Chapter 4 Newton’s Laws of Motion - Physics and …people.physics.tamu.edu/wwu/docs/P201S2017NotesExamReview2.pdf · Chapter 4 Newton’s Laws of Motion . Type of Forces (that we

Newton’s Second Law:

in vector form σ റ𝐹 = 𝑚 റ𝑎

in component form σ𝐹𝑥 = 𝑚𝑎𝑥 σ𝐹𝑦 = 𝑚𝑎𝑦

in equilibrium and static situations 𝑎𝑥 = 0; 𝑎𝑦 = 0

Strategy for Solving Newton’s Law Problems

● Isolate the bodies in a system.

●Analyze all the forces acting on each body and draw one free-body diagram for each body.

● Based on the free-body diagram, set up a most convenient x-y coordinate system.

● Break each force into components using this coordinates.

● For each body, sum up all the x-components of the forces to an equation: σ𝐹𝑥 = 𝑚𝑎𝑥.

● For each body, sum up all the y-components of the forces to an equation: σ𝐹𝑦 = 𝑚𝑎𝑦.

● Use these equations to solve for unknown quantities.

Chapter 4 Newton’s Laws of Motion

Type of Forces (that we will deal with in this course)

Contact forces

● Normal Force 𝑛 ● Friction Force റ𝑓𝑘normal to the tangent to the

contacting surface contacting surface

● Tension Force 𝑇 ● Spring Force റ𝐹𝑠𝑝𝑟(in a rope) opposite to the

along the rope deformation

Action-at-a-Distance Forces (non-contacting)

● Gravitational Force റ𝐹𝐺(along the line connecting the two centers of mass)

റ𝑓𝑘

𝑛 𝑇

റ𝐹𝐺

• An example of superposition of forces:

• In general, the resultant, or vector sum of forces, is:

y

x

𝜃 = 𝑡𝑎𝑛−1𝑅𝑦

𝑅𝑥direction

Superposition of Forces: Resultant and Components of Force Vectors

Example: Box of mass m on a frictionless incline

Given: m and q

Find: n and a

Solutions:

Weight: W = mg Wx = mgsinq Wy = – mgcosq

x-axis: mgsinq = ma a = gsinq

y-axis: n – mgcosq = 0 n = mgcosq

n

q

y

x

q

mg

Another Question: Near the bottom of the incline, the box is given an initial velocity of

a known magnitude v0 pointing up the incline. What distance will it slide before it turns

around and slides downward?

Solution: The acceleration is given above a = gsinq

Use the kinematic equation 𝑣22 − 𝑣1

2 = 2𝑎(𝑥2 − 𝑥1)(𝑥2 − 𝑥1) = −𝑣1

2/2𝑎 = −𝑣02/2gsinq

The distance that it will slide up the incline is 𝑣02/2gsinq.

A Review of Uniform Circular Motion----Section 3.4

Velocity: tangent to the circle with constant magnitude 𝑣1 = 𝑣2 = 𝑣

Acceleration: arad = 𝑣2

𝑅, always pointing toward the center of the circle.

known as the centripetal acceleration

A new concept

Period (T, in s): Time for the object to complete one circle.

New relationships between v, R, and T:

Velocity 𝑣 =2π𝑅

𝑇

Acceleration: arad = 𝑣2

𝑅=

4π2𝑅

𝑇2

Another new concept

Frequency (f, in s-1, or, Hz): revolutions per second.

Relationship between T and f: f = 1/T

at the top

at the bottom

6.2 Motion in a Vertical Circle

Example 6.5 Dynamics of a Ferris wheel

at a constant speed, page 158

Given: m = 60.0 kg

R = 8.00 m

T = 10.0 s (v = 2pR/T)

Find the normal force:

(a) at the top (nT)

(b) at the bottom (nB)

(a) At the top

𝑛𝑇 −𝑚𝑔 = 𝑚𝑎𝑇 = 𝑚(−𝑣2

𝑅) 𝑛𝑇 = 𝑚𝑔 −𝑚

𝑣2

𝑅= 𝑚(𝑔 −

𝑣2

𝑅)

(b) At the bottom

𝑛𝐵 −𝑚𝑔 = 𝑚𝑎𝐵 = 𝑚(𝑣2

𝑅) 𝑛𝑻 = 𝑚𝑔 +𝑚

𝑣2

𝑅= 𝑚(𝑔 +

𝑣2

𝑅)

A roller-coaster car of mass m = 100 kg is initially

resting on an incline at a height H = 20.0 m above the

bottom of the loop. It slides down and loops inside a

circular track of radius R = 5.00 m. Neglect all friction.

(a) What is its speed at the bottom of the loop (Point A)?

Set up the y-axis. Apply the conservation of energy:

𝑚𝑔𝐻 +1

2m(0)2 = 𝑚𝑔 0 +

1

2m(𝑣𝐴)

2

𝑣𝐴 = 2𝑔𝐻

(b) At Point A, draw a free-body diagram. What is the

normal n that the track exerts on the car?

𝑛 −𝑚𝑔 = 𝑚𝑎𝐴 = 𝑚𝑣𝐴2

𝑅= 𝑚

2𝑔𝐻

𝑅

𝑛 = 𝑚𝑔 + 2𝑚𝑔𝐻/𝑅(c) Speed at the top of the loop (Point B)?

𝑚𝑔𝐻 +1

2m(0)2 = 𝑚𝑔 2𝑅 +

1

2m(𝑣𝐵)

2

𝑣𝐵 = 2𝑔𝐻 − 4𝑔𝑅

H

R

A

B

y

o

(d) Free-body diagram and normal

force at Point B?

− 𝑛 −𝑚𝑔 = 𝑚𝑎𝐵 = 𝑚 −𝑣𝐵2

𝑅

= −𝑚(2𝑔𝐻 − 4𝑔𝑅)/𝑅

𝑛 = 𝑚(2𝑔𝐻 − 5𝑔𝑅)/𝑅

(a) Minimum H for car not to fall off

the track at Point B?

𝑛 = 0, 𝑚𝑔 = 𝑚𝑣𝐵2

𝑅, 𝐻 =

5

2𝑅

mg

A

n

mg

B

n

• If a satellite is in a circular orbit with speed vorbit, the gravitational force provides the centripetal force needed to keep it moving in a circular path.

Gravitation and Circular Satellite Orbit

The orbital speed of a satellite

𝐺𝑚𝑚𝐸

𝑟2= 𝐹𝑔 = 𝐹𝑟𝑎𝑑 = 𝑚

𝑣2

𝑟

→ 𝑣𝑜𝑟𝑏𝑖𝑡 =𝐺𝑚𝐸

𝑟

The period of a satellite

𝑣 =2𝜋𝑟

𝑇

𝑇 =2𝜋𝑟

𝑣= 2𝜋𝑟

𝑟

𝐺𝑚𝐸=2𝜋𝑟3/2

𝐺𝑚𝐸

• The weight of an object near the surface of the earth is:

• With this we find that the acceleration due to gravity near

the earth's surface is:

6.4 Weight and Gravitation Acceleration near the surface of the Earth

Example: Problem 7, Exam II, Fall 2016

(a) A satellite of mass 80.0 kg is in a circular orbit around a spherical planet Q of radius 3.00×106 m. The

satellite has a speed 5000 m/s in an orbit of radius 8.00×106 m. What is the mass of the planet Q?

(b) Imagine that you release a small rock from rest at a distance of 20.0 m above the surface of the

planet. What is the speed of the rock just before it reaches the surface?

Given:

● About the satellite (ms = 80.0 kg, rorbit = 8.00×106 m,

v = 5000 m/s)

●About the planet Q(RQ = 3.00×106 m)

Find: (a) The mass of the planet Q (mQ)

(b) Speed of a rock after falling h = 20.0 m.

rorbit

RQx

(a) 𝐺ms𝑚𝑄

𝑟𝑜𝑟𝑏𝑖𝑡2 = 𝐹𝑔 = 𝐹𝑟𝑎𝑑 = ms

𝑣2

rorbit

𝑚𝑄 =rorbit𝑣

2

G

(b) First, find the gravitational acceleration 𝑔𝑄near the surface of the planet Q.

ms𝑔𝑄 = 𝐺ms𝑚𝑄

𝑅𝑄2 𝑔𝑄 = 𝐺

𝑚𝑄

𝑅𝑄2

Then, apply the kinematic equation

𝑣22 = 𝑣1

2 + 2𝑔𝑄ℎ

to find v2 with v1 = 0.

• More generally, work is the product of the component of the force in the direction of displacement and the magnitude s of the displacement.

• Forces applied at angles must be resolved into components.

• W is a scalar quantity that can be positive, zero, or negative.

• If W > 0 (W < 0), energy is added to (taken from) the system.

Equivalent Representations: 𝑊 = 𝐹||𝑠 = 𝐹 ∗ 𝑐𝑜𝑠∅ 𝑠 = 𝐹 𝑠 ∗ 𝑐𝑜𝑠∅ = 𝐹𝑠||

Example 7.2 on Page 186: Sliding on a Ramp

Package of mass m slides down a frictionless incline of height h and angle b.

Find: The total work done by all the relevant forces.

Work done by the gravity force (the weight):

Component of the weight parallel to displacement = mgsinb.

Work done: (mgsinb)s = mg[s(sinb)] = mgh

Work done by the normal force = 0

since the component of the normal force parallel to displacement = 0

Total work done by all the relevant forces = mgh

Applications of Force andResultant Work

x

y

What if the net amount of work done on an object is not zero?

• The work-energy theorem:

• The kinetic energy K of a particle with mass m moving with speed 𝑣 is 𝐾 =1

2𝑚𝑣2. It is the

energy related to the motion of the particle.

• During any displacement of the particle, the net amount of work done by all the external

force on the particle is equal to the change in its kinetic energy.

• Although the kinetic energy K is always positive, Wtotal may be positive, negative, or zero (energy added to, taken away, or left the same, respectively).

• If Wtotal = 0, then the kinetic energy does not change and the speed of the particle remains constant.

7.3 Work and Kinetic Energy

Examples of Force and Potential Energy

Force Type Work Done by the Force (i → f ) Potential Energy

Gravity

𝐹𝑔 = 𝑚𝑔 𝑊 = 𝑚𝑔𝑦𝑖 −𝑚𝑔𝑦𝑓 = 𝑈𝑖 − 𝑈𝑓 𝑈𝑔𝑟𝑎𝑣 = 𝑚𝑔𝑦

Spring (elastic)

𝐹𝑠𝑝𝑟𝑖𝑛𝑔 = − 𝑘𝑥 𝑊 =1

2𝑘𝑥𝑖

2 −1

2𝑘𝑥𝑓

2 = 𝑈𝑖 − 𝑈𝑓 𝑈𝑒𝑙 =1

2𝑘𝑥2

7.6 Conservation of Energy

For a conservative force, the work done is related to the change in the potential energy:

𝑊 = 𝑈𝑖 − 𝑈𝑓.

Apply the work-energy theorem, we obtain:

𝑈𝑖 − 𝑈𝑓 = 𝐾𝑓 − 𝐾𝑖 or 𝑈𝑖 + 𝐾𝑖 = 𝑈𝑓+ 𝐾𝑓.

The expression in red is the conservation of energy, when the force is a conservative force.

The total mechanical energy E is the sum of the potential energy and kinetic energy:

𝐸 = 𝑈 + 𝐾

With gravity force: 𝐸 = 𝑈𝑔𝑟𝑎𝑣 + 𝐾 = 𝑚𝑔𝑦 +1

2𝑚𝑣2

Conservation of energy: 𝑚𝑔𝑦𝑖 +1

2𝑚𝑣𝑖

2 = 𝑚𝑔𝑦𝑓 +1

2𝑚𝑣𝑓

2

With spring force: 𝐸 = 𝑈𝑒𝑙 + 𝐾 =1

2𝑘𝑥2 +

1

2𝑚𝑣2

Conservation of energy:1

2𝑘𝑥𝑖

2 +1

2𝑚𝑣𝑖

2 =1

2𝑘𝑥𝑓

2 +1

2𝑚𝑣𝑓

2

With both g&s forces: 𝐸 = 𝑈𝑔𝑟𝑎𝑣 + 𝑈𝑒𝑙 + 𝐾 = 𝑚𝑔𝑦 +1

2𝑘𝑦2 +

1

2𝑚𝑣2

Conservation of energy: 𝑚𝑔𝑦𝑖 +1

2𝑘𝑦𝑖

2 +1

2𝑚𝑣𝑖

2 = 𝑚𝑔𝑦𝑓 +1

2𝑘𝑦𝑓

2 +1

2𝑚𝑣𝑓

2

Useful Expressions for Applications

When both conservative and nonconservative forces act on an object

(𝑈𝑔𝑟𝑎𝑣,𝑖 + 𝑈𝑒𝑙,𝑖+1

2𝑚𝑣𝑖

2) +𝑊𝑜𝑡ℎ𝑒𝑟 = (𝑈𝑔𝑟𝑎𝑣,𝑓+𝑈𝑒𝑙,𝑓 +1

2𝑚𝑣𝑓

2)

8.2 Conservation of Momentum

External Forces: Forces acting on objects in the system by objects outside the system.

Net External Force: the sum of the external forces.

Conservation of Momentum:

If the net external force is zero, the momentum of the system is conserved:

റ𝑝𝐴,𝑖 + റ𝑝𝐵,𝑖 +⋯ = റ𝑝𝐴,𝑓 + റ𝑝𝐵,𝑓 +⋯.

In the component form: 𝑝𝐴,𝑖.𝑥 + 𝑝𝐵,𝑖,𝑥 +⋯ = 𝑝𝐴,𝑓,𝑥 + 𝑝𝐵,𝑓,𝑥 +⋯

𝑝𝐴,𝑖,𝑦 + 𝑝𝐵,𝑖,𝑦 +⋯ = 𝑝𝐴,𝑓,𝑦 + 𝑝𝐵,𝑓,𝑦 +⋯

Consider a system consisting of a number of particles (A, B, … )

Internal Forces: Forces acting on particles in the system by particles in the system.

Net Internal Force: the sum of the internal forces, which must be zero.

Examples 8.3 and 8.4 on page 226

Example 8.5 on page 227

8.3 Inelastic Collision

Momentum is conserved but not the Kinetic Energy

Completely inelastic collision along a straight line:

𝑚𝐴𝑣𝐴,𝑖,𝑥 +𝑚𝐵𝑣𝐵,𝑖,𝑥 = (𝑚𝐴 +𝑚𝐵)𝑣𝑓,𝑥

8.4 Elastic Collision

Both the Momentum and the Kinetic Energy are conserved

Consider the elastic collision of two particles along a straight line.

Conservation of Momentum: 𝑚𝐴𝑣𝐴,𝑖 +𝑚𝐵𝑣𝐵,𝑖 = 𝑚𝐴𝑣𝐴,𝑓 +𝑚𝐵𝑣𝐵,𝑓

Conservation of Kinetic Energy:1

2𝑚𝐴𝑣𝐴,𝑖

2 +1

2𝑚𝐵𝑣𝐵,𝑖

2 =1

2𝑚𝐴𝑣𝐴,𝑖

2 +1

2𝑚𝐵𝑣𝐵,𝑖

2

8.5 Impulse

When a constant force റ𝐹 acts on an object over a period of time ∆𝑡, the impulse of the force is

റ𝐽 = റ𝐹 𝑡𝑓 − 𝑡𝑖 = റ𝐹∆𝑡

If the force is not a constant, then റ𝐽 = റ𝐹𝑎𝑣𝑔∆𝑡

Since റ𝐹𝑎𝑣𝑔 = 𝑚 റ𝑎𝑎𝑣𝑔 = 𝑚∆𝑣

∆𝑡=

∆ റ𝑝

∆𝑡,

we have റ𝐽 = ∆ റ𝑝 = റ𝑝𝑓 − റ𝑝𝑖

Center of Mass Position

● General definition of the center of mass position:

Vector form റ𝑟𝑐𝑚 =𝑚𝐴 റ𝑟𝐴+𝑚𝐵 റ𝑟𝐵+𝑚𝐶 റ𝑟𝐶+⋯

𝑚𝐴+𝑚𝐵+𝑚𝐶+⋯

Component form 𝑥𝑐𝑚 =𝑚𝐴𝑥𝐴+𝑚𝐵𝑥𝐵+𝑚𝐶𝑥𝐶+⋯

𝑚𝐴+𝑚𝐵+𝑚𝐶+⋯

𝑦𝑐𝑚 =𝑚𝐴𝑦𝐴+𝑚𝐵𝑦𝐵+𝑚𝐶𝑦𝐶+⋯

𝑚𝐴+𝑚𝐵+𝑚𝐶+⋯

x

y

o

റ𝑟𝐴

റ𝑟𝐵

റ𝑟𝐶