The Laws

A section in the chapter

of the study of Dynamics

of motion

Program Line-up

Take Aim Get Moving Building Momentum Let’s get impulsive What a pair! Conserve that momentum Rounding up

Taking Aim

State Newton’s 3 Laws of motion in their complete forms

Describe the qualities of momentum and impulse

Solve impulse and momentum problems State the principle of conservation of linear

momentum

Get Moving

Newton’s First Law of Motion in your own words…

No… really, what is Newton’s First Law?

Also known as the Law of Inertia

When no net resultant force acts upon an object, then if that object is at rest, it will stay at rest and if it is moving, it will continue to move

in a straight line with constant velocity

Building Momentum (1/2)

From before:

But that is not complete…

F ma t

v

mvF

t

Momentum

(p)

Building Momentum (2/2)

Momentum is a vector Momentum is in the same direction as

velocity Force need not be in the same direction as

the momentum

ROADBLOCK!!!

Roadblock 1:

Air being pushed downwards by the blades of a helicopter travels at a velocity of vair m/s. Assuming the cross-section of the air being pushed away by the blades is A m2, what is the average force that the blades are exerting on air? State any assumptions made.

Solution to Roadblock 1 (1/3)

Assume that air being pushed out by the rotor blades takes the shape of a cylinder

Cross-sectional area, A m2

Length of cylinder, l m

Solution to Roadblock 1 (2/3)

Volume of air being pushed away by the rotor blades per second can be written as:

Now Newton’s Second Law of motion states:

13

smvA

t

Vair

air

t

mvF

Solution to Roadblock 1 (3/3)

v remains constant but m changes Therefore:

NAvvAvvt

mF airairairair

airaverage

2

BINGO!!!

Let’s get IMPULSIVE (1/3)

Newton’s Second Law:

Therefore:t

pF

Change in momentum, also known as Impulse

ptF

This is the Impulse-momentum theorem

Let’s get IMPULSIVE (2/3)

Force vs Time graph:

Area under the curve gives Tells nothing of the initial and final momentum

p

Force/N

Time/s

Let’s get IMPULSIVE (3/3)

Impulse is a Vector

Direction of Impulse:

ROADBLOCK!!!

Initial Momentum

Final Momentum

Impulse

Roadblock 2

A baseball, mass m kg is moving horizontally at a velocity of v m/s when it is struck by a baseball bat. It leaves the bat horizontally at a velocity of v m/s in the opposite direction. (a) Find the impulse of the force exerted on the ball. (b) Assuming that the collision lasts for x ms, what is the average force?

Solution to roadblock 2 (1/3)

Let the initial direction that baseball is travelling in be positive. Therefore, initial momentum, pinitial, is:

As the baseball is travelling horizontally in the opposite direction, the final momentum, pfinal, is:

1 kgmsmvpinitial

1 kgmsmvp final

Solution to roadblock 2 (2/3)

Therefore impulse is:

By the impulse-momentum theorem:

And given time of impact, x ms,

11 2 kgmsmvkgmsmvmvppp initialfinal

ptF

t sx

1000=x ms =

Solution to roadblock 2 (3/3)

Therefore the average force, Fave, is:

N

x

mvN

x

mv

t

pF ave

200010002

YEAH!!!!

What a pair (1/2)

What a pair (2/2)

Newton’s Third law of motion:

ROADBLOCK!!!

Every action will produce an equal and opposite

reaction

Roadblock 3

In roadblock 1, what is the force that air exerts on the rotor?

Give other examples of action-reaction pairs that are useful

Solution to Roadblock 3

Force on rotor blades will be equal in magnitude but opposite in direction to that of the average force on air

Other examples:– Jet engines– Walking– Swimming– Fans

Conserve that momentum (1/6)

The principle of conservation of momentum:

Extension of Newton’s Second and Third Laws of Motion

The total momentum of a closed system is constant if no external

resultant forces act on it.

Conserve that momentum (2/6)

Consider two isolated particles m1 and m2 before and after they collide. Before the collision, the velocities of the two particles are v1i and v2i; after collision, the velocities are v1f and v2f.

Before collision

During collision

After collision

Conserve that momentum (3/6)

Free body diagram:

v1im1

v2i

m2

F1 F2

v2f

v1f

Conserve that momentum (4/6)

Applying the Impulse-Momentum Theorem to m1:

Likewise, for m2

1 1 1 1 1f iF t m v m v

2 2 2 2 2f iF t m v m v

Conserve that momentum (5/6)

By Newton’s Third Law:

Time of collision same for both masses:

By Newton’s Second Law:

21 FF

tFtF 21

ifif vmvmvmvm 22221111

Conserve that momentum (6/6)

From which we find:

ffii vmvmvmvm 22112211

Total initial momentum = Final total momentum

Rounding up

Dynamics:Newton’s Laws of Motion

First Law

Inertia

Force required to change state of motion

Second Law

Force proportional to rate of change of momentum

Momentum

Impulse

Third Law

Principle of conservation of linear momentum

Impulse-momentum theorem