phy2053 lecture 11 conservation of energy - department of physics · pdf file ·...

PHY2053 Lecture 11Conservation of Energy

Conservation of EnergyKinetic Energy

Gravitational Potential Energy

• Symmetry - fundamental / descriptive property of the Universe itself [“vacuum”]

• Laws of Physics are the same at any point in space [“translational invariance”]

• Conservation of Momentum [Ch 7]

• Laws of Physics are the same at any point in time [“time invariance”]

• Conservation of Energy [today’s lecture]

PHY2053, Lecture 11, Conservation of Energy

Symmetries in Physics

2

Colloquial:“Symmetric”

Physics term:“Parity”


More practical aspect• there are different, mathematically equivalent ways to

formulate Newton’s laws• all these calculations predict certain quantities will be

conserved for a “closed” system (0 net external force)• energy, momentum, angular momentum ..• existence of conserved quantities simplifies otherwise

complicated calculations • Key concepts:• learn to recognize and exploit conserved quantities• conserved quantities derived from Newton’s laws• solutions immediately satisfy Newton’s laws

3


Energy Conservation• term “closed system” means: no net external force is

acting upon any element of the system

• The total energy of a closed system does not change over time: total energy before = total energy after

• textbook implies that the Universe is a closed system“The total energy in the Universe is unchanged by any physical process”

• next: define change of energy (work), energy itself

4

Concept of Work• colloquial meaning of work:

effort which produces a result.• analogy in terms of mechanics:• Effort → Force, F• Result → Displacement ∆r• interested in displacement due to force

• angular term cos(θ) projects force ↔ displacement

• SI unit: Joule [ J ]; relation to calorie: 1 cal = 4.2 JPHY2053, Lecture 11, Conservation of Energy 5

θ

∆r

F

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi = 0 (6)

X

i

Fi�r cos(�i) = 0 (7)

�rX

i

Fi cos(�i) = 0 (8)

X

i

Wi = (9)

1


Work: signed scalar quantity• Work can be positive, negative, and zero depending

on the orientation of the force to the displacement

6

θF

∆r

θ = 90°

F ∆r θF ∆r

θ < 90°cosθ > 0W > 0

θ = 90°cosθ = 0W = 0

θ > 90°cosθ < 0W < 0

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi = 0 (6)

X

i

Fi�r cos(�i) = 0 (7)

�rX

i

Fi cos(�i) = 0 (8)

X

i

Wi = (9)

Ugrav = �Gm1 m2

r(10)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi = 0 (6)

X

i

Fi�r cos(�i) = 0 (7)

�rX

i

Fi cos(�i) = 0 (8)

X

i

Wi = (9)

Ugrav = �Gm1 m2

r(10)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi = 0 (6)

X

i

Fi�r cos(�i) = 0 (7)

�rX

i

Fi cos(�i) = 0 (8)

X

i

Wi = (9)

Ugrav = �Gm1 m2

r(10)

1


Total Work in a Closed System • start with total work on a particular object

7

• recall the definition of a closed system

• vector sum, has to be zero in all directions

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi = 0 (6)

X

i

Fi�r cos(�i) = 0 (7)

�rX

i

Fi cos(�i) = 0 (8)

X

i

Wi = (9)

Ugrav = �Gm1 m2

r(10)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi = 0 (6)

X

i

Fi�r cos(�i) = 0 (7)

�rX

i

Fi cos(�i) = 0 (8)

X

i

Wi = (9)

Ugrav = �Gm1 m2

r(10)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi = 0 (6)

X

i

Fi�r cos(�i) = 0 (7)

�rX

i

Fi cos(�i) = 0 (8)

X

i

Wi = (9)

Ugrav = �Gm1 m2

r(10)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi = 0 (6)

X

i

Fi�r cos(�i) = 0 (7)

�rX

i

Fi cos(�i) = 0 (8)

X

i

Wi = (9)

Ugrav = �Gm1 m2

r(10)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi = 0 (6)

X

i

Fi�r cos(�i) = 0 (7)

�rX

i

Fi cos(�i) = 0 (8)

X

i

Wi = (9)

Ugrav = �Gm1 m2

r(10)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi

= 0 (6)

X

i

Fi

�r cos(�i

) = 0 (7)

�rX

i

Fi

cos(�i

) = 0 (8)

X

i

Wi

= (9)

Ugrav

= �Gm1 m2

r(10)

W = Fx

�x = max

�x (11)

2 ax

�x = v2f,x

� v2i,x

(12)

ax

�x =v2

f,x

2�

v2i,x

2(13)

W = max

�x = m

v2

f,x

2�

v2i,x

2

!

= mv2

f,x

2�m

v2i,x

2(14)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi

= 0 (6)

X

i

Fi

�r cos(�i

) = 0 (7)

�rX

i

Fi

cos(�i

) = 0 (8)

X

i

Wi

= (9)

Ugrav

= �Gm1 m2

r(10)

W = Fx

�x = max

�x (11)

2 ax

�x = v2f,x

� v2i,x

(12)

ax

�x =v2

f,x

2�

v2i,x

2(13)

W = max

�x = m

v2

f,x

2�

v2i,x

2

!

= mv2

f,x

2�m

v2i,x

2(14)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi

= 0 (6)

X

i

Fi

�r cos(�i

) = 0 (7)

�rX

i

Fi

cos(�i

) = 0 (8)

X

i

Wi

= (9)

Ugrav

= �Gm1 m2

r(10)

W = Fx

�x = max

�x (11)

2 ax

�x = v2f,x

� v2i,x

(12)

ax

�x =v2

f,x

2�

v2i,x

2(13)

W = max

�x = m

v2

f,x

2�

v2i,x

2

!

= mv2

f,x

2�m

v2i,x

2(14)

K = mv2

2(15)

W = Kf

�Ki

= �K (16)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi

= 0 (6)

X

i

Fi

�r cos(�i

) = 0 (7)

�rX

i

Fi

cos(�i

) = 0 (8)

X

i

Wi

= (9)

Ugrav

= �Gm1 m2

r(10)

W = Fx

�x = max

�x (11)

2 ax

�x = v2f,x

� v2i,x

(12)

ax

�x =v2

f,x

2�

v2i,x

2(13)

W = max

�x = m

v2

f,x

2�

v2i,x

2

!

= mv2

f,x

2�m

v2i,x

2(14)

K = mv2

2(15)

W = Kf

�Ki

= �K (16)

1

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi

= 0 (6)

X

i

Fi

�r cos(�i

) = 0 (7)

�rX

i

Fi

cos(�i

) = 0 (8)

X

i

Wi

= (9)

Ugrav

= �Gm1 m2

r(10)

W = Fx

�x = max

�x (11)

2 ax

�x = v2f,x

� v2i,x

(12)

ax

�x =v2

f,x

2�

v2i,x

2(13)

W = max

�x = m

v2

f,x

2�

v2i,x

2

!

= mv2

f,x

2�m

v2i,x

2(14)

K = mv2

2(15)

W = Kf

�Ki

= �K (16)

1PHY2053, Lecture 11, Conservation of Energy

Kinetic Energy, Definition

8

• consider impact of work on the velocity of an object• start from 1D motion, works in all three (x, y, z)

Work Energy Theorem


Example #1: Mass DriverA mass driver is a device which uses magnetic fields to accelerate a container (mass). Predicted commercial uses include launching people and cargo to bases on the Moon. The common way to specify mass drivers is to quote the kinetic energy that an object will have when leaving the driver, if it started from rest. For a 1 MJ mass driver, compute the muzzle velocity of

a) a 0.5 kg projectileb) a 50 kg projectile

9

PHY2053, Lecture 11, Conservation of Energy 10

Mass driver notes pt 1


Mass driver notes pt 2


Gravitational Potential EnergyNear Earth

• near Earth, the usual orientation of coordinate systems is so that the positive y axis points “up”• the force of gravity has only one component,

in the y-direction: Fy = −mg • only y displacement, ∆y matters for computing work:

W = FG,y×∆y = −mg × ∆y• consider a vertical shot upwards, vf = 0• W = ∆K = Kf − Ki = 0 − ½mvf2, also = −mg × ∆y• gravity did negative work, “removing” kinetic energy

12


Energy Conservation Law

• where did the kinetic energy go? temporarily stored in gravitational field• define potential energy ∆Ugrav = −Wgrav = mg × ∆y• computes how much kinetic energy could be

released if we let gravity work across ∆y

• work-energy theorem: W = ∆K; ∆K − W = 0• ∆K + ∆U = 0 → ∆( K + U ) = 0• sum of kinetic and potential energy does not change• define E = K + U, then E is constant in time

13


Choice of Zero Point, Near Earth

• Due to conservation of energy, only changes in potential energy are really relevant for kinematics• The absolute value of potential energy at a point in

space is arbitrary - up to an additive constant• We have the freedom to pick a convenient point in

space and declare that the potential energy at that point equals 0 J• All other potential energies are then computed

relative to that point, based on ∆U = U(y) − U(0)• U(y) = ∆U + U(0) = mg × ∆y + 0 = mg × (y − 0)

14


Example #1: RollercoasterA roller-coaster is barely moving asit starts down a ramp of height h. The first figure it encounters is aloop of radius R. How high must theramp be so that the roller-coaster never loses contact with the rails?

15

Rh


Rollercoaster notes pt 1


Rollercoaster notes pt 2

Comment: Given that the total height of the loop is 2R, this is not really much taller than the loop itself. The ratio of the height of the ramp and the height of the loop is 2.5R / 2R = 1.25 - the ramp has to be only 25% taller than the loop for the rollercoaster to clear the highest point in the loop and stay in contact with the rails.


More Realistic: Dissipative (Non-conservative) Forces

• friction converts mechanical energy into heat• heat does not “store” mechanical energy• therefore, there is no point in defining a “heat” or

“frictional” potential energy • friction always opposes motion, so Wfriction < 0

• extend the law of energy conservation to account for non-conservative forces:

(Ki + Ui) + WNC = (Kf + Uf)

18

• derivation requires math beyond baseline calculus

• for gravitational potential at planetary scales, there already exists a “usual” convention:• potential energy infinitely far away from a planet is = 0• convention: an object with positive total energy can

“escape” a planet (will not fall back to the planet)• allows easy computation of “escape” velocities for

objects starting from any R from the planet’s center


Gravitational Potential Energy, Planetary Scales

19

X⇥F = m⇥ ⇥a (1)

F = Gm1 m2

r2(2)

⇥F2,1 (3)

⇥F1,2 (4)

W = F �r cos(�) (5)

X

i

⇥Fi = 0 (6)

X

i

Fi�r cos(�i) = 0 (7)

�rX

i

Fi cos(�i) = 0 (8)

X

i

Wi = (9)

Ugrav = �Gm1 m2

r(10)

1


Example #2: Hyperbolic CometA comet not bound to the Sun will only pass by the Sun once. It will trace a hyperbolic trajectory through the Solar system. Compute the minimum velocity of a hyperbolic comet when it is roughly 1 A.U. away from the Sun. The mass of the Sun is MS = 2×1030

kg. 1 Astronomical Unit is the distance from the Earth to the Sun, 150 million km. Does the velocity depend on the mass of the comet?

20


Hyperbolic Comet notes

Next Lecture:

Hooke’s Law, Elastic Potential Energy

Power

phy2053 lecture 11 conservation of energy - department of physics · pdf file ·...

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