introductory physics-i part 3particle.korea.ac.kr/class/2011/introphy/introphysics-iii.pdf · 1 kg...

Introductory Physics-IPart 3

Eunil WonDepartment of Physics

Korea University

Fundamentals of Physics by Eunil Won, Korea University 2

Energy and WorkEnergy?

Work? : human labor (wikipedia)


Energy and Work

Energy: a scalar quantity that is associated with a state (I know this is too vague)

Kinetic energy: K =1

2mv

2

SI unit of kinetic energy (and every other type of energy): Joule (J)

1 Joule = 1 J = 1 kg m2/s2

Work is energy transferred to or from an object by means of a force acting on the object


Work and Kinetic EnergyConsider a bead that can slide along a frictionless wire, which is stretched along a horizontal x axis

A constant force, directed at an angle to the wire, accelerates the bead

!F!

Fx = max

!d

!v0

: displacement of the bead

: initial velocity

We know from the basic constant-acceleration equations v2

= v2

0 + 2axd

1

2mv

2!

1

2mv

2

0 = Fxd

The work W done on the bead by the force (the energy transfer due to the force) is

W = Fxd W = Fd cos ! W =!F ·

!d


Work-Kinetic energy TheoremA force does positive (negative) work if it has a vector component in the same (opposite) direction as the displacement

Work-Kinetic Energy Theorem:

SI unit for work: Joule (same as kinetic energy unit)

!K = Kf ! Ki = W!

change in the kineticenergy of a particle

"

=

!

net work done onthe particle

"

!

kinetic energy afterthe net work is done

"

=

!

kinetic energybefore the net work

"

+

!

the network done

"

Kf = Ki + W


How “large” is one Joule?

1 kg of a rock, from 1 m high, falling on your foot:

W = mgh = 1 kg · 9.8 m/s2 · 1 m ≈ 10 J

So you will fill as much as 10 Joule of pain on your foot!

1 m

1 kg

g = 9.8 m/s2


Sample ProblemIn 1896 in Waco, Texas, William Crush parked two locomotives at opposite ends of a 6.4 km-long track, fired them up, tied their throttles open, and then allowed them to crash head-on at full speed in front of 30,000 spectators. Hundreds of people were hurt by flying debris: several were killed. Assuming each locomotive weighted 1.2x106 N and its acceleration was a constant 0.26 m/s2, what was the total kinetic energy of the two locomotives just before the collision?

v2 = v20 + 2a(x− x0)

v0 = 0 and (x− x0) = 3.2× 103 m

v2 = 0 + 2(0.26 m/s2)(3.2× 103 m)

or

v ∼= 40.8 m/s

mass?

speed?

m =1.2× 106 N

9.8 m/s2= 1.22× 105 kg

K = 2× 1

2mv2 = 2× 1

2(1.22× 105 kg)(40.8 m/s)2

= 2.0× 108 J


Work Done by a Gravitational ForceThe work done by the gravitational force:

Wg = mgd cos !

In the case of the figure left, the object moves upward:

Wg = mgd cos 1800

= !mgdThe negative sign indicates that the gravitational force transfers energy (mgd) from the kinetic energy of the object

If the object moves downward:

Wg = mgd cos 00

= +mgdThe positive sign indicates that the gravitational force transfers energy (mgd) to the kinetic energy of the object


Work Done by a Gravitational ForceWork Done in Lifting and Lowering an Object

Suppose we lift an object by applying a vertical force, our applied force does positive work Wa on the object

!K = Kf ! Ki = Wa + Wg

!KThe change in the kinetic energy of the object

If object is stationary before and after the lift, Wa = !Wg

And it becomes Wa = !mgd cos !


Work Done by a Spring ForceThe Spring Force

Relaxed state

!F = !k!d (Hooke’s law)

k : spring constant (force constant)

The negative sign indicates the spring force is always opposite to the displacement

Note: a spring force is a variable force

!F = !F (x)

(This is the first time to see this type of force)


Work Done by a Spring Force

Since the spring force is a variable force, we cannot assume

The angle = 0 and the work done within each segment (j) can be written as

The net work Ws done

by spring, from xi to xf is

the sum of all works:

W = Fd cos !

! Fj!x

Ws =!

Fj!xIn the limit as each segment goes to zero,

Ws =

!xf

xi

Fdx

Ws =1

2kx

2

i !1

2kx

2

f Ws = !

1

2kx

2

Using Hooke’s law (F=-kx)

We get

If xi=0 and xf=x,

(work by a spring force)

The work done by a spring force

Ws =

! xf

xi

(!kx)dx = !k

! xf

xi

x dx

= (!1

2k)[x2]

xfxi = (!

1

2k)(x2

f ! x2

i )


Work Done by a Spring ForceThe work done by an applied force

Now suppose that we displace the block along the x axis while continuing to apply a force to it. Then!Fa

!K = Kf ! Ki = Wa + Ws

If the block is stationary before and after the displacement,

Wa = !Ws


Work Done by a General Variable ForceConsider one-dimensional variable force (see figure)

Fj,avg

!Wj = Fj,avg!x

: average value of F(x) within the j-th interval (constant)

work done by the force in the j-th interval:

The total work done from xi to xf is then

W =!

!Wj =!

Fj,avg!x

W = lim!x!0

!Fj,avg!x

W =

!xf

xi

F (x)dx

We let Δx approach zero:

then we get the integral of the function F(x) (The usual trick in calculus)

(This becomes the area between the F(x) curve and the x axis)


Work Done by a General Variable ForceThree dimensional analysis:

consider now a particle in a three-dimensional force !F = Fxi + Fy j + Fz k

Now let the particle move through an incremental displacement

d!r = dxi + dyj + dzk

dW = !F · d!r = Fxdx + Fydy + FzdzThen the increment of work dW is

Since we know how to calculate this component by component, we get

W =

! rf

ri

dW =

! xf

xi

Fxdx +

! yf

yi

Fydy +

! zf

zi

Fzdz


Work Done by a General Variable ForceWork-Kinetic Energy Theorem with a Variable Force

consider now a particle of mass m, moving along the x axis and acted on a net force F(x) that is directed along that axis

The work done on the particle by this force is: W =

!xf

xi

F (x)dx =

!xf

xi

ma dx

ma dx = mdv

dtdx

= mdv

dx

dx

dtdx

= mdv

dxvdx

= mv dv

Since (chain rule here)

We get W =

! vf

vi

mv dv = m

! vf

vi

v dv

=1

2mv

2

f !

1

2mv

2

ior W = Kf - Ki = ΔK


PowerPower: the time rate at which work is done by a force

Average power:

Instantaneous power:

Pavg =W

!t

P =

dW

dt

SI unit of power : watt (W) 1 W = 1 J/s

P =

dW

dt=

F cos !dx

dt= F cos !

!

dx

dt

"

= Fv cos !

For a particle that is moving along a straight line and is acted on by a constant force directed at some angle φ to that line:

P =!F · !vIn general case we write:


Sample ProblemA block of mass m = 0.40 kg slides across horizontal frictionless spring with speed v=0.50 m/s. It then runs into and compresses a spring of spring constant k=750 N/m.

When the block is momentarily stopped by the spring, by what distance d is the spring compressed?

Kf −Ki = −12kd2

d = v

�m

k= 0.50 m/s

�0.40 kg750N/m

= 1.2× 10−2 m = 1.2 cm


Potential Energy

Potential energy is the energy that can be associated with the configuration

ex) gravitational potential energy elastic potential energy

These two states has two different potential energy

Work and Potential Energy

The change ΔU in gravitational potential energy is defined to equal the negative of the work done on the tomato by gravitational force

!U = !W


Conservative and Nonconservative Forces

Conservative force:

Nonconservative force:

Work done by a force to change configuration is same but negative to the work done to restore the configuration

ex) gravitational force

A force that is not conservative

ex) kinetic frictional energy (thermal energy cannot be transferred back to kinetic energy)


Path Independence of Conservative Forces

The net work done by a conservative force on a particle moving around every closed path is zero

1

2mv

2

0

1

2mv

2

0

The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle

Wab,1 + Wba,2 = 0

Wab,1 = !Wba,2 = Wab,2


Potential Energy ValuesConsider a particle with a conservative force

Gravitational Potential Energy:

W =

!xf

xi

F (x)dx !U = !

!xf

xi

F (x)dxor

!U = !

! yf

yi

(!mg)dy = mg

! yf

yi

dy = mg(yf ! yi)

U(y) = mgy (if we take Ui = yi = 0)

Elastic Potential Energy:

U(x) =1

2kx

2 (if we take Ui = xi = 0)

!U = !

! xf

xi

(!kx)dx = k

! xf

xi

x dx =1

2kx

2

f !

1

2kx

2

i


Conservation of Mechanical EnergyThe mechanical Energy Emec Emec = K + U

When a conservative force does work W on an object, it transfers energy between K and U

!K = W = !!U

K2 ! K1 = !(U2 ! U1)

K2 + U2 = K1 + U1

!Emec = !K + !U = 0

from the top, we find that

and becomes

Principle of conservation of mechanical energy:


Conservation of Mechanical Energy

K + U is constant all the time

ex) a pendulum


Potential Energy Curve

We had this before: !U = !

!xf

xi

F (x)dx

and it becomes at the differential limit: F (x) = !

dU(x)

dx(obtaining the force from the potential)

ex) elastic potential U(x) =1

2kx

2

U(y) = mgy F = !mg

F (x) = !kx

gravitational potential


Potential Energy CurvePotential Energy Curve

unstable equilibrium(U = 3J)

neutral equilibrium(U = 4J)

stable equilibrium(U = 1J)


Work Done on a System by an External Force

Work is energy transferred to or from a system by means of an external force acting on that system

Definition of work with external force:

No friction involved:

W = !Emec

Work done on the system is equal to the change in the mechanical energy


Work Done on a System by an External ForceFriction involved: A constant force F pulls a block, increasing

the block’s velocity from v0 to v

From Newton’s 2nd law, we write

acceleration is constant as forces are constant, so we can use the following equation

v2

= v2

0 + 2ad

F ! fk = ma

a =1

2d(v2

! v2

0)

Fd =1

2mv2

!

1

2mv2

0 + fkd

and,

Fd = !K + fkd

Fd = !Emec + fkd

it becomes

(if we include the vertical motion as well)

!Eth(increase in thermal energy by sliding)


Conservation of EnergyThe total energy E of a system can change only by amounts of energy that are transferred to or from the system

Isolated system: there can be no energy transfers to or from the isolated system

The total energy E of an isolated system cannot change

W = !E = !Emec + !Eth + !Eint

!Emec + !Eth + !Eint = 0

(internal Energy)


The Center of Mass

The motion of the baseball bat looks complicated? Yes, but there is a special point moves like a particle (parabolic path) called the center of mass


The Center of Mass

The motion of the baseball bat looks complicated? Yes, but there is a special point moves like a particle (parabolic path) called the center of mass

For a system of two particles shown in the figure left, the center of mass is defined as

xcom =m1x1 + m2x2

m1 + m2


The Center of Mass


The Center of Mass

For a system of n particles along the x axis:(M: total mass of the system)

M = m1 + m2 + ... + mn


The Center of Mass

xcom =m1x1 + m2x2 + ... + mnxn

M

=1

M

n!

i=1

mixi


M = m1 + m2 + ... + mn


The Center of Mass


M

=1

M

n!

i=1

mixi


M = m1 + m2 + ... + mn

If the particles are distributed in three dimensions, the center of mass must be identified by three coordinates


The Center of Mass


M

=1

M

n!

i=1

mixi


M = m1 + m2 + ... + mn

xcom =1

M

n!

i=1

mixi, ycom =1

M

n!

i=1

miyi, zcom =1

M

n!

i=1

mizi,



The Center of Mass


M

=1

M

n!

i=1

mixi


M = m1 + m2 + ... + mn

xcom =1

M

n!

i=1

mixi, ycom =1

M

n!

i=1

miyi, zcom =1

M

n!

i=1

mizi,


!rcom =1

M

n!

i=1

mi!riEquivalent single vector equation:


The Center of Mass


The Center of MassSolid Bodies

An ordinary object contains many particles (atoms) that we treat it as a continuous distribution of matter. The coordinates of the center of mass are defined as




xcom =1

M

!x dm ycom =

1

M

!y dm zcom =

1

M

!z dm




xcom =1

M

!x dm ycom =

1

M

!y dm zcom =

1

M

!z dm

! =

dm

dV=

M

V

If the object has the uniform density (ρ = constant)




xcom =1

M

!x dm ycom =

1

M

!y dm zcom =

1

M

!z dm

! =

dm

dV=

M

V

If the object has the uniform density (ρ = constant)

xcom =1

V

!x dV ycom =

1

V

!y dV zcom =

1

V

!z dV

The coordinates of the center of mass become:


Newton’s 2nd Law for a System of ParticlesFor the system of particles, the vector equation that governs the motion of the center of mass of such a system of particle is:

The center of mass of the fragments would continue to follow the original parabolic path




!Fnet = M!acom




!Fnet = M!acom

Proof) From the definition of rcom, we write

M!rcom = m1!r1 + m2!r2 + ... + mn!rn !rcom =1

M

n!

i=1

mi!ri




!Fnet = M!acom

M!acom = m1!a1 + m2!a2 + ... + mn!an

Differentiating the above with respect to time twice leads to



M

n!

i=1

mi!ri




!Fnet = M!acom

M!acom = m1!a1 + m2!a2 + ... + mn!an

Differentiating the above with respect to time twice leads to

M!acom = !F1 + !F2 + ... + !Fn

= !Fnet

From Newton’s 2nd law we know that thusmi!ai =!Fi



M

n!

i=1

mi!ri


Linear Momentum


Linear Momentum

!p = m!vLinear Momentum of a particle:

(We will discuss angular momentum later)


Linear Momentum



The time rate of change of the momentum of a particle is equal to the net force acting on the particle and is in the direction of that force

!Fnet =

d!p

dt


Linear Momentum




!Fnet =

d!p

dt

Linear Momentum of a System of Particles:

!P = !p1 + !p2 + ... + !pn

= m!v1 + m!v2 + ... + m!vn


Linear Momentum




!Fnet =

d!p

dt

Linear Momentum of a System of Particles:

!P = !p1 + !p2 + ... + !pn

= m!v1 + m!v2 + ... + m!vn

!P = M!vcom

From the argument in the previous slide (differentiating with respect to time only once)

!Fnet =

d!P

dt

or (system of particles)


Conservation of Linear MomentumSuppose that the net force acting on an isolated system of particles is zero



!Fnet = 0d!P

dt= 0 !P = constant



!Fnet = 0d!P

dt= 0 !P = constant

In words, if no net force acts on an isolated system of particles, the total linear momentum of the system cannot change

Law of conservation of linear momentum:

It can also be written as:

!Pi =!Pf

!

total linear momentum

at some initial time ti

"

=

!

total linear momentum

at some later time tf

"

or


Momentum and Kinetic Energy in CollisionsIf the total kinetic energy of the system of two colliding bodies is unchanged by the collision, it is called an elastic collision

ex) collision of billiard balls

If the kinetic energy of the system is not conserved, it is called an inelastic collision

In a closed, isolated system containing a collision, the total linear momentum cannot change, whether the collision is elastic or inelastic

(another statement of the law of conservation of linear momentum)

ex) collision of two cars


Inelastic Collisions in One Dimension

We assume the two bodies form our system, which is closed and isolated

!

total momentum !Pi

before the collision

"

=

!

total momentum !Pf

after the collision

"

!p1i + !p2i = !p1f + !p2f

m1v1i + m2v2i = m1v1f + m2v2f

or

completely inelastic collision (two particles stick together):

m1v1i = (m1 + m2)V

V =m1

m1 + m2

v1ior

(V is less than v1i)


Inelastic Collisions in One Dimension

In a closed, isolated system, the velocity of the center of mass cannot be changed by a collision (no net external force)

after the collision

!P = M!vcom = (m1 + m2)!vcom

!P = !p1i + !p2i

!vcom =

!P

m1 + m2

=!p1i + !p2i

m1 + m2

before the collision

from the above:

(constant)


Elastic Collisions in One Dimension: total kinetic energy of the system does not change after a collision

Stationary targetm1v1i = m1v1f + m2v2f

1

2m1v

2

1i =1

2m1v

2

1f +1

2m2v

2

2f

(linear momentum)

(kinetic energy)

m1(v1i ! v1f ) = m2v2f

m1(v1i ! v1f )(v1i + v1f ) = m2v2

2f

above two equations can be rewritten as

dividing the latter from the former we get

(v1i + v1f ) = v2f

eliminating v2f from the above

m1v1i ! m1v1f = m2v1i + m2v1f v1f =m1 ! m2

m1 + m2

v1i

v2f =2m1

m1 + m2

v1isimilarly we get


Elastic Collisions in One Dimension

v1f =m1 ! m2

m1 + m2

v1i

v2f =2m1

m1 + m2

v1i

we obtained

Equal mass: m1 = m2 v1f = 0 and v2f = v1ithen

ex)

A massive target: m2 ! m1 then v1f ! "v1i and v2f !

2m1

m2

v1i

A massive projectile: m1 ! m2 v1f ! v1i and v2f ! 2v1ithengolf ball

bowling ball


Elastic Collisions in One Dimension

Moving Target: m1v1i + m2v2i = m1v1f + m2v2f

1

2m1v

2

1i +1

2m2v

2

2i =1

2m1v

2

1f +1

2m2v

2

2f

m1(v1i ! v1f ) = !m2(v2i ! v2f )

m1(v1i ! v1f )(v1i + v1f ) = !m2(v2i ! v2f )(v2i + v2f )

(linear momentum)

(kinetic energy)

rewriting above equations we get

dividing the latter from the former we get (v1i + v1f ) = (v2i + v2f )

m1(v1i ! v1f ) + m2v2i = m2v2f

v1f =m1 ! m2

m1 + m2

v1i +2m2

m1 + m2

v2i v2f =2m1

m1 + m2

v1i +m2 ! m1

m1 + m2

v2i

since

similar way

m2(v1i + v1f ) = m2v2i + m1(v1i ! v1f ) + m2v2i

Finally solving for v1f:


System with Varying Mass: A RocketWe discuss now a system with varying mass. An example would be a rocket: the most confusing problem in my class

In a) M: mass of the rocket v: speed of the rocket at an arbitrary time t

In b) M + dM: mass of the rocket (dM is negative quantity) v + dv : speed of the rocket at an arbitrary time t+dt -dM : mass of the exhaust products U : velocity of the exhaust products

Linear momentum of the system must be conserved during dt: (assume no gravitational force)

Pi = Pf

Mv = !dM U + (M + dM)(v + dv)

(we observe the rocket from the ground)


System with Varying Mass: A Rocket

vrel: relative speed between the rocket and the exhaust products

In symbols, (v + dv) = vrel + U or

U = v + dv ! vrel

Mv = !dM U + (M + dM)(v + dv)

Substituting this result for U into the below equation (that we derived just before)

Mv = !dM(v + dv ! vrel) + (M + dM)(v + dv)

= !dMv ! dMdv + dMvrel + Mv + Mdv + dMv + dMdv

!dMvrel = Mdv

!

dM

dtvrel = M

dv

dt

or

!

velocity of rocketrelative to us

"

=

!

velocity of rocketrelative to products

"

+

!

velocity of productsrelative to us

"


System with Varying Mass: A Rocket

We replace dM/dt (the rate at which the rocket loses mass) by -R (R: positive mass rate of fuel consumption)

!

dM

dtvrel = M

dv

dt

Then, the previous equation becomes

Rvrel = Ma

thrust

Velocity?

dv = !vrel

dM

M! vf

vi

dv = !vrel

! Mf

Mi

dM

M

vf ! vi = vrel lnMi

Mf

introductory physics-i part 3particle.korea.ac.kr/class/2011/introphy/introphysics-iii.pdf · 1 kg...

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