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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
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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
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• 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
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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.
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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
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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)
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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
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• 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
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• 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
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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.
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• 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||
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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
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Applications of Force andResultant Work
x
y
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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
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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
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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
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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
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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)
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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
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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
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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
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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