introductory physics -...

Introductory Physics

Week 6 @K301

2015/05/22

Week 6 @K301 Introductory Physics

Part I

Summary of week 5


Summary of week 5

We learned

how to solve a differential equation ẋ+ ax = g(t)

how to use Taylor series expansion to obtain anapproximation

and utilized them to analyse the vertical motion under uniformgravity.


Vertical motion under uniform gravity

Question 1

A particle is projected vertically upward with speed v0 underuniform gravity ~g with no air resistance.

The equation of motion is

mz̈ = mv̇ = −mgBy solving this differential equation (and applying the initialconditions), we obtain v(t) and z(t).

v = −gt+ v0

z = −12gt2 + v0t


Vertical motion under uniform gravity

Question 2

A particle is projected vertically upwards with speed v0 underuniform gravity ~g. The air exerts a drag force −k~v, where k isa positive constant.


mz̈ = mv̇ = −mg − kv

v̇ +k

mv = −g

We need to solve this differential equation.


Solving differential equation

formula

The general solution of differential equation

ẏ + ay = g(t)

is a sum of the general solution of

ẏ + ay = 0

and a particular solution of

ẏ + ay = g(t)


Solving differential equation

formula

The general solution of differential equation

ẏ + ay = 0

isy = Ce−at,

where C is the integration constant.


Vertial motion under uniform gravity

We want to solve the differential equation v̇ +k

mv = −g.

The general solution of v̇ +k

mv = 0 is v = Ce(−k/m)t.

We can find a particular solution of v̇ +k

mv = −g. It is

v = −mg/k.

Thus, the general solution of v̇ + kmv = −g is

v = Ce(−k/m)t − mgk

Using the initial condition v = v0 when t = 0, we obtain

v = v0e(−k/m)t − mg

k(1− e(k/m)t)


Vertial motion under uniform gravity

ż =(v0 +

mg

k

)e(−k/m)t − mg

k

By integrating with respect to t

z =(v0 +

mg

k

)(−mk

)e(−k/m)t − mg

kt+D

Using the inital condition z = 0 when t = 0, we obtain

z = (v0 +mg

k)(−m

k)e(−k/m)t − mg

kt+ (v0 +

mg

k)m

k

=m

k(v0 +

mg

k)(1− e(−k/m)t)− mg

kt


the maximum height

The maximum height zmax is achieved when ż = 0. Aftersome calculation, we obtain

zmax =v20g

(mg

kv0−(mg

kv0

)2ln

(1 +

kv0mg

))

When there is no air resistance, zmax =v202g

.

We expect when the air resistance is small, zmax ∼v202g

.

How can we confirm it?


Taylor series

Definition

When a function f(x) is infinitely differentiable in aneighbourhood of a, the Taylor series of f(x) is

f(x) = f(a) + f ′(a)(x− a) + 12!f ′′(a)(x− a)2 + · · ·

=∞∑n=0

f (n)

n!(x− a)n

Approximation using the first few terms of a Taylor series hasmany applications in physics and other fields of science.


maximum height when kv0/mg � 1

Using formulae

d

dxln(x) =

1

xd2

dx2ln(x) = − 1

x2

d3

dx3ln(x) =

2

x3

We can obtain

f(1 + x) = f(1) + f ′(1)x+f ′′(1)

2!x2 +

f ′′′(1)

3!x3 + · · ·

ln(1 + x) = ln 1 + x− 12x2 +

1

3x3 − · · ·


maximum height when kv0/mg � 1

zmax =v20g

(mg

kv0−(mg

kv0

)2ln

(1 +

kv0mg

))

=v20g

(mg

kv0−(mg

kv0

)2(kv0mg− 1

2

(kv0mg

)2+

1

3

(kv0mg

)3− · · ·

))

=v20g

(1

2− 1

3

(kv0mg

)+ · · ·

)=v202g

(1− 2

3

(kv0mg

)+ · · ·

)Indeed, when the air resistance is small (kv0/mg � 1), the

maximum height zmax ∼v202g

.


ProjectileConstrained rectilinear motion

Part II

Motion under uniform gravity(cont.)



Projectile without air resistance

Question 3

A particle of mass m is projected with speed v0 in a directionmaking an angle θ with the horizontal. The particle is underuniform gravity, and no drag force. Find the subsequentmotion.

Take Cartesian coordinate system (x, z).The equation of motion is

md~v

dt= −mg~ez,

with the initial condition ~v = v0 cos θ ~ex + v0 sin θ~ez at t = 0.




In Cartesian coordinate system, we can think the motion of xand z direction independently.

~r = x~ex + z~ez

~̇r = ẋ~ex + x~̇ex + ż~ez + z~̇ez

Because ~̇ex = ˙̇ez = 0,

~̇r = ẋ~ex + ż~ez

Similarly,

~̈r = ẍ~ex + z̈~ez




The equation of motion

mdv

dt= −mg~ez,

mdvxdt

= 0, mdvzdt

= −mg

Using the initial conditions,

vx = v0x = v0 cos θ, vz = v0z − gt = v0 sin θ − gt

x = (v0 cos θ)t, z = (v0 sin θ)t−1

2gt2



the path of the projectile

To find the path of the projectile, we eliminate t from theequations

x = (v0 cos θ)t, z = (v0 sin θ)t−1

2gt2

t =x

v0 cos θ

z = (tan θ)x− g2v20 cos

2 θx2

This is an inverted parabola.



the range of the projectile

The projectile will reach the ground when z = 0.

z = (tan θ)x− g2v20 cos

2 θx2 = x

(tan θ − g

2v20 cos2 θx

)= 0

x =

0,2v20 cos2 θ tan θg

=2v20 sin θ cos θ

g=v20 sin 2θ

g

The maximum range of the projectile is achived when θ = π/4.



Digression 1: Canon’s range

It is Galileo who solved the problem of the projectile under theuniform gravity without air resistance (published in ”Two NewSciences” 1638).He also compiled a table of cannon’s range and the angle ofelevation for his patron, the state of Venice. Since the fall ofConstantipole (1453), the usage of cannons had widely spreadto Europe, and often was a deciding factor on war. His tablemust be very valuable for the state of Venice.



the Monkey and the Hunter

A hunter with a gun found a monkey hanging in a tree. Themonkey releases its grip the instant the hunter fires his gun.Where should the hunter aim?We consider only uniform gravity. Air registance are neglected.

(a) aim high (over the monkey’s head)(b) directly at the monkey(c) aim low (below the monkey)




A hunter with a gun found a monkey hanging in a tree. Themonkey releases its grip the instant the hunter fires his gun.Where should the hunter aim?We consider only uniform gravity. Air registance are neglected.

(a) aim high (over the monkey’s head)(b) directly at the monkey(c) aim low (below the monkey)




(Loading Video...)


MIT-Physics-Demo-Monkey-and-Gun.mp4Media File (video/mp4)



The equations of motion for the bullet are

ẍP = 0, ÿP = −g

with initial conditions of

ẋP = v0 cos θ, xP = 0, ẏP = v0 sin θ, yP = 0

The equations of motion for the monkey are

ẍM = 0, ÿM = −g


ẋM = 0, xM = l cos θ, ẏM = 0, yM = l sin θ




The equations to describe the motion of the bullet P

xP = (v0 cos θ)t, yP = (v0 sin θ)t−1

2gt2

The equations to describe the motion of the monkey M

xM = l cos θ, yM = l sin θ −1

2gt2




xP = xM on t =l cos θ

v0 cos θ=

l

v0.

At that time,

yP = (v0 sin θ)l

v0− 1

2gt2

= l sin θ − 12gt2 = yM

We proved that on t =l

v0, both xP = xM and yP = yM are

satisfied. The bullet hits the monkey.




Isn’t it obvious?




Let us write ~r = x′~ex′ + y~ey

Then because ~̇ex′ = ~̇ey = 0, (just as in the case of Cartesiancoordinate system)

~̇r = ẋ~ex′ + ẏ~ey, ~̈r = ẍ~ex′ + ÿ~ey

This means we can independently consider the change of x′

and y.




The equations of motion for the bullet are

ẍ′P = 0, ÿP = −g


ẋ′P = v0, x′P = 0, ẏP = 0, yP = 0

The equations of motion for the monkey are

ẍ′M = 0, ÿM = −g


ẋ′M = 0, x′M = l, ẏM = 0, yM = 0




Because both the differential equations for yP and yM aresame (ÿP = ÿM = −g), and their inital conditions are same(ẏP = ẏM = 0, yP = yM = 0), It is obvious that yP = yMholds for any t.

Since x′P = v0t and x′M = l, the bullet hit the monkey (x

′P

becomes equal to x′M) when t = l/v0.



Constrained rectilinear motion

block on a slope

A block of mass m is placed on the inclined surface of a fixedrigid wedge of angle θ.

If there were no other forces than the uniform gravity, theblock would move vertically downwards.




There are contact forces which come into play only whenbodies are in physical contact.One is normal reaction force, which is normal to the slopeof the wedge. The other is frictional force, which is parallelto the slope. The normal reaction force is an example ofconstraint forces, which are not prescribed beforehand, butenforce a specified geometrical constraint.





md~v

dt= −mg~ez − F~ex′ +N~ez′

We take components of the vector equation in the x′ and z′

directions (parallel and normal to the slope).Noting ~ez = − sin θ~ex′ + cos θ~ez′ , we obtain

mdv

dt= mg sin θ − F 0 = N −mg cos θ

The second equation determines the normal reactionN = mg cos θ.




When the friction force is neglected, we say the surface issmooth. In this case,

mdv

dt= mg sin θ

The block slides down the slope with the constant accelerationg sin θ.

If x′ = v = 0 when t = 0,

v = (g sin θ)t

x′ =1

2(g sin θ)t2




The surface is smooth, and the block was at rest when t = 0.When the block slides down by l, what is its velocity?

l =1

2(g sin θ)t2l → tl =

√2l

g sin θ

vl = (g sin θ)

√2l

g sin θ=√

2gl sin θ =√2gh,

where h = l sin θ is the change of the height of the block.


Work, Kinetic Energy and Potential EnergyEnergy conservation in rectilinear motion

Part III

Energy and Work



Field

A force ~F is said to be a field if it depends only on theposition of the particle (~r = x~ex + y~ey + z~ez), not its velocityor the time.

A scalar function U is said to be a field if it is a function of(only) position.



Examples of force field

Gravitational Force ~F = −GMmr2

~er is a force field. The

force depends only on the position of the particle ~r = r~er.

Uniform gravity ~F = −mg~ez is a force field. The force isa constant function of ~r, and does not depends on anyother things.

Air drag force ~F = −k~v is not a force field. The forcedepends on the velocity of the particle.



work and kinetic energy

A particle P of mass m is under a force ~F . The equation ofmotion is

md~v

dt= ~F

take the scalar product with ~v,

m~v · d~vdt

=d

dt(1

2m~v · ~v) = ~F · ~v

dK

dt= ~F · ~v

where K =1

2m~v · ~v




Definition

The scalar quantity

K =1

2m~v · ~v = 1

2m|~v|2

is called the kinetic energy.




If the particle’s kinetic energies are K1 and K2 at times t1 andt2,

K2 −K1 =∫ t2t1

dK

dtdt =

∫ t2t1

~F · ~vdt

Definition

The scalar quantity

W =

∫ t2t1

~F · ~vdt

is called the work done by the force ~F .




Energy Principle

K2 −K1 = W

The increase of the kinetic energy in a given time interval isequal to the total work done by applied force(s) in the timeinterval.



Units for work and energy

Work and energy has the SI unit of the joule.

1J = 1kg ·m2/s2 = 1N ·m

Other units used for work and energy include calory.

1 cal = 4.184J

1 cal is approximately equal to the amount of energy neededto raise the temperature of one gram of water by one degreeCelsius.Confusingly, 1 kcal is sometimes written as 1Cal in foodindustry (often in the US)... be aware you convert these units.



Work in rectilinear motion

If a particle P moves along x-axis under the force field F (x),the work done by F (x) is

W =

∫ t2t1

F (x)vdt =

∫ t2t1

F (x)dx

dtdt =

∫ x2x1

F (x)dx



Potential energy in rectilinear motion

If F can be written as F (x) = −dV (x)dx

using its indefinite

integral −V (x), then

K2 −K1 =∫ x2x1

Fdx =

∫ x2x1

−dVdx

dx = −∫ x2x1

dV

= V (x1)− V (x2)V (x) is called the potential energy function of the force F .



Potential energy and its reference point

The potential energy V (x) of force F (x) is the indefinitefunction of −F (x). This means V (x) is uniquely determinedexcept an integration constant, which can be determined forour convenience.

For example, when we determined the constant so thatV (x) = 0, we say we took x = 0 as its reference point.

The difference of the integration constant does not changeforce F (x).



Example: uniform gravity

The uniform gravity F (z) = −mg

The potential energy V (z) = −∫F (z)dz = mgz + C

(Take the reference point at the ground)

Take V (0) = 0 → V (z) = mgz(Take the reference point at z = z0)

Take V (z0) = 0 → V (z) = mgz −mgz0You can confirm that

F (z) = −dV (z)dz

= −mg

regardless of the choice of the reference point.



Mechanical energy

When F (x) = −dV (x)dx

,

K2 −K1 = V (x1)− V (x2)K2 + V (x2) = K1 + V (x1)

E = K + V is called the mechanical energy of the particle.



Conservation of mechanical energy

Energy conservation in rectilinear motion

When a particle undergoes rectilinear motion in a force field,the mechanical energy (the sum of its kinetic and potentialenergy) remains constant in the motion.




When we dropped a particle from height h0, what is theparticle’s speed at the ground?

Because the uniform gravity is a conservative force, themechanical energy remains constant.Let us take the ground as the reference point of the potentialenergy of the uniform gravity. Then

initial mechanical energy = final mechanical energy

0 +mgh0 =1

2mv2f + 0

vf = −√2gh0

(the minus sign is to show the particle is moving downward.)




Another way to solve the problem.

While the particle dropping from height h0 to the ground, thework done by the gravitational force is mg · h0.The change of the kinetic energy is equal to the work done byapplied force.

1

2mv2f − 0 = mgh0

vf = −√

2gh0

(the minus sign is to show the particle is moving downward.)




You can solve the problem by making the equation of motion.

v̇ = −gv = −gt (initial condition v = 0 at t = 0)

z = −12gt2 + h0 (initial condition z = h0 at t = 0)

Now we obtain the time tf when the particle hit the ground

0 = −12gt2f + h0 → tf =

√2h0g

vf = −g

√2h0g

= −√

2gh0



Not the case: resistance force

The resistance force in medium is not a function ofparticle’s position, but a function of its velocity. In this case,we can not define its potential energy. In other words, theconservation of mechanical energy does not hold true.

The force which exerts on a particle dropping under uniformgravity with linear resistance is F = −mg − kvIn this case, the work done by the gravitational force is notfully converted to the particle’s kinetic energy, but part of thework is converted to heat.



Work done by constraint force

In most of idealized problems, the constraint force is alwaysperpendicular to the velocity of the particle (~F · ~v = 0). In thiscase, the work done by the constraint force is zero.

Energy conservation in constrained motion

When a particle moves under conservative force, nd is subjectto constraint forces that do not work, the mechanical energy(the sum of its kinetic and potential energy) remains constantin the motion.



Example: block on a slope

The surface is smooth, and the block is at rest when t = 0.When the block slides down by l, what is its velocity?

Since the normal reaction force does no work, and there is nofriction force, the energy conservation applies in the system.Let us take the reference point to the original position

0 + 0 =1

2mv2 −mgh

v =√

2ghWeek 6 @K301 Introductory Physics


Example: Downhill Skier

The constraint force is often not known beforehand, so we cannot solve the equation of motion.When a skier descends a bumpy slope, the magnitude and thedirection of the normal reaction force changes so that the skieris kept on the slope. But when the slope is smooth, theconstraint force does no work, so the energy conservationapplies. The speed of the skier after losing attitude h isv =√2gh regardless of the shape of the smooth slope.


Summary of week 5Motion under uniform gravity (cont.)ProjectileConstrained rectilinear motion

Energy and WorkWork, Kinetic Energy and Potential EnergyEnergy conservation in rectilinear motion

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