introductory physics -...

Introductory Physics

Week 9 @K301

2015/06/12

Week 9 @K301 Introductory Physics

Part I

Summary of week 8


Summary of week 8

We studied

potential energy of simple harmonic motion

energy diagram of simple harmonic motion

bound and unbound motion

Then we studied

uniform circular motion in polar coordinate system


Potential energies

Potential energy function V (x) of force F (x) is a function

which satisfy F (x) = −dV (x)

dx.

name force potential energy

uniform gravity F (z) = −mg V = mgz(reference point at z = 0)

restoring force F (x) = −kx V =1

2kx2

(reference point at x = 0)


Energy diagram


Bound motion and unbound motion


Stable equilibrium


Polar coordinate system

~r = r~er

~er = cos θ~ex + sin θ~ey

~eθ = − sin θ~ex + cos θ~ey

d~erdθ

= − sin θ~ex + cos θ~ey = ~eθ

d~eθdθ

= − cos θ~ex − sin θ~ey = −~er

~er =d~erdθ

dθ

dt= θ~eθ

~eθ =d~eθdθ

dθ

dt= −θ~er


Uniform circular motion in polar coordinate system

A particle is moving with a constant speed v in theanti-clockwise direction around a circle center O and radius r.The particle is at the point B(r, 0) when t = 0.

the distance from the centeris kept constant.

r = 0

the speed is constant.

|~v| = rθ = const.

→ d

dt|~v| = rθ + rθ = 0

→ θ = 0


Uniform circular motion in polar coordinate system

Calculate ~v and ~a under conditions r = 0 and θ = 0.

~r = r~er

~v = r~er + r~er = rθ~eθ

~a = rθ~eθ + rθ~eθ + rθ~eθ

= rθ(−θ~er)

= −rθ2~er = −v2

r~er


Uniform circular motion

Cartesian coordinate system

~r = r cos θ~ex + r sin θ~ey

~v = −v sin θ~ex + v cos θ~ey

~a = −v2

rcos θ~ex −

v2

rsin~ey

= −v2

r~er


~r = r~er

~v = rθ~eθ

~a = −rθ2~er

= −v2

r~er

Uniform circular motion

In uniform circular motion, the direction of the accelerationvector is toward the circle’s center, and its magnitude isconstant.

|~a| = a = rθ2 =v2

r


Part II

Circular Motion (cont.)


velocity and acceleration in polar coordinate system

We would like to obtain the formula for the velocity andacceleration in polar coordinate system. Note that

~er =d~erdθ

dθ

dt= θ~eθ

~eθ =d~eθdθ

dθ

dt= −θ~er

We differenciate ~r with respect to t

~r = r~er

~v = r~er + r~er = r~er + rθ~eθ

~a = r~er + r~er + rθ~eθ + rθ~eθ + rθ~eθ

= r~er + rθ~eθ + rθ~eθ + rθ~eθ + rθ(−θ~er)= (r − rθ2)~er + (rθ + 2rθ)~eθ


velocity and acceleration formula

Cartesian coordinate system

~r = x~ex + y~ey

~v = x~ex + y~ey

~a = x~ex + y~ey


~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

In this coordinate system,

~er = − cos θ~ez + sin θ ~ex ~eθ = sin θ~ez + cos θ ~ex

~er = θ ~eθ ~eθ = −θ ~er

The acceleration of the particle is

~r = l ~er

~r = l ~er + lθ ~eθ = lθ ~eθ

~r = (lθ + lθ)~eθ + lθ(−θ)~er= (lθ)~eθ − (lθ2)~er


The simple pendulum

The equation of motion is

m{−(lθ2)~er + (lθ)~eθ} = −mg~ez − T ~er= (mg cos θ − T )~er −mg sin θ~eθ

So, we have two quations

−mlθ2 = mg cos θ − T, mlθ = −mg sin θ

By solving the second equation we can describe the motion ofthe pendulum. The first equation determines the tension T(constraint force) of the string.


The simple pendulum

The equation

θ +(gl

)sin θ = 0

is a non-linear differential equation.Non-linear differential equation is difficult to solve, and weoften need computational methods to evaluate the solution ofthe equation.


The simple pendulum

d sin θ

dθ= cos θ,

d2 sin θ

dθ2=

d

dθcos θ = − sin θ

d3 sin θ

dθ3= − cos θ,

d4 sin θ

dθ4= sin θ, · · ·

Taylor series expansion of the function sin θ around θ = 0 is,

sin θ = sin 0 + (cos 0)θ +1

2!(− sin 0)θ2

+1

3!(− cos 0)θ3 +

1

4!(sin 0)θ4 +

1

5!(cos 0)θ5 + · · ·

= θ − 1

3!θ3 +

1

5!θ5 + · · ·


The simple pendulum

When θ � 1, sin θ ∼ θ. In this case, the equation of motioncan be approximated as

θ +(gl

)θ = 0,

which is a linear differential equation.

θ = C cos(ωt+ δ)

(ω =

√g

l

)The period of motion is

T =2π

ω= 2π

√l

g


The simple pendulum

If the pendulum was initially at the position of θ = θ0, and itwas at rest(this means the initial conditions θ = θ0 and θ = 0 at t = 0),

θ = θ0 cosωt = θ0 cos

√g

lt

θ

P

mg

T

eθ

er

O

ez

ex


The simple pendulum

The pendulum was initially at the position of θ = θ0 and itwas at rest.

θ = θ0 cosωt = θ0 cos

√g

lt

θ = −ωθ0 sinωt

The speed of the pendulum at the bottom (θ = 0) is,

|v| = |lθ| = | − lωθ0 sinωt|

at the bottom, θ = 0. This means ωt = π/2

|v| = lωθ0 =√glθ0


The energy conservation in the simple pendulum

The tension T is always perpendicular to the velocity of thependulum, and does no work. We can use the energyconservation law.

0 +1

2mv2 = mgl(1− cos θ0) + 0

|v| =√

2gl(1− cos θ0)

This is exact answer.

When θ � 1, cos θ ∼ 1− 12θ2

|v| ∼

√2gl

(1− 1 +

1

2θ2

0

)=√glθ0

This agrees to the result obtained in the previous slide.


Stable equilibrium position of the simple pendulum

θ

P

mg

T

eθ

er

O

ez

ex

mg

T

P

When P is released from rest, P re-mains at original position. P is in equi-librium position.

The potential energy of the pendulumis V = mgl(1− cos θ), where l is thelenglth of the string.V have its minimum value at θ = 0.Thus θ = 0 is the stable equilibriumposition


the lamp of Galileo

It is said that whenGalileo was attend-ing a mass in thecathedral of Pisa,he noticed the pe-riod of the chan-delier’s movementis independent ofits amplitude (It issaid that that wasyear 1583).


Galileo’s laws of pendulum

Galileo later said that

period varies with the square root of length

period is independent of amplitude

period is independent of mass

for a given length, all periods are the same.

But we know that these are only approximately true.

Around 1660, Christiaan Huygens showed that a cycloid curve,not a circular arc will make isochronous pendulum.

x = c(θ + sin θ), z = c(1− cos θ)

I will leave you to show that if a particle moves along thecycloid curve under uniform gravity, its period is independentof its amplitude.


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.


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 ~t is a unit vector,

V (~r + δ~t)− V (~r)

= V (x+ δtx, y + δty, z + δyz)− V (x, y, z)

=∂V

∂xδtx +

∂V

∂yδty +

∂V

∂zδtz

= gradV · δ~t


meaning of gradient

V (~r + δ~t)− V (~r)

δ= gradV · ~t

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

Centrall 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


Bounded and unbounded motions

The conservation of mechanical energy

1

2mv2 + V (~r) = E

means that the position of the particle is restricted to theplaces which satisfy

V (~r) ≤ E.

If the above condition limits the position of the particle to alocalized area, the motion of the particle is called boundedmotion. If the particle can excape to infinitely far places, themotion is unbounded motion.


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


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