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Physics 111: Lecture 25 Today’s Agenda

Physics 111: Lecture 25 Today’s Agenda. Recap of last lecture Using “initial conditions” to solve problems The general physical pendulum The torsion pendulum Energy in SHM Atomic Vibrations Problem: Vertical Spring Problem: Transport Tunnel SHM Review. k. s. 0. k. m. m. s. 0.

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Physics 111: Lecture 25 Today’s Agenda

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  1. Physics 111: Lecture 25Today’s Agenda • Recap of last lecture • Using “initial conditions” to solve problems • The general physical pendulum • The torsion pendulum • Energy in SHM • Atomic Vibrations • Problem: Vertical Spring • Problem: Transport Tunnel • SHM Review

  2. k s 0 k m m s 0 SHM and Springs Force: Solution: s = A cos(t + )

  3. k m x 0 Velocity and Acceleration Position: x(t) = A cos(t + ) Velocity: v(t) = -A sin(t + ) Acceleration: a(t) = -2A cos(t + ) by taking derivatives, since: xMAX = A vMAX = A aMAX = 2A

  4. Lecture 25, Act 1Simple Harmonic Motion • A mass oscillates up & down on a spring. Its position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration? y(t) (a) (c) t (b)

  5. Lecture 25, Act 1Solution • The slope of y(t) tells us the sign of the velocity since • y(t) and a(t) have the opposite sign since a(t) = -w2 y(t) a < 0v > 0 a < 0v < 0 y(t) (a) (c) t (b) a > 0v > 0 The answer is (c).

  6. k m x Example • A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm. At t = 0 its speed is maximum, and is v = +2 m/s. • What is the angular frequency of oscillation ? • What is the spring constant k? vMAX =A  = Also: k = m2 Sok = (2 kg) x (20 s -1) 2 = 800 kg/s2 = 800 N/m

  7. x(t) = A cos(t + ) v(t) = -A sin(t + ) a(t) = -2A cos(t + )    k sin cos m x 0 Initial Conditions Use “initial conditions” to determine phase ! Suppose we are told x(0) = 0 , and x is initially increasing (i.e. v(0) = positive): x(0) = 0 = A cos()  = /2 or -/2 v(0) > 0 = -A sin()  < 0 So  = -/2

  8. A x(t) t   k m -A x 0 Initial Conditions... So we find  = -/2!! x(t) = A cos(t -/2) v(t) = -A sin(t -/2) a(t) = -2A cos(t -/2) x(t) = A sin(t) v(t) = A cos(t) a(t) = -2A sin(t)

  9. m Lecture 25, Act 2Initial Conditions • A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describes its velocity and acceleration as a function of time? (a) v(t) = -vmax sin(wt) a(t) = -amax cos(wt) k y (b) v(t) = vmax sin(wt) a(t) = amax cos(wt) d t = 0 (c) v(t) = vmax cos(wt) a(t) = -amax cos(wt) 0 (both vmax and amax are positive numbers)

  10. m Lecture 25, Act 2Solution Since we start with the maximum possibledisplacement at t = 0 we know that: y = dcos(wt) k y d t = 0 0

  11. z  L m d mg Review of Simple Pendulum • Using =I and sin    for small    I We found where Which has SHM solution  = 0 cos(t + )

  12. z L/2  x CM L d mg Review of Rod Pendulum • Using =I and sin   for small    I We found where Which has SHM solution  = 0 cos(t + )

  13. General Physical Pendulum Physical Pendulum • Suppose we have some arbitrarily shaped solid of mass M hung on a fixed axis, and that we know where the CM is located and what the moment of inertia I about the axis is. • The torque about the rotation (z) axis for small  is (sin   ) = -Mgd-MgR z-axis R  x CM   d Mg where  = 0 cos(t + )

  14. Lecture 25, Act 3Physical Pendulum • A pendulum is made by hanging a thin hoola-hoop of diameter D on a small nail. • What is the angular frequency of oscillation of the hoop for small displacements? (ICM = mR2 for a hoop) pivot (nail) (a) (b) (c) D

  15. So Lecture 25, Act 3Solution Hoop Pendulum • The angular frequency of oscillation of the hoop for small displacements will be given by (see Lecture 25 notes) Use parallel axis theorem: I = Icm + mR2 = mR2 + mR2 = 2mR2 pivot (nail) cm x R m

  16. wire   I Torsion Pendulum • Consider an object suspended by a wire attached at its CM. The wire defines the rotation axis, and the moment of inertia I about this axis is known. • The wire acts like a “rotational spring.” • When the object is rotated, the wire is twisted. This produces a torque that opposes the rotation. • In analogy with a spring, the torque produced is proportional to the displacement: = -k

  17. wire   I Torsion Pendulum... Torsion Pendulum • Since = -k=Ibecomes where This is similar to the “mass on spring” except I has taken the place of m (no surprise).

  18. U K E U s -A 0 A Energy in SHM • For both the spring and the pendulum, we can derive the SHM solution by using energy conservation. • The total energy (K + U) of a system undergoing SHM will always be constant! • This is not surprising since there are only conservative forces present, hence K+U energy is conserved.

  19. U U K E U x x -A 0 A SHM and quadratic potentials • SHM will occur whenever the potential is quadratic. • Generally, this will not be the case: • For example, the potential betweenH atoms in an H2 molecule lookssomething like this:

  20. U(x) = U(x0 ) + U(x0 ) (x- x0 ) + U (x0 ) (x- x0 )2+.... U U x0 x Definex = x - x0 andU(x0 ) = 0 Then U(x) = U (x0 )x2 x  SHM and quadratic potentials... However, if we do a Taylor expansion of this function about the minimum, we find that for smalldisplacements, the potential IS quadratic: U(x0) = 0 (since x0 is minimum of potential)

  21. SHM and quadratic potentials... U(x) = U (x0)x2 Letk = U (x0) Then: U(x) = kx2 U U x0 x x  SHM potential!!

  22. Problem: Vertical Spring • A mass m = 102 g is hung from a vertical spring. The equilibrium position is at y = 0. The mass is then pulled down a distance d = 10 cm from equilibrium and released at t = 0. The measured period of oscillation is T = 0.8 s. • What is the spring constant k? • Write down the equations for the position, velocity, and acceleration of the mass as functions of time. • What is the maximum velocity? • What is the maximum acceleration? k y 0 -d m t = 0

  23. So: Problem: Vertical Spring... • What is k ? k y 0 -d m t = 0

  24. k y 0 -d m t = 0 Problem: Vertical Spring... • What are the equations of motion? • At t = 0, • y = -d = -ymax • v = 0 • So we conclude: y(t) = -d cos(t) v(t) = d sin(t) a(t) = 2d cos(t)

  25. Problem: Vertical Spring... y(t) = -d cos(t) v(t) = d sin(t) a(t) = 2d cos(t) t 0   k y xmax = d = .1m vmax = d = (7.85 s-1)(.1m) = 0.78 m/s amax = 2d = (7.85 s-1)2(.1m) = 6.2 m/s2 0 -d m t = 0

  26. Transport Tunnel • A straight tunnel is dug from Urbana through the center of the Earth and out the other side. A physics 111 student jumps into the hole at noon. • What time does she get back to Urbana?

  27. Transport Tunnel... where MRis the mass inside radius R FG R RE MR but

  28. Transport Tunnel... FG R RE MR Like a mass on a spring with

  29. So: Transport Tunnel... Like a mass on a spring with FG R RE plug in g = 9.81 m/s2 and RE = 6.38 x 106 m get  = .00124 s-1 and so T = = 5067 s 84 min MR

  30. Transport Tunnel... • So she gets back to Urbana 84 minutes later, at 1:24 p.m.

  31. Transport Tunnel... • Strange but true: The period of oscillation does not require that the tunnel be straight through the middle!! Any straight tunnel gives the same answer, as long as it is frictionless and the density of the Earth is constant.

  32. Transport Tunnel... • Another strange but true fact: An object orbiting the earth near the surface will have a period of the same length as that of the transport tunnel. a =2R 9.81 = 2 6.38(10)6 m  = .00124 s-1 so T = = 5067 s 84 min

  33. k m s 0 s L Simple Harmonic Motion: Summary k s 0 m Force: Solution: s = A cos(t + )

  34. Recap of today’s lecture • Recap of last lecture • Using “initial conditions” to solve problems (Text: 14-1) • The general physical pendulum (Text: 14-3) • The torsion pendulum • Energy in SHM (Text: 14-2) • Atomic Vibrations • Problem: Vertical Spring (Text: 14-3) • Problem: Transport Tunnel • SHM Review • Look at textbook problems Chapter 14: # 51, 53, 57, 58, 65, 125

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