review: using f=ma. relevant concepts: “inertia” and force a) inertial reference frame b)...

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  • Slide 1
  • Review: using F=ma
  • Slide 2
  • Relevant concepts: inertia and force a) Inertial reference frame b) accelerated frame of the car - fictitious force
  • Slide 3
  • Rolling friction
  • Slide 4
  • Slide 5
  • Conical Pendulum (e.g. )
  • Slide 6
  • Car traveling in a banked curve Design the curve with no friction in equilibrium in the vertical direction. in uniform circular motion in the horizontal direction a component of the normal force supplies the centripetal force. The angle of bank is Note: The banking angle is independent of the mass of the vehicle. If the car rounds the curve at less than the design speed, friction is necessary to keep it from sliding down the bank. If the car rounds the curve at more than the design speed, friction is necessary to keep it from sliding up the bank.
  • Slide 7
  • Car traveling in a horizontal (Flat) Curve uniform circular motion in the horizontal direction. in equilibrium in the vertical direction. The force of static friction supplies the centripetal force. The maximum speed at which the car can negotiate the curve is: Note: this does not depend on the mass of the car. Section 6.1
  • Slide 8
  • Ferris Wheel Uniform circular motion with constant speed v (controlled by the motor) Under gravity, the child feels apparent weight differently at top and bottom Section 6.1
  • Slide 9
  • Ferris Wheel (2) At the bottom of the loop, the upward force (the normal) experienced by the object is greater than its weight. Section 6.1 At the top of the circle, the force exerted on the object is less than its weight.
  • Slide 10
  • Non-uniform Circular Motion Section 6.2 If the speed also changes in magnitude there is non-zero tangential acceleration
  • Slide 11
  • Vertical Circle with Non-Uniform Speed Section 6.2
  • Slide 12
  • Slide 13
  • Slide 14
  • Slide 15
  • Lecture 4 Work and Energy
  • Slide 16
  • Outline Work done by a constant force Projection and scalar product of vectors Force that results in positive work. negative work? forces that do no work? Work done by varying force with displacement Work-energy theorem
  • Slide 17
  • 6.1 Work Done by a Constant Force
  • Slide 18
  • Slide 19
  • Work, cont. W = F r cos A force does no work on the object if the force does not move through a displacement. The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application. Section 7.2
  • Slide 20
  • 6.1 Work Done by a Constant Force
  • Slide 21
  • Example 1 Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m.
  • Slide 22
  • 6.1 Work Done by a Constant Force
  • Slide 23
  • Example 3 Accelerating a Crate The truck is accelerating at a rate of +1.50 m/s 2. The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it?
  • Slide 24
  • 6.1 Work Done by a Constant Force The angle between the displacement and the friction force is 0 degrees.
  • Slide 25
  • Scalar Product of Two Vectors Section 7.3
  • Slide 26
  • Scalar Product, cont The scalar product is commutative. The scalar product obeys the distributive law of multiplication. Section 7.3
  • Slide 27
  • Dot Products of Unit Vectors Using component form with vectors: In the special case where Section 7.3
  • Slide 28
  • 6.2 The Work-Energy Theorem and Kinetic Energy Consider a constant net external force acting on an object. The object is displaced a distance s, in the same direction as the net force. The work is simply
  • Slide 29
  • 6.2 The Work-Energy Theorem and Kinetic Energy DEFINITION OF KINETIC ENERGY The kinetic energy KE of and object with mass m and speed v is given by
  • Slide 30
  • 6.2 The Work-Energy Theorem and Kinetic Energy THE WORK-ENERGY THEOREM When a net external force does work on and object, the kinetic energy of the object changes according to
  • Slide 31
  • 6.2 The Work-Energy Theorem and Kinetic Energy Example 4 Deep Space 1 The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If the 56.0-mN force acts on the probe through a displacement of 2.4210 9 m, what is its final speed?
  • Slide 32
  • 6.2 The Work-Energy Theorem and Kinetic Energy
  • Slide 33
  • Slide 34
  • In this case the net force is
  • Slide 35
  • 6.2 The Work-Energy Theorem and Kinetic Energy Conceptual Example 6 Work and Kinetic Energy A satellite is moving about the earth in a circular orbit and an elliptical orbit. For these two orbits, determine whether the kinetic energy of the satellite changes during the motion. W0 W