Introductory Physics - University of Tokyoradphys4.c.u-tokyo.ac.jp/~matsuday/lectures/peak/20150619-Physics... · The acceleration formula in polar coordinate system is di cult to

Introductory Physics

Week 10 @K301

2015/06/19

Week 10 @K301 Introductory Physics

Part I

Summary of week 9


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.


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”.


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


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.


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.


Part II

Fictituous Forces


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?


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)


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


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.


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.


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


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.


Part III

Work and Energy in 3-dim motion


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 .


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.


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.


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


Line integral

The length of circumference is

L =

∫Pdl =

∫ 2π

0

Rdθ = 2πR

Rθ

R

dl = Rdθ


Surface integral

The area of circle is

S =

∫SdA =

∫ R

0

∫ 2π

0

rdrdθ = 2π1

2R2 = πR2


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


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


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 .


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


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


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


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


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)


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).


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.


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.


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


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


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.


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



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



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.


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.


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


Example: Escape speed

If the particle’s velocity at r = R is larger than the escapespeed, the motion of the particle is unbound motion.

r

V

Obound motion

unbound motion

R


Documents

Introductory Physics - University of Tokyoradphys4.c.u-tokyo.ac.jp/~matsuday/lectures/peak/20150619-Physics... · The acceleration formula in polar coordinate system is di cult to