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1 Calculus with Algebra and Trigonometry II Lecture 20 More physics applications Apr 14, , / 14

2 Motion in two dimensions A particle s motion can be described by specifying how the coordinates of the particle vary as functions of time. The path of the particle is then a curve in parametric form with t as the parameter. x = x(t) y = y(t) The velocity of the particle will have two components corresponding the the velocity in the x direction x (t) and the velocity in the y direction, y (t). The speed is the length of the hypotenuse speed = (x (t)) 2 + (y (t)) 2 14, / 14

3 The acceleration will have two components also The horizontal acceleration will be x (t) and the vertical acceleration is y (t) The direction of the acceleration will be in the direction the particle is turning. For example for the particle shown above the acceleration is up and to the left. The magnitude of the acceleration is given by Pythagoras a = (x (t)) 2 + (y (t)) 2 14, / 14

4 Motion in a circle Consider a particle moving counter clockwise in a circle of radius r. The parametric equations are x = r cos(ωt) y = r sin(ωt) The quantity ω is called the angular velocity it is the rate the angle changes with respect to time. The time the particle takes to make one revolution is called the period, T. Since one revolution is 2π radians then ωt = 2π 14, / 14

5 The components of velocity are x (t) = ωr sin(ωt) y (t) = ωr cos(ωt) and the speed is v = x 2 + y 2 = The components of acceleration are ( ωr sin(ωt)) 2 + (ωr cos(ωt)) 2 = rω x (t) = ω 2 r cos(ωt) = ω 2 x(t) x (t) = ω 2 r sin(ωt) = ω 2 y(t) so the acceleration points toward the center of the circle, it is called centripetal in physics. The magnitude of the centripetal acceleration is a = ( ω 2 x) 2 + ( ω 2 y) 2 = rω 2 = v 2 r 14, / 14

6 Newton discovers Universal gravitation Newton used the expression for the centripetal acceleration and some data about the moon s orbit to deduce the form of the law of gravity. The radius of the moon s orbit, r m, is m and the period for the moon, T m, is 27.3 days = sec, so the acceleration of the moon is ( ) 2π 2 ( ) a m = rω 2 = r m = π 2 T m = m/s 2 Comparing this with acceleration at the earth s surface g a m = , / 14

7 The radius of earth, r e = m. r m = r e = 60 so we have g a m = ( rm r e ) 2 gr 2 e = a m r 2 m Newton deduced from this that the acceleration at a distance r from the earth s center would satisfy ar 2 = constant = G(mass of earth) where G is a universal constant ( in SI units). The final formula is a = GM r 2 where M is the mass of the attracting body. 14, / 14

8 Newton s second law In the Principia Newton explained his theory of mechanics. His most important equation is that the net force acting on a body is equal to the mass of the body times the bodies acceleration F = ma If the force depends only on time, the equation is easily solved for the motion using the fundamental theorem. a = F (t) m v(t) v(0) = x(t) x(0) = t 0 t 0 v(s) ds F (s) m ds 14, / 14

9 Work-Energy theorem If the force depends only on the position then Newton s second has the form ma = F (x) Multiply the equation by v and integrate from t 0 to t t t 0 F (x)v dt = t t 0 mav dt Suppose we have the initial conditions x(t 0 ) = x 0 and v(t 0 ) = v 0. Now dv = v (t) dt = a(t) dt dx = x (t) dt = v dt The equation becomes x x 0 F (x) dx = v v 0 mv dv 14, / 14

10 The integral of the left hand side of the equation is called the work Work = x x 0 F (x) dx The integral on the right hand side of the equation becomes t t 0 mav dt = v v 0 mv dv = 1 2 mv mv 2 0 The quantity K = 1 2 mv 2 is called the kinetic energy. So the equation states Work done by the force = change in kinetic energy This is the Work-Energy theorem Calculus with Algebra and Trigonometry II Lecture 20More physics Apr 14, applications / 14

11 Conservation of energy If we define the potential energy U(x) by U(x) = F (x) dx then the Work-Energy theorem becomes U(x 0 ) U(x) = 1 2 mv mv 2 0 or 1 2 mv 2 + U(x) = 1 2 mv U(x 0 ) The is the principle of conservation of energy Kinetic Energy + Potential Energy = constant Calculus with Algebra and Trigonometry II Lecture 20More physics Apr 14, applications / 14

12 Example In a free fall ride at an amusement park, the rider is winched up to a position at the same height as the tower. The rider is attached to the top of the tower by a cable with a length of 15 m. The rider is released from rest,. What is the rider;s speed when the cable is vertical (take g=10 m/s 2 ) Calculus with Algebra and Trigonometry II Lecture 20More physics Apr 14, applications / 14

13 Since the force = mg the potential energy is U(x) = ( mg) dx = mgx Initially the speed is zero v 0 = 0 so conservation of energy gives Solving for v 1 2 mv 2 + mgx = mgx 0 v 2 = 2g(x 0 x) = 2(10)(15) = 300 v = 10 3 m/s 2 Calculus with Algebra and Trigonometry II Lecture 20More physics Apr 14, applications / 14

14 Escape from earth s gravity Using Newtons Universal gravitation law F = Gmm ( e r 2 U(r) = Gmm ) e r 2 dr = Gmm e r the force is negative since it is attractive. If the particle just barely escapes the earths gravity its eventual velocity will be zero and it will be far enough away that the potential energy will be zero. If our rocket starts from the earth s surface energy conservation gives v = 2Gme r e = 1 2 mv 2 Gmm e r e = 0 ( )( = m Calculus with Algebra and Trigonometry II Lecture 20More physics Apr 14, applications / 14

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