Kinematics Dynamics

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Kinematics Dynamics
CPI Lecture 04
Exam I

• Todays lecture will cover Textbook Sections 4.3,
  4.7

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Review

• Kinematics Description of Motion
• position
• displacement
• Velocity v Dx / Dt
• average
• instantaneous
• Acceleration a Dv / Dt
• average
• Instantaneous

05
3
Preflight 4.1
3 93 4
(A) (B) (C)

• Which x vs t plot shows positive acceleration?

07
4
Equations for Constant Acceleration(text, page
80)

• x x0 v0t 1/2 at2
• v v0 at
• v2 v02 2a(x-x0)

• ?x v0t 1/2 at2
• ?v at
• v2 v02 2a ?x

Lets derive this
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5
Equations for Constant Acceleration(text, page
80)

• x x0 v0t 1/2 at2
• v v0 at
• v2 v02 2a(x-x0)

• ?x v0t 1/2 at2
• ?v at
• v2 v02 2a ?x

12
6
Kinematics Example

• A car is traveling 30 m/s and applies its breaks
  to stop. Assuming constant acceleration of -6
  m/s2, how long does it take for the car to stop,
  and how far does it travel before stopping?

17
7
Pre-Flight 4.3

• An car accelerates uniformly from rest. If it
  travels a distance D in time t then how far will
  it travel in a time 2t ?
• 1. D/4 2. D/2 3. D 4. 2D
  5. 4D

Demo
Follow up question If the car has speed v at
time t then what is the speed at time 2t ? 1.
v/4 2. v/2 3. v 4. 2v
5. 4v
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8
Newtons Second Law Fma
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Newtons Second Law Fma

• A tractor T is pulling a trailer M with a
  constant acceleration. If the forward
  acceleration is 1.5 m/s2. Calculate the force on
  the trailer (m400Kg) due to the tractor
  (m500Kg).

X direction
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ACT

• A force F acting on a mass m1 results in an
  acceleration a1.The same force acting on a
  different mass m2 results in an acceleration a2
  2a1. What is the mass m2?

(1) 2m1 (2) m1 (3) 1/2 m1

• Fma
• F m1a1 m2a2 m2(2a1)
• Therefore, m2 m1/2
• Or in wordstwice the acceleration means half the
  mass

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Example

• A tractor T (m500Kg) is pulling a trailer M
  (m400Kg). It starts from rest and pulls with
  constant force such that after 10 seconds it has
  moved 30 meters to the right. Calculate the
  horizontal force on the tractor due to the
  ground.

X direction Tractor SF ma Fw T
mtractora Fw T mtractora
Combine Fw mtrailera mtractora Fw
(mtrailermtractor ) a
X direction Trailer SF ma T mtrailera
35
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Example

• A tractor T (m500Kg) is pulling a trailer M
  (m400Kg). It starts from rest and pulls with
  constant force such that after 10 seconds it has
  moved 30 meters to the right. Calculate the
  horizontal force on the tractor due to the
  ground.

Combine Fw mtrailera mtractora Fw
(mtrailermtractor ) a
Acceleration Dx v0t ½ a t2 a 2 Dx / t2
0.6 m/s2
FW 540 Newtons
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Pre-Flight 4.5

• Compare the average horizontal force on a car
  accelerating from 0 to 60 mph in 6 seconds to
  that when it reaches 60 mph in 12 seconds.
• F6 F12
• 2. F6 2 F12
• 3. F6 4 F12

F m a F m (vf-vi)/t ½ the time requires twice
the force
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Pulley Example

• Two boxes are connected by a string over a
  frictionless pulley. Box 1 has mass 1.5 kg, box 2
  has a mass of 2.5 kg. Box 2 starts from rest 0.8
  meters above the table, how long does it take to
  hit the table.

• Compare the acceleration of boxes 1 and 2
• 1) a1 gt a2 2) a1 a2 3) a1 lt a2

1) T - m1 g m1 a1 2) T m2 g -m2 a1 using
a1 -a2 2) T m2 g -m2 a1 1) m2 g -m2 a1 - m1
g m1 a1 a1 (m2 m1)g / (m1m2)
1
2
45
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Pulley Example

• Two boxes are connected by a string over a
  frictionless pulley. Box 1 has mass 1.5 kg, box 2
  has a mass of 2.5 kg. Box 2 starts from rest 0.8
  meters above the table, how long does it take to
  hit the table.

• Compare the acceleration of boxes 1 and 2
• 1) a1 gt a2 2) a1 a2 3) a1 lt a2

a1 (m2 m1)g / (m1m2) a 2.45 m/s2 Dx v0t
½ a t2 Dx ½ a t2 t sqrt(2 Dx/a) t 0.81
seconds
1
2
48
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Summary of Concepts

• Constant Acceleration

• x x0 v0t 1/2 at2
• v v0 at
• v2 v02 2a(x-x0)

• F m a
• Draw Free Body Diagram
• Write down equations
• Solve