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Status: Unit 4 - Circular Motion and Gravity

Status: Unit 4 - Circular Motion and Gravity. Uniform circular motion, angular variables, and the equations of motion for angular motion (3-9, 10-1, 10-2) Applications of Newton’s Laws to circular motion (5-2, 5-3, 5-4) Harmonic Motion (14-1, 14-2)

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Status: Unit 4 - Circular Motion and Gravity

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  1. Status: Unit 4 - Circular Motion and Gravity • Uniform circular motion, angular variables, and the equations of motion for angular motion (3-9, 10-1, 10-2) • Applications of Newton’s Laws to circular motion (5-2, 5-3, 5-4) • Harmonic Motion (14-1, 14-2) • The Universal Law of Gravitation, Satellites, and Kepler’s Law’s (Chapter 6) Physics 253

  2. Something New: Oscillations and Vibrations • These are seen everywhere, at every length scale • Oscillating cesium atoms as clocks 10-10m • Tuning forks (10-3m) • Your car on a washboard road (10m) • Lake Michigan in a seiche (105m) • The sun itself (109m) • All but the first are explained fully with Newtonian mechanics. Physics 253

  3. Physics 253

  4. Solar Vibration Modes Physics 253

  5. Oscillation of a Spring • A repeating vibration or oscillation is an example of periodic motion. • A mass connected to spring is one of the easiest systems with periodic motion to contemplate. • The spring is massless • We ignore friction • The spring is not deformed • At the equilibrium position the mass simply rests with v=0. • If the spring is compressed or stretched the spring tries to move the mass back to the equilibrium point. The force is opposite the displacement. • This is an example of a restoring force – always present in oscillations or vibrations! Physics 253

  6. Oscillation of a Spring • More precisely we have Hooke’s Law. • The equilibrium position is x=0 • The minus sign means the restoring force is always opposite the displacement • In the figure if x is positive the spring is trying to pull it back and F is negative • If x is negative the spring is trying to push it away and F is positive. • The proportionality is called the “spring constant”. • This is a “new” force to be included in Newton’s second law. • And since the force is not constant, the equations of motion for constant acceleration do not apply. Physics 253

  7. Oscillation of a Spring: • Consider the mass at displacement x=+A, • The stretched spring exerts a restoring force towards the equilibrium point. However, the mass gains speed and passes the equilibrium point. • The spring now compressed pushes back, slows, and finally stops the spring at displacement x=-A. • It the is pushed through the equilibrium point. and slowed by the stretched spring displacement. • The mass slows and stops at displacement x=+A. • In the absence of friction the motion repeats and we have an oscillation. Physics 253

  8. Robert Hooke 1635 - 1703 An accomplished scientist of his time Physics 253

  9. Oscillating Spring Nomenclature & Metrics • Displacement, x: distance x from equilibrium. • Amplitude, A : Maximum displacement or distance, unit typically meters. • Cycle: On complete motion from an initial point back through the equilibrium position and back. Usually from A to –A to A. • Period, T: time for one cycle, unit typically seconds • Frequency, f: number of cycles per second, unit the hertz. f= 1/T and T = 1/f. Physics 253

  10. A family of total mass 200kg piles into their 1200kg car and the springs compress 0.030 m. What is the overall spring constant for the suspension system? If they add 100kg of luggage how far will the car lower? We start we F=-kx and just focus on the magnitude F=kx. The families weight (200kg*9.8m/s2) gives us the force that compresses the springs by 0.030cm Now that we have spring constant we can easily calculate the displacement for a given force: Example 1: Car Springs Physics 253

  11. To measure k one only need measure the displacement for a given force. This is most easily done by hanging a know mass from a vertical string. Once in equilibrium the 2nd Law gives: For m=1.0kg, x=0.01m: Ex 2: Measuring the Spring Constant Physics 253

  12. Simple Harmonic Motion • A simple harmonic oscillator (SHO): Any vibrating system for which the net restoring force is directly proportional to the negative displacement and exhibits simple harmonic motion (SHM). • We can use Newton’s second law and approach it much like we did terminal velocity last lesson to find the position, velocity, and acceleration as a function of time. Physics 253

  13. Which is the equation of motion for the simple harmonic oscillator. This is a differential equation. Basically these equations are solved with a big bag of tricks, one of which is GUESSING! We can do a little better by motivating our guess… Physics 253

  14. Physics 253

  15. The sinusoidal functions appear plausible. We simply try one to see if it satisfies the equation of motion. Substituting Our solution has three variables: the amplitude, A the phase, f, the frequency, w (constrained by the diff. eq.). Physics 253

  16. Actually the amplitude and phase are determined by the initial conditions. • Let’s imagine the mass starts at distance x0 and is released at rest. • By assumption at t=0 • x=x0 • v=0. • So the amplitude is just given by the initial position. • And in this case the phase ascribed to the initial velocity Physics 253

  17. Let’s try these initial conditions, at t=0 • x=0 • v is positive Physics 253

  18. The phase angle or f, just indicates how long after t=0 the peak of full amplitude is reached. • It corresponds to a displacement of the curve at time t. This can be seen by looking for the 1st maxima: Physics 253

  19. More SHM Metrics. • The SHO repeats its motion after a period T which is analytically described by the arguement of the sinusoidal functions repeating every 2p or wt=2p. • This can be rearranged to w=2p/T = 2pf. We’ve seen this before! • w is the angular frequency • f is the frequency of motion and also the natural frequency. • The equations of motion can be re-expressed as: • The frequency, angular frequency, and period do not depend on the amplitude, only the mass and spring constant. For instance • the greater the spring constant or the stiffer the spring the higher the frequency • the greater the mass the lower – both intuitive. Physics 253

  20. Example 3: Back to the Car • What’s the period and frequency of the car after hitting a speed table? • We have the mass of the car and we derived spring constant so it’s really just plug-and-chug. Which is the reason changing speed helps when your car bounces on I88 – it avoids the natural frequency Physics 253

  21. Physics 253

  22. A loudspeaker cone vibrates at 262Hz at t=0, x=A=.00015m. What equation describes the cone? What is the maximum velocity and acceleration and position at t=0.001s? The general solution for this SHM is We already have A, require f=0 for a maximum at t=0, and can derive w with 2pf = 2p(262Hz) =1650rad/s, so We know that And finally at t=0.001s: Example 4: A Loudspeaker Physics 253

  23. A vertical spring stretches 0.15m with a mass of 0.300kg. The spring is stretched 0.100 more and released. Determine k, w, A, the maximum velocity and acceleration, T, f, x(t), and v at t=0.150s. First off we just use the result from Ex 2 to calculate k=mg/x= (0.300kg)(9.80m/s2)/0.150m= 19.6N/m. Then w=sqrt(k/m)= sqrt((19.6N/m)/0.300kg)=8.08s-1 A=0.100m since the spring is held motionless at release. Example 5: Full Treatment of a Spring Physics 253

  24. vmax=wA=(8.08s-1)(0.100m)= • amax=w2A=(8.08s-1)2(0.100m)= 6.53m/s2 • T=2psqrt(m/k)= 6.28sqrt(0.300kg/19.6N/m)= 0.777s • f=1/t=1/0.777s=1.29 Hz • Eq.: At t=0 we have x=x0=-A=-0.100m (assuming that +x is up): • Now the velocity is just given by the derivative: Physics 253

  25. Relation to Circular Motion • Springs and simple harmonic are related to circular motion through the frequency and angular frequency. • http://www.kineticbooks.com/email/whiteboard.html Physics 253

  26. Schedule • Next Quiz a week from Wednesday : Wednesday March 7th. • Units 3 and 4: • Force and Laws of Motion • Circular Motion and Gravity. • Homework and Extra Credit for Units 3 and 4 due March 7th. Physics 253

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