Chap A

Chap. 4: Work and Energy

R i s h i k e s h V a i d y a

Theoretical Particle Physics

Office: 3265

[email protected]

Physics Group, B I T S Pilani

September 4, 2010

Outline

1 Work Energy Theorem

2 Potential Energy

3 Non Conservative Force

4 Conservation Laws and Particle Collisions

A Small Story and a Lesson

A worried father of a BITSian who was not

doing well, asked two of his son’s friends

to keep an eye on him.

Friend A (from Humanities) kept

Detailed record of his expenditure

His interactions, clubs, department

activities

His Attendance

Affluence of Inkahol


A worried father of a BITSian who was not

doing well, asked two of his son’s friends

to keep an eye on him.

Friend B (from Physics)

Didn’t bother much about the details like

friend A and yet at the end of the

semester gave a better explanation of his

friend’s CGPA.


Two complementary approaches to

mechanics

Take the bull by the horn: The

dynamical methods of Chap.2

Tame the Bull: Exploit conservation

laws as in Chap. 3 and 4

The Fundamental Problem of Mechanics

Given all forces, solve

~F = m~a

to find ~r(t).

Problem!

What we know is F(x) and not F(t).

Solve half the problem!

md~v

dt= ~F(~r)

Equation of Motion in One Dimension

md2x

dt2= F(x) or m

dv

dt= F(x)

Integrate with respect to x

m

∫ xb

xa

dv

dtdx =

∫ xb

xa

F(x)dx

Change variable from x to t using

differentials:

dx =

(dx

dt

)

dt = vdt


md2x

dt2= F(x) or m

dv

dt= F(x)

Integrate with respect to x

m

∫ xb

xa

dv

dtdx=m

∫ tb

ta

dv

dtvdt

=m

∫ tb

ta

d

dt

(1

2v2

)

dt =1

2mv2

∣∣∣∣

ta

tb

=1

2mv2

b − 1

2mv2

a


md2x

dt2= F(x) or m

dv

dt= F(x)

Work-Energy Theorem:

1

2mv2

b − 1

2mv2

a =

∫ xb

xa

F(x)dx


md2x

dt2= F(x) or m

dv

dt= F(x)

Work-Energy Theorem: Motion in 3-D

1

2mv2

b − 1

2mv2

a =

∫ b

a

~F · d~r

How Useful is Work-Energy Theorem

Evaluation of∫ b

a~F · d~r is tentamount to

knowing the trajectory beforehand !Two cases where Work-EnergyTheorem is useful.

◮ Conservative Forces: For many

interesting forces∫ b

a~F · d~r depends

only on end-points !◮ Constrained Motion: External

Constraints pre-determine the

trajectory.

How Useful is Work-Energy Theorem

Regardless of whether forces are

conservative or not the Work-

Energy theorem is always true

Prob. 4.2 A Block of mass M slides

along a horizontal table with speed v0. At

x = 0 it hits a spring with spring constant

k and begins to experience a friction force.

The coefficient of friction is variable and is

given by µ = bx, where b is constant. Find

the loss in mechanical energy when the

block has first come momentarily to rest.

Prob. 4.2

Example 4.6 A pendulum consists of a

light rigid rod of length l, pivoted at one

end and with mass attached at the other

end. The pendulum is released from rest

at an angle φ0. What is the velocity of m

when the rod is at angle φ.

Example 4.6

Prob.4.5 Mass m whirls on a frictionless

table, held to circular motion by a string

which passes through a hole in the table.

The string is slowly pulled through the

hole so that the radius of the circle

changes from �l1 to �l2. Show that the work

done in pulling the string equals the

increase in kinetic energy of the mass.

Prob.4.5 Mass m whirls on a frictionless

table, held to circular motion by a string

which passes through a hole in the table.

The string is slowly pulled through the

hole so that the radius of the circle

changes from �l1 to �l2. Show that the work

done in pulling the string equals the

increase in kinetic energy of the mass.

The fundamental mystery here is ... how can

you pull radially and still end up changing the

angular velocity of the mass.

Solution 4.5 For cicular motion:

Fr = mv21

�l1

= m �l1ω21

If Fr is increased, m will move to smaller r:

Fr(r) = mv2(r)

r= mrω2(r)

Evaluate:

W = −∫

�l2

�l1

Fr(r)dr = −∫

�l2

�l1

mrω2(r)dr

~a = (r − rθ2)r + (rθ + 2rθ)︸︷︷︸

=0

θ

rω + 2rω=0 [θ = ω]1

ω

dω

dt=−2

1

r

dr

dt∫ ω(r)

ω1

dω

ω=−2

∫ r

�l1

dr

r⇒ ω(r) = �l

2

1

ω1

r2

W = −∫ �l2

�l1

Fr(r)dr = −∫ �l2

�l1

mrω2(r)dr

W=−m�l

4

1ω21

∫�l2

�l1

dr

r3=

1

2m�l

4

1ω21

1

�l

2

2

− 1

�l

2

1

=1

2m

�l

4

1ω21

�l

2

2

− �l

2

1ω21

W=1

2m

[

(�l2ω2)2 − (�l1ω1)

2] [

ω2 �l

2

2 = ω1 �l

2

1

]

=K2 − K1

Prob. 4.7 A ring of mass M, hangs from a

thread, and two beads of mass m slide on

it without friction. The beads are released

simultaneously from the top of the ring

and slide down opposite side. Show that

the ring will start to rise if m > 3M/2, and

find the angle at which this occurs.

Forces are different for cosθ > 2/3 and

cos θ < 2/3

cos θ, N and T vs. θ (m = 2M)

N and T vs. θ: Magnified view

2N cos θ vs. θ

Accelaration of the ring vs. θ

Prob. 4.10

A block of mass M on a horizontal

frictionless table is connected to a spring

(spring constant k). The block is set in

motion so that it oscillates about its

equilibrium position with a certain

amplitude A0. The period of motion is

T0 = 2π√

M/k.

Prob. 4.10

(a) A lump of sticky putty of mass m is

dropped onto the block. the putty

sticks without bouncing. The putty hits

M at the instant when the velocity of

M is zero. Find1 The new period

2 The new amplitude3 The change in mechanical energy of the

system

(b) Repeat part a, assuming that the sticky

putty hits M at the instant when M has

its maximum velocity.

Potential Energy

For a conservative Force Field:

∫ ~rb

~ra

~F · d~r= func. of (~rb) − func. of (~ra)

Wba =−U(~rb) + U(~ra)

Wba =Kb − Ka [Work − energyTh.]

Wba =−Ub + Ua = Kb − Ka

Ua + Ka = Ub + Kb = E

What Does Potential Energy Tell us About Force?

Ub − Ua = −∫ xb

xa

F(x)dx

U(x + ∆x) − U(x) = ∆U

∆U=−∫ x+∆x

x

F(x)dx

∆U≈−F(x)(x + ∆x − x) = −F(x)∆x

F(x)≈−∆U

∆xor F(x)

lim ∆x→0= −dU

dx

Potential Energy Determins Stability of a System

Harmonic Oscillator: U = kx2/2


Simple Pendulum:

U(θ) = mg�l(1 − cos θ) and dUdθ

= mg�l sin θ


General Stability Criteria:

Rock me Spin me yet I Stand Tall:

The Amazingly Stable Teeter-Toy

A Sports Car

Stability requires low center of mass and

hence the peculiar design of a sports car.

Non-Conservative Forces

~F = ~Fc + ~Fnc

Wtotalba =

∫ b

a

~F · d~r

Wtotalba =

∫ b

a

~Fc · d~r +

∫ b

a

~Fnc · d~r

Kb − Ka =−Ub + Ua + Wncba

Kb + Ub − (Ka + Ua) = Wncba

Eb − Ea = Wncba

Power: Time rate of doing Work

P =dW

dt= ~F · d~r

dt= ~F ·~v

Units:

[S.I.] 1 W=1J/s [CGS]1 erg/s = 10−7W

[English] 1 hp=550 lb ft/s ≈ 746W

Power: Time rate of doing Work

Typical Power Consumption

Man running upstairs: 1/2 to 1 hp for

30 s

A husky man can do work over a period

of 8 hours only at a rate of

0.2 hp ≈ 1000 Kcal

Per person energy use: India (0.7 kW),

Germany (6 kW), USA (11.4 kW)

Prob. 4.18 A 160 lb man leaps into the

air from a crouching position. His center

of gravity rises 1.5 ft before he leaves the

ground, and it then rises 3 ft to the top of

his leap. What power does he develop

assuming that he pushes the ground with

constant force?

Sol.4.18

P=W/T (W = work done by N)

W=N · 1.5 (c.g. rises by 1.5ft)

N=mg + ma or N = 160 +160

32a

a=v2

2s= 64

[

v =√

2gs′ =√

2 · 32 · 3 = 8√

3]

N=480 lb W = 720 lb.ft T = v/a =√

3/8

P = W/T = 3325lb.ft/s ≈ 6hp

Prob. 4.19

In the preceeding problem take

F = F0 cos ωt where F0 is the peak force,

and the contact with ground ends at

ωt = π/2. Find the peak power that the

man develops during the jump.

Sol.4.19

P(t)=N(t)v(t) [N(t) = −F(t)]

N(t) − mg=ma(t)

m

∫ v(t)

0

dv=

∫ t

0

(F0 cos ωt − mg) dt

v(t)=F0

mωsin ωt − gt [F0, ω?]

x(t)=F0

mω2(1 − cos ωt) − 1

2gt2

Sol.4.19

Boundary conditions:

v(t = π/2ω)=8√

3 =F0

mω− gπ

2ω

x(t = π/2ω)=1.5 ft =F0

mω2− gπ2

8ω2

Sol.4.19

ω = 9.96s−1 F0 = 832 lb t =π

2ω= 0.16s

P(t)=F(t)v(t)

F0

2mω

F0 sin 2ωt︸︷︷︸

1

− 2mgωt cos ωt︸︷︷︸

2

A reasonable approximation: F0 >> mg

then 1st >> 2nd

Sol.4.19

P(t)≈ F20

2mωsin 2ωt

For Pmax. :dP

dt=0

dP

dt=

F20

mcos 2ωt = 0

Pmax. = P|t= π4ω

=F2

0

2mωsin

(

2ωπ

4ω

)

=F2

0

2mω

Sol.4.19

Check:

d2P

dt2

∣∣∣∣∣t= π

4ω

=−F2

0

m2ω sin 2ωt

∣∣∣∣∣t= π

4ω

< 0

Sol.4.19

Prob. 4.20 A uniform rope of

mass λ per unit length is

coiled on a smooth horizontal

table. One end is pulled

straight up with constant

speed v0. (a) Find the force

exerted on the end of the rope as a

function of height y. (b) Compare

the power delivered to the rope

with the rate of change of the

rope’s total mechanical energy.

Sol.4.20

Fext. =Mdv

dt− vrel.

dM

dt

F − Mg=v0

dM

dt[v = v0 vrel. = −v0]

F=λyg + λv20

P=Fv0 = λygv0 + λv30

E=1

2mv2

0 + mgy

2=

1

2λyv2

0 + λgy2

2dE

dt=

1

2λv3

0 + λyv0g

dEdt

is 12λv3

0 short of P because that is the amount

dissipated as heat.

What do collisions teach us?

~F = m~a is a double-edge sword

Collisions studies gave most profound

knowledge about fundamental Physics

Constraints from energy and

momentum conservation severe enough

to extract vital information about

scattering

LHC@CERN Geneva

Fermilab@Illinois USA

Classical Collisions

A + B → C + D

Mass is conserved:

mA + mB → mC + mD

Momentum is conserved:

pA + pB = pC + pD

K.E. may or may not be conserved

Classical Collisions

A + B → C + D

Types of Collisionssticky: K.E. decreases

KA + KB > KC + KD

Explosive: K.E. increases

KA + KB < KC + KD

Elastic: K.E. is conserved

KA + KB = KC + KD

Prob. 4.23 A small ball of

mass m is placed on top of the

a “superball” of mass M, and

the two balls are dropped to

the floor from height h.How

high does the small ball rise

after the collision? Assume

that the collision is perfecly

elastic, and that m << M.

Sol. 4.23 Let v1 and v2 be the initial

velocities of m and M before collision and

v′1 and v′

2, after collision. In a two body

one dimensional collision:

v′1 =

(m − M)v1 + 2Mv2

m + M

v′2 =

(M − m)v2 + 2mv1

m + M

Here: v1 = −√2gh, v2 =

√2gh, and

M >> m

Sol. 4.23

v′1 =

3M · √2gh

M= 3

√

2gh

If m rises to height h′ after collision:

1

2m1v

′21 = m1gh′

h′ =v

′21

2g= 9h

Problem 4.9 A “superball” of mass m bounces

back and forth between two surfaces with speed v0.Gravity is neglected and collisions are perfectly

elastic.

(a) Find the average force F on each wall.

(b) If one surface is moving uniformly toward the

other with speed V << v0, the bounce rate willincrease due to shorter distance between

collisions, and because the ball’s speed increaseswhen it bounces from the moving surface. Find

the F in terms of separation of surfaces, x.

(c) Show that work needed to push the surface

from l to x equals gain in kinetic energy of theball.

Problem 4.29

Sol. 4.29

(a) Momentum transfer to wall in 1

bounce: m(2v0)

No. of bounce per unit time: v0/2�l

F = 2mv0

v0

2�l

=mv2

0

�l

(b) F(x) = mv(x)2

x

Integrate a(x) = dv(x)dt

to find v(x).

Sol. 4.29

(b) ∆v after each bounce ∆v = 2V

(Sling-shot effect)

This change happens in time ∆t = 2x/v

dv

dt=

Vv

x∫ v

v0

dv′

v′ =

∫ t

0

Vdt

x= −

∫ x

�l

dx′

x′

v(x) =v0 �l

x⇒ F =

mv(x)2

x=

m�l

2v20

x3

Sol. 4.29

(c) Work done in moving the surface:

W=−∫ x

�l

F(x′) dx′ = −m�l

2v20

∫ x

�l

dx′

x′3

=m�l

2v20

2

[

1

x2− 1

�l

2

]

=m�l

2v20

2x2− 1

2mv2

0

=1

2mv(x)2 − 1

2mv2

0

=Change in K.E. of ball

Sling-Shot (Gravity-Assist) and NASA’s Cassini

Collisions and Center of Mass Coordinates

How does it help?

Total momentum in C-system is zero

Initial and final velocities lie in the

same plane

Each particle is scattered through the

same angle θ in the plane of scattering.

For elastic collisions, the speed of each

particle is same before & after the

collision


m1 & m2 having velocityv1 & v2.

V =m1v1 + m2v2

m1 + m2


v1c =v1 − V

=m2

m1 + m2

(v1 − v2)

v2c =v2 − V

=− m1

m1 + m2

(v1 − v2)


The momenta inC-system:

p1c =m1v1c

=m1m2

m1 + m2

(v1 − v2)

=µv

p2c =−µv

0=p1c + p2c

C-System Plane of Scattering

m1 and m2 striking with

v1 & v2.

Initial velocities in L &

C-system.


m1 and m2 striking with

v1 & v2. Final velocities in L &

C-system (The plane in

general is different)


C-system plane of

scattering.

For elastic scattering,

v1c = v′1c and v2c = v′

2c.

Velocity vectors simple

rotate.

If m2 is at rest

V=m1

m1 + m2

v1

v1c =v1 − V

=m2

m1 + m2

v1

v2c =−V

=− m1

m1 + m2

v1

Limitations on Lab scattering angle

λ = m1m2

= Vv1c

< 1

θ : 0 → π θ1 : 0 → π


λ = m1m2

= Vv1c

= 1

θ : 0 → π θ1 : 0 → π/2


λ = m1m2

= Vv1c

> 1

θ : 0 → π θ1 : 0 → θmax.1

Look at solved example 4.19 to understand

limitations on laboratory scattering angle.

It will also help you solve problem 4.30

Chap A

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