Kinematics Dynamics

1
Kinematics Dynamics
Physics 101 Lecture 04

• Uniform acceleration
• Todays lecture will cover Textbook Chapter 4

Hour Exam 1 Monday Oct 1, 7 PM Lectures 1-8 Sign
up online
2
Review

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

07
3
Preflight 4.1
(A) (B) (C)

• Which x vs t plot shows positive acceleration?

Graph A displays no acceleration as the slope
remains constant (constant velocity). Graph B
shows an increase in velocity as the slope
increses positively. Graph C reveals a decrease
in velocity as the the slope is decreasing.
08
4
Newtons Second Law SFma
5
Newtons Second Law SFma
Net Force determines the acceleration
6
Newtons Second Law SFma
Net Force determines the acceleration
position and velocity depend on history
7
Equations for Constant Acceleration(text, page
108)

• 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
10
8
Equations for Constant Acceleration(text, page
108)

• 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

14
9
Kinematics Example

• A car traveling 30 m/s applies its breaks and
  stops after a distance of 150 m. How fast was
  the car going after it had traveled ½ the
  distance (75 meters) ?
• A) v 8 m/s B) v 15 m/s C) v 21 m/s

Dx related to square of v (but history matters)
18
10
Acceleration ACT

• A car accelerates uniformly from rest. If it
  travels a distance D in time t then how far will
  it travel in a time 2t ?
• A. D/4 B. D/2 C. D D. 2D
  E. 4D

Demo
Follow up question If the car has speed v at
time t then what is the speed at time 2t ? A.
v/4 B. v/2 C. v D. 2v
E. 4v
24
11
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?

(A) 2m1 (B) m1 (C) 1/2 m1

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

29
12
Example

• A tractor T (m300Kg) is pulling a trailer M
  (m400Kg). It starts from rest and pulls with
  constant force such that there is a positive
  acceleration of 1.5 m/s2. Calculate the pulling
  force of the tractor.

F 1050 N
Newtons 3rd law gt TM
X direction Tractor (T) SF ma F M
mtractora F M mtractora
Combine F mtrailera mtractora Fw
(mtrailermtractor ) a
X direction Trailer (M) SF ma T mtrailera M
35
13
Net Force ACT

• Compare Ftractor the NET force (SF) on the
  tractor, with Ftrailer the NET force (SF) on the
  trailer from the previous problem.
• A) Ftractor gt Ftrailor
• B) Ftractor Ftrailor
• C) Ftractor lt Ftrailor

SF m a Ftractor mtractor a (300
kg) (1.5 m/s2) 450 N Ftrailer
mtrailer a (400 kg) (1.5 m/s2)
600 N
38
14
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
• A) a1 gt a2 B) a1 a2 C) a1 lt a2

1) T m1 g m1 a1 2) T m2 g - m2 a1 T m2
g m2 a1 m2 g m2 a1 m1 g m1 a1 a1 (m2
m1)g / (m1m2)
1
2
42
15
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
• A) a1 gt a2 B) a1 a2 C) 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
46
16
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

• Ch 2 Problems 1,5,7,11,13,17,29,49,69
• Ch 3 Problems 5,13,33,47,57,65,67
• OLD HOUR EXAMS http//online.physics.uiuc.edu/cou
  rses/phys101/fall07/practice/index.html