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Topic B: Momentum

CONTENT

1. Newton’s laws of motion

2. Equivalent forms of the equation of motion

2.1 Force, impulse and energy

2.2 Derivation of the equations of motion for particles

2.3 Examples

3. Conservation of momentum

3.1 Momentum in collisions

3.2 Restitution

3.3 Oblique collisions

4. Friction

4.1 Types of friction

4.2 Coefficient of friction

4.3 Object on a plane

5. Systems of particles and finite objects

5.1 Dynamics of a system

5.2 The centre of mass

5.3 Important properties of the centre of mass

5.4 Finding the centre of mass

5.5 Some useful centres of mass

Philosophiae Naturalis Principia Mathematica

Lex I: Corpus omne perseverare in statu suo quiescendi vel movendi

uniformiter in directum, nisi quatenus a viribus impressis cogitur statum

illum mutare.

Lex II: Mutationem motus proportionalem esse vi motrici impressae, et

fieri secundum lineam rectam qua vis illa imprimitur.

Lex III: Actioni contrariam semper et æqualem esse reactionem: sive

corporum duorum actiones in se mutuo semper esse æquales et in partes

contrarias dirigi.

Every body perseveres in its state of rest, or of uniform motion in a right line,

unless it is compelled to change that state by forces impressed thereon.

The alteration of motion is ever proportional to the motive force impressed; and is

made in the direction of the right line in which that force is impressed.

To every action there is always opposed an equal reaction: or the mutual actions

of two bodies upon each other are always equal, and directed to contrary parts.

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NEWTON’S LAWS OF MOTION

Law 1. Every particle continues in a state of rest or

uniform motion in a straight line unless

acted on by a resultant external force.

Law 2. When acted upon by an unbalanced force F

a particle of mass 𝑚 accelerates at a rate a

such that F = 𝑚a.

Law 3. The forces of action and reaction between

interacting particles are equal in magnitude,

opposite in direction and collinear.

NEWTON’S LAWS OF MOTION

• Strictly apply only to particles

‒ apply to finite objects by summation

• Law 1 requires the concept of inertial frames

‒ accelerating (e.g. rotating) reference frames yield additional apparent forces

• Law 3 (“equal and opposite reactions”) - use with care!

‒ e.g. magnetic forces between non-parallel current-carrying wires

‒ a more general version is “total momentum is conserved”

I I

F F(=BIL)

I

F

I

F

u

In inertial frame In rotating frame

FORMS OF THE EQUATION OF MOTION

Force-momentum:

force = rate of change of momentum

Impulse (force × time):

impulse = change of momentum

Energy (force × distance):

work done = change of kinetic energy

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FORMS OF THE EQUATION OF MOTION

force = mass acceleration

force = rate of change of momentum

Rate form:

Impulse form:

impulse = change of momentum

Energy form:

work done = change of kinetic energy

F = 𝑚a

F =d(𝑚v)

d𝑡

නF d𝑡 = (𝑚v)𝑒𝑛𝑑 − (𝑚v)𝑠𝑡𝑎𝑟𝑡

නF • dx = (1

2𝑚𝑣2)𝑒𝑛𝑑 − (

1

2𝑚𝑣2)𝑠𝑡𝑎𝑟𝑡

WORK-ENERGY

force = mass × acceleration

work done = change of kinetic energy

F = 𝑚dv

d𝑡

නF • dx = න𝑚dv

d𝑡• dx

= න𝑚dv

d𝑡• vd𝑡

= න 12𝑚

d

d𝑡v • v d𝑡

= නd

d𝑡(12𝑚𝑣

2) d𝑡

= 12𝑚𝑣

2𝑠𝑡𝑎𝑟𝑡

𝑒𝑛𝑑

නF • dx = 12𝑚𝑣

2𝑒𝑛𝑑

− 12𝑚𝑣

2𝑠𝑡𝑎𝑟𝑡

ADVANTAGES OF WORK-ENERGY FORM

work done = change of kinetic energy

● Scalar – not vector – equation

● Just “start” and “end” – no need for intermediate values

● Gives velocities without having to integrate acceleration

● Most forces depend on position, not time

● Some “work” conveniently written in terms of potential energy

නF • dx = 12𝑚𝑣2

𝑒𝑛𝑑− 1

2𝑚𝑣2

𝑠𝑡𝑎𝑟𝑡

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EXAMPLE

Two boxes with masses 25 kg and 35 kg are placed on a

smooth frictionless surface. The boxes are in contact and

a force of 30 N pushes horizontally on the smaller box.

What is the force that the smaller box exerts on the

larger box?

25 kg35 kg30 N

EXAMPLE

Masses are connected by light, inelastic cables passing round

light, smooth pulleys in different configurations, as shown below.

Find the direction and magnitude of the acceleration of each

mass, and the tension in each cable, for each configuration.

(Give your answers in kg-m-s units, as multiples of 𝑔.)

20kg

30kg

30kg

20kg

(a) (b)

CONSERVATION OF MOMENTUM

Corollary (or more modern version) of Newton’s Third Law

When two bodies A and B collide:

force exerted by A on B is equal and opposite to the force

exerted by B on A

change in momentum of B equal and opposite to change in

momentum of A

total momentum (A + B) conserved.

A

B

Before After

uB

uA

vB

vA

𝑚𝐴u𝐴 +𝑚𝐵u𝐵

total momentum before

= 𝑚𝐴v𝐴 +𝑚𝐵v𝐵

total momentum after

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EXAMPLE

A vehicle A of mass 500 kg travelling at 30 m s–1 collides with a

vehicle B of mass 900 kg travelling at 21 m s–1 along the same

line. After the collision the vehicles remain stuck together for a

short distance.

What is the speed and direction of the combined assembly:

(a) if the vehicles are originally travelling in the same direction?

(b) if the vehicles are originally travelling in opposite directions?

RESTITUTION

When objects collide, they compress, and kinetic

energy is replaced by elastic potential energy. A B

When they separate, kinetic energy is recovered.

Internal friction leads to mechanical energy being lost (as heat).

Model: coefficient of restitution, e

In general (if no external forces act):

Perfectly elastic collision (all kinetic energy recovered):

Perfectly inelastic collision (objects stick together):

0 ≤ 𝑒 ≤ 1

𝑒 = 1

𝑒 = 0

𝑒 =relative velocity of separation

relative velocity of approach

RESTITUTION

Before

AftervA

A B

vB

uA uB

Total momentum:

Restitution:

𝑚𝐴𝑢𝐴 +𝑚𝐵𝑢𝐵 = 𝑚𝐴𝑣𝐴 +𝑚𝐵𝑣𝐵

𝑣𝐵 − 𝑣𝐴 = 𝑒 𝑢𝐴 − 𝑢𝐵

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EXAMPLE

A ball of mass 0.6 kg and speed 12 m s–1 collides with

a second ball of mass 0.2 kg moving in the opposite

direction with speed 18 m s–1.

Assuming that all motion takes place along a straight

line, what is the outcome of the collision if the

coefficient of restitution e is:

(a) 0.5

(b) 0.8

OBLIQUE COLLISIONS

• Apply restitution perpendicular to the surface of contact:

Before Afterx

BAA B

• Conserve total momentum in any direction, including 𝑥:

• Conserve individual momentum components (and hence

velocity components) parallel to contact surfaces

𝑣𝐵𝑥 − 𝑣𝐴𝑥 = 𝑒 𝑢𝐴𝑥 − 𝑢𝐵𝑥

𝑚𝐴𝑢𝐴𝑥 +𝑚𝐵𝑢𝐵𝑥 = 𝑚𝐴𝑣𝐴𝑥 +𝑚𝐵𝑣𝐵𝑥

ቅ𝑣𝑦 = 𝑢𝑦𝑣𝑧 = 𝑢𝑧

for both objects

EXAMPLE

A ball is dropped vertically a distance of 5 m onto a

plane surface sloping at 35° to the horizontal.

How far down the slope is its second bounce if the

coefficient of restitution 𝑒 is:

(a) 1.0;

(b) 0.5.

5 m

35o

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FRICTION

● Tangential force generated between surfaces in contact that

opposes relative motion.

● May or may not be sufficient to prevent relative motion.

● If sliding does occur, friction does negative work and

mechanical energy is lost as heat.

● For rolling rather than sliding, friction causes the rotation.

● Models where friction is neglected are called ideal.

● Many devices rely on friction; e.g. conveyor belts, belt drives,

brakes, clutch plates, screws.

TYPES OF FRICTION

● Dry friction (this course)

● Fluid friction (hydraulics courses)

● Internal friction

U(y)

y

MODELLING FRICTION

Friction between solid surfaces is:

● tangential to the surfaces in contact;

● direction opposed to relative motion or tendency to motion;

● dependent on surfaces and normal reaction N.

Up to the point of sliding:

Once sliding occurs:

μ𝑠 is the coefficient of static friction

μ𝑘 is the coefficient of kinetic friction

In most cases we won’t make a distinction: just write μ

𝐹 ≤ 𝐹𝑚𝑎𝑥 = μ𝑠𝑁

𝐹 = μ𝑘𝑁

N

F

P

F

Fmax Ns

Nk

mg

N

F P

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FRICTION – REMEMBER!

● Direction: opposite to resultant force in the

absence of friction (the “motive force”);

● Magnitude: only given by μ𝑁 when moving.

P

F

Fmax Ns

Nk

EXAMPLE

(a) A block of mass 5 kg can be pushed up a slope of 1 in 3

at constant speed by a force of 45 N applied horizontally.

Deduce the coefficient of friction, μ.

(b) If, instead, the 45 N force is applied parallel to the slope,

what is the block’s acceleration up the slope?

GRAVITY-DRIVEN MOTION ON A SLOPE

An object on a slope may:

‒ topple over

‒ stay still

‒ slide

Does it topple?

– Consider the line of action through the centre of gravity.

If it doesn’t topple:

No sliding if:

N

F

mg

𝑁 = 𝑚𝑔 cos θ 𝐹𝑚𝑎𝑥 = μ𝑠𝑁 = μ𝑠𝑚𝑔cos θ

𝑑𝑜𝑤𝑛𝑠𝑙𝑜𝑝𝑒 𝑤𝑒𝑖𝑔ℎ𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 ≤ 𝐹𝑚𝑎𝑥

𝑚𝑔sin θ ≤ 𝜇𝑠𝑚𝑔 cos θ

tanθ ≤ μ𝑠

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EXAMPLE

A crate of mass 30 kg is propelled along a horizontal surface

at steady speed by a horizontal force of magnitude 𝑃 = 120 N.

(a) What is the coefficient of friction between crate and

ground?

For the same coefficient of friction, what minimum force must

be applied to move the crate if it is directed:

(b) at 25° below the horizontal;

(c) at 25° above the horizontal.

30 kgP

30 kg

P

30 kg

P25

o25

o

(a) (b) (c)

EXAMPLE

Two cars, A of mass 800 kg and B of mass 1200 kg, collide at a 90° crossroads as

shown. The collision is completely inelastic and after the collision the cars continue

to move as a single body, sliding a distance 18 m at angle 30º to the original

direction of A before stopping.

(a) If car A was travelling at 30 m s–1 before the collision, find the speed of car B

before the collision and the speed of the combined vehicles after the collision.

(b) Assuming that all car wheels lock at the point of impact, find the total

coefficient of kinetic friction between the tyres and the ground.

800 kg30 m/s

1200kg

30o

18 mA

B

FINITE OBJECTS, NOT PARTICLES

● Newton’s Laws apply to point particles.

● Real objects consist of many particles … acted upon by:

‒ internal forces (equal and opposite in pairs);

‒ external forces.

● Important properties: total mass 𝑀 and centre of mass x

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DYNAMICS OF A SYSTEM

m1m4

m3

m2

m5

F

Individual particle:

Sum over all particles:

Motion of centre of mass:

where:

f𝑖 = 𝑚𝑖

d2x𝑖

d𝑡2

f𝑖 =𝑚𝑖

d2x𝑖

d𝑡2=

d2

d𝑡2𝑚𝑖x𝑖

F =d2

d𝑡2(𝑀x)

F = 𝑀d2x

d𝑡2

𝑀 =𝑚𝑖

𝑀x =𝑚𝑖 x𝑖

MOMENTS

moment = quantity distance

CENTRE OF MASS

Lots of masses 𝑚1, 𝑚2, 𝑚3, ... at points x1, x2, x3, ...

vs

one mass 𝑀 at point x

Require: same total mass

same total moment of mass

𝑀 =𝑚𝑖

𝑀x =𝑚𝑖x𝑖

x =σ𝑚𝑖x𝑖

𝑀=𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑚𝑎𝑠𝑠

𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠

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INDIVIDUAL COORDINATES

𝑥 =σ𝑚𝑖𝑥𝑖𝑀

, 𝑦 =σ𝑚𝑖𝑦𝑖𝑀

, 𝑧 =σ𝑚𝑖𝑧𝑖𝑀

MOMENT ABOUT THE CENTRE OF MASS

The net moment about the centre of mass is zero.

𝑚𝑖(x𝑖 − x) =𝑚𝑖x𝑖 −𝑚𝑖x

=𝑚𝑖x𝑖 −𝑀x

= 0

IMPORTANT PROPERTIES (I)

For a system of particles:

Momentum:

● The centre of mass moves like a single particle of mass 𝑀under the resultant of the external forces .

Energy:

● In uniform gravity, total gravitational potential energy is the

same as a single particle of mass 𝑀 at the centre of mass.

● Total kinetic energy of a system is:

‒ kinetic energy of the centre of mass, 1

2𝑀𝑉2; plus

‒ kinetic energy of the system relative to the centre of mass.

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IMPORTANT PROPERTIES (II)

Rotation:

● For a rigid body, the motion relative to the centre of

mass is pure rotation:

● Total angular momentum of a system is:

‒ angular momentum of the centre of mass; plus

‒ angular momentum relative to the centre of mass.

● The rotational equation of motion

torque = rate of change of angular momentum

holds for the resultant of all external torques about a

point which is either:

‒ fixed; or

‒ moving with the centre of mass.

V V

2

21 MVKE = 2

212

21 IMVKE ω+=

FINDING THE CENTRE OF MASS

1. First principles:

2. Geometric centre:

(centroid)

3. Symmetry G

4. Addition or subtraction of simpler elements

m1m2

m1

m2

add subtract

x =σ𝑚𝑖x𝑖

𝑀=𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑚𝑎𝑠𝑠

𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠

x =σ𝑣𝑖x𝑖𝑉

=𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑣𝑜𝑙𝑢𝑚𝑒

𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒

EXAMPLE

Three particles of mass 3, 3 and 4 units are at rest

on a smooth horizontal plane at points with position

vectors i + 4j, 5i + 10jand 3i + 2j respectively.

Find the position of their centre of mass.

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EXAMPLE

Find the position of the centre of mass of the L-shaped

lamina shown.

0.5

2.0

0.5

1.5

EXAMPLE

A uniform lamina is formed by cutting quarter-circles of radius 0.5 m and

0.2 m from the top-right and bottom-left corners respectively of a rectangle

whose original dimensions were 0.6 m by 0.8 m.

(a) Find the area of the lamina.

(b) Find the position of the centre of mass relative to the top-left corner, A.

(c) If the lamina is allowed to pivot freely in a vertical plane about corner

A, find the angle made by side AB with the vertical when hanging in

equilibrium.

(The centre of mass of a quarter-circular

lamina of radius 𝑅 is4

3𝜋𝑅 from either

straight side.)

A

B

0.5 m

0.2 m

0.8 m

0.6 m

EXAMPLE

Find the position of the centre of mass of a

uniform solid hemisphere of radius 𝑅.

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USEFUL CENTRES OF MASS

Triangle (2-d):

Pyramid (any base; 3-d):

Semi-circular arc (1-d):

Semi-circular area (2-d):

Hemispherical shell:

Hemispherical solid (3-d):

y

h

y

R

𝑦 =1

3ℎ

𝑦 =1

4ℎ

𝑦 =2

π𝑅

𝑦 =4

3π𝑅

𝑦 =1

2𝑅

𝑦 =3

8𝑅

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