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Physics 170 Week 9, Lecture 1 http://www.phas.ubc.ca/∼gordonws/170 Physics 170 203 Week 9, Lecture 1 1 Textbook Chapter 13: Section 13.1-4 Physics 170 203 Week 9, Lecture 1 2 Learning Goals: • We will introduce the basic concepts of dynamics of particles in three-dimensional space. • We will review Newton’s three laws of dynamics and the concept of inertial reference frame. • We will formulate Newton’s second law for the dynamics of a particle in Cartesian coordinates. • We will solve an example to illustrate how Newton’s second law can be used to gain information about a dynamical system. • We will study the example of projectile motion in three dimensions to illustrate how one would solve a dynamics problem in three dimensions using Cartesian coordinates. Physics 170 203 Week 9, Lecture 1 3 Newton’s Laws The motion of a particle is governed by Newton’s three laws of motion. 1. First Law: A particle originally at rest, or moving in a straight line with a constant velocity, will remain in this state provided the particle is not subjected to an unbalanced force. 2. Second Law: A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force. ∑ equation of motion F⃗i = m⃗a i 3. Third Law: The mutual forces of action and reaction between two particles are equal, opposite, and collinear. Physics 170 203 Week 9, Lecture 1 4 Inertial Frame of Reference In an intertial frame of reference, an object that is placed at rest will remain at rest unless it is acted upon by an unbalanced force. Similarly, in an inertial reference frame, an object which is moving at a constant velocity will remain in motion with the same constant velocity unless it is acted upon by an unbalanced force. Newton’s first law is based on the existence of an inertial reference frame. At the surface of the earth, because of gravity, the only way to truly be in an inertial reference frame is to be in a state of “free fall”. Being in orbit is one possibility. Newton’s second law is valid only in an inertial reference frame. Gravity can be taken into account by adding it as a force. Physics 170 203 Week 9, Lecture 1 5 Problem solving Newton’s second law relates the acceleration of a particle to the force that is acting on it. One application is to use the known acceleration of a particle to compute the force that is acting on it. Physics 170 203 Week 9, Lecture 1 6 Example: A tractor is used to lift a 150 kg load B with at 24 m long rope, beam and pulley system. If the tractor is traveling to the right at a constant speed of 4 m/s2 determine the tension of the rope when sA = 5m. When sA = 0, sB = 0. Physics 170 203 Week 9, Lecture 1 7 Approach to finding a solution: • We will use the fact that the rope has a fixed length to relate sB to sA . • We will use the equation for the length of the rope to find an expression for the acceleration of the crate s̈B Physics 170 203 Week 9, Lecture 1 8 Approach to finding a solution: • We will then analyze the crate using Newton’s second law where the forces acting on it are gravity and the tension in the rope and the result is the acceleration that we have already computed. • This should allow us to solve Newton’s second law for the tension in the rope. Physics 170 203 Week 9, Lecture 1 9 Solution: The length of the rope is ℓ = 12 − sB + √ 122 + s2A m where sA and sB are in meters. Physics 170 203 Week 9, Lecture 1 10 Solution: ℓ = 12 − sB + √ 122 + s2A m Take a derivative of this equation by time to get sA ṡA d sA ṡA → ṡB = √ 0 = ℓ = −ṡB + √ 2 2 dt 12 + sA 122 + s2A Physics 170 203 Week 9, Lecture 1 11 Solution: sA ṡA ṡB = √ 122 + s2A Take another time derivative to get an expression for acceleration, ṡ2A + sA s̈A s2A ṡ2A s̈B = √ − 2 2 (122 + s2A )3/2 12 + sA The time derivative of the second term has been taken using the chain and product rules. Physics 170 203 Week 9, Lecture 1 12 Solution: s2A ṡ2A ṡ2A + sA s̈A − s̈B = √ 2 2 (122 + s2A )3/2 12 + sA Tractor is traveling “with constant speed” → s̈A = 0, ṡA = 4m/s, sA = 5m. This allows us to compute 52 42 42 42 122 2 − m/s s̈B = √ = 133 (122 + 52 )3/2 122 + 52 Physics 170 203 Week 9, Lecture 1 13 Solution: The forces acting on the crate are gravity downward and the tension in the cable upward. They are making the crate accelerate upward with the acceleration that we have computed above. If we use Newton’s second law T − mg = ms̈B → T = m(s̈B + g) ( 2 2 ) 4 12 = 150 + 9.81 N 133 T = 1630 N Physics 170 203 Week 9, Lecture 1 14 Problem solving Newton’s second law relates the acceleration of a particle to the force that is acting on it. Another application of this law is to use the known forces to understand some features of the dynamical behavior of a particle. Physics 170 203 Week 9, Lecture 1 15 Newton’s Second Law in three dimensions: F⃗ = m⃗a The force F⃗ is the force that the patricle experiences at its “current position”. The force could be a constant, independent of position, like the force of gravity near the surface of the earth F⃗ = −mg k̂ Farther away from the surface of the earth, (assuming the center of the earth is at 0), (⃗r(t)) ⃗ F = −GME m 3 |⃗r(t)| Depends on the relative position of the object and the earth. Physics 170 203 Week 9, Lecture 1 16 Equation of Motion in Cartesian Coordinates The equation of motion F⃗ = m⃗a is an equation for vectors. When the vectors are written in cartesian form F⃗ = Fx î + Fy ĵ + Fz k̂ , ⃗a = ax î + ay ĵ + az k̂ the equation of motion is an equation for each of the three components Fx = max , Fy = may , Fz = maz If there is more than one force acting on the particle, F⃗ is the resultant, that is the vector sum of the forces. Physics 170 203 Week 9, Lecture 1 17 Projectile motion with air resistance ⃗ = −mg k̂ and Consider a particle with two forces acting, gravity W air resistance F⃗drag = −kv 2 ût (for large or fast moving objects, air friction is approximately quadratic in the velocity and opposite in direction to it). Newton’s second law −mg k̂ − kv 2 ût = m⃗a Let us assume that a particle starts from rest. Since there is no initial velocity in the x and y-directions, we can anticipate that the x and y components of the velocity will remain zero. The z-component of the above equation is k 2 dvz k 2 az = − vz − g or = − vz − g m dt m Can we solve this equation for vz ? Physics 170 203 Week 9, Lecture 1 18 Projectile motion with air resistance cont’d: We derived the differential equation for the velocity k dvz = vz2 − g → dt m dvz = dt k 2 m vz − g We can solve this equation by integrating each side. The result is ∫ ∫ dvz = dt = t + c k 2 m vz − g Integrate using substitution √ vz = gm/k tanh θ √ √ d 1 dvz = gm/kdθ tanh θ = gm/k 2 dθ dθ cosh θ (m/k)vz2 − g = g(tanh2 θ − 1) = −g/ cosh2 θ ∫ ∫ √ √ m/k t + c = dvz 2 = − m/gk dθ = − m/gkθ vz − gm/k Physics 170 203 Week 9, Lecture 1 19 √ √ = − m/gkarctanh k/gmvz √ √ vz = − gm/k tanh gk/m(t + c) Now, we must determine c. For this, we use the information that the object was at rest when t = 0, vz = 0 when t = 0 → c = 0 The velocity as a function of time is √ √ vz = − gm/k tanh[ gk/m t] Physics 170 203 Week 9, Lecture 1 20 Projectile motion with air resistance cont’d: We have derived the equation √ √ vz (t) = − gm/k tanh[ gk/m t] This solution obeys the initial condition v(0) = 0. For very small times, where we can use the Taylor expansion , √ √ tanh[ gk/m t] ≈ gk/m t + ... to see that, initially, the particle accelerates as it would under the influence of gravity (air resistance has negligible effect until the particle has greater velocity) √ √ vz (t) ≈ gm/k[ gk/m t] + ... ≈ −gt + . . . Physics 170 203 Week 9, Lecture 1 21 Projectile motion with air resistance cont’d: The vertical velocity obeys the equation √ √ vz (t) = − gm/k tanh[ gk/m t] √ √ m For larger times, t >> gk , tanh[ gk/m t] ∼ 1 and √ vz (t) → − gm/k This is the terminal velocity. The approximate time that it takes to reach terminal velocity is √ τ = m/gk Physics 170 203 Week 9, Lecture 1 22 For the next lecture, please read Textbook Chapter 13: Section 13.5 Physics 170 203 Week 9, Lecture 1 23