
)If the particle to escape the planet, the right side of theequation must be positive for any r.
Week 10 @K301 Introductory Physics
Example : Escape speed
Thus the initial speed v0 needs to satisfy
v20 −
2GM
R≥ 0
The minimum speed for escaping the planet is
vmin =
√2GM
R
This is called the escape speed.
For Moon: M = 7.35× 1022kg, R = 1740km→ vmin ∼ 2.4km/s
For Earch: M = 5.97× 1024kg, R = 6360km→ vmin ∼ 11.2km/s
Week 10 @K301 Introductory Physics