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Circular Motion and Forces: Lecture 6, Chapter 4 Review

This lecture covers the topic of circular motion, including different reference frames and forces acting on objects. It also discusses equilibrium and non-equilibrium forces using Newton's laws. Assignments and exam details are provided.

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Circular Motion and Forces: Lecture 6, Chapter 4 Review

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  1. Lecture 6 Chapter 4 • Discuss circular motion • Discern differing reference frames and understand how they relate to particle motion Chapters 5 & 6 • Recognize different types of forces and know how they act on an object in a particle representation • Identify forces and draw a Free Body Diagram • Solve problems with forces in equilibrium (a=0) and non-equilibrium (a≠0) using Newton’s 1st & 2nd laws. Assignments: HW3 (Chapters 4 & 5), finish reading Chapter 6 Exam 1: Thurs. Oct. 7 from 7:15-8:45 PM Chapters 1-7

  2. Concept Check Q1. You drop a ball from rest, how much of the acceleration from gravity goes to changing its speed? A. All of it B. Most of it C. Some of it D. None of it Q2. A hockey puck slides off the edge of the table, at the instant it leaves the table, how much of the acceleration from gravity goes to changing its speed? A. All of it B. Most of it C. Some of it D. None of it

  3. Circular Motion (quick “review”) s vT r q • Angular position q • Arc traversed s = r q • Tangential speed |vT | = Ds / Dt & if Dt  0 ds / dt = r dq /dt • Angular velocity w = dq /dt = | vT | / r • Radial acceleration | ar | = vT2 / r = w2 r

  4. Uniform Circular Motion (UCM, a=0) Period (T): The time required to do one full revolution, 360° or 2p radians s vt r q Frequency (f): 1/T, number of “cycles” per unit time Angular velocity or angular speed w = 2p f = 2p/T Positive  counter clockwise Negative  clockwise

  5. Mass-based separation with a centrifuge Before After How many g’s (1 g is ~10 m/s2)? ar = vT2 / r = w2 r f = 6000 rpm = 100 rev. per second is typical with r = 0.10 m ar = (2p 102)2 x 0.10 m/s2 ar = 4 x 104 m/s2 or ca. 4000 g’s !!! bb5

  6. g’s with respect to humans • 1 g Standing • 1.2 g Normal elevator acceleration (up). • 1.5-2g Walking down stairs. • 2-3 g Hopping down stairs. • 3.5 g Maximum acceleration in amusement park rides (design guidelines). • 4 g Indy cars in the second turn at Disney World (side and down force). • 4+ g Carrier-based aircraft launch. • 10 g Threshold for blackout during violent maneuvers in high performance aircraft.   • 11 g Alan Shepard in his historic sub orbital Mercury flight experience a maximum force of 11 g. • 20 g Colonel Stapp’s experiments on acceleration in rocket sleds indicated that in the 10-20 g range there was the possibility of injury because of organs moving inside the body. Beyond 20 g they concluded that there was the potential for death due to internal  injuries. Their experiments were limited to 20 g. • 30 g The design maximum for sleds used to test dummies with commercial restraint and air bag systems is 30 g. Comment: In automobile accidents involving rotation severe injury or death can occur even at modest speeds

  7. Circular Motion s vT r q • Radial acceleration| ar | = vT2 / r = w2 r • Angular acceleration a = d2q /dt = | aT | / r s = s0+ vT0Dt + ½ aTDt2

  8. Example • A horizontally mounted disk 2.0 meters in diameter (1.0 m in radius) spins at constant angular speed such that it first undergoes (1) 10 counter clockwise revolutions in 5.0 seconds and then, again at constant angular speed, (2)2 counter clockwise revolutions in 5.0 seconds. • 1 What is T the period of the initial rotation? T = time for 1 revolution = 5 sec / 10 rev = 0.5 s

  9. Example • A horizontally mounted disk 2 meters in diameter spins at constant angular speed such that it first undergoes 10 counter clockwise revolutions in 5 seconds and then, again at constant angular speed, 2 counter clockwise revolutions in 5 seconds. • 2 What is w the initial angular velocity? w = dq /dt = Dq /Dt w = 10 • 2π radians / 5 seconds = 12.6 rad / s( also 2 p f = 2 p / T )

  10. Example • A horizontally mounted disk 2 meters in diameter spins at constant angular speed such that it first undergoes 10 counter clockwise revolutions in 5 seconds and then, again at constant angular speed, 2 counter clockwise revolutions in 5 seconds. • 3 What is the tangential speed of a point on the rim during this initial period? | vT | = ds/dt = (r dq) /dt = r w | vT | = r w = 1 m • 12.6 rad/ s = 12.6 m/s

  11. Example s vt r q • A horizontally mounted disk 1 meter in radius spins at constant angular speed such that it first undergoes 10 counter clockwise revolutions in 5 seconds and then, again at constant angular speed, 2 counter clockwise revolutions in 5 seconds. • 4 Sketch the q (angular displacement) versus time plot. q = q0 + w0Dt

  12. Sketch of q vs. time q = qo + wDt q = 20prad + (4p/5) 5 rad q = 24 rad 30p q = qo + wDt q = 0 + 4p 5rad 20p q (radians) 10p 0 10 5 time (seconds)

  13. Example • 5 What is the average angular velocity over the 1st 10 seconds?

  14. Sketch of q vs. time q = qo + wDt q = 20prad + (4p/5) 5 rad q = 24 rad 30p q = qo + wDt q = 0 + 4p 5rad 20p q (radians) 10p 0 10 5 time (seconds) 5Avg. angular velocity = Dq / Dt = 24 p /10 rad/s

  15. Example • 6 If now the turntable starts from rest and uniformly accelerates throughout and reaches the same angular displacement in the same time, what must be the angular acceleration a ? Key point ….. a is associated with tangential acceleration (aT).

  16. Tangential acceleration? aT 1 r 2 s vt r q • 6 If now the turntable starts from rest and uniformly accelerates throughout and reaches the same angular displacement in the same time, what must the “tangential acceleration” be? q = qo + woDt + Dt2 (from plot, after 10 seconds) 24 p rad = 0 rad + 0 rad/s Dt + ½ (aT/r) Dt2 48 p rad 1m / 100 s2 = aT • 7 What is the magnitude and direction of the acceleration after 10 seconds?

  17. Non-uniform Circular Motion For an object moving along a curved trajectory, with varying speed Vector addition: a = ar + aT(radial and tangential) aT ar a

  18. Tangential acceleration? s vt r q aT = 0.48 p m / s2 and vT = 0 + aTDt = 4.8 p m/s = vT ar = vT2 / r = 23 p2 m/s2 • 7 What is the magnitude and direction of the acceleration after 10 seconds? Tangential acceleration is too small to plot!

  19. Do different observers see the same physicsRelative motion and frames of reference • Reference frame S is stationary • Reference frame S’ is moving at vo This also means that S moves at – vo relative to S’ • Define time t = 0 as that time when the origins coincide

  20. Relative Velocity • The positions, r and r’, as seen from the two reference frames are related through the velocity, vo, where vo is velocity of the r’ reference frame relative to r • r’ = r – voDt • The derivative of the position equation will give the velocity equation • v’ = v – vo • These are called the Galilean transformation equations • Reference frames that move with “constant velocity” (i.e., at constant speed in a straight line) are defined to be inertial reference frames (IRF); anyone in an IRF sees the same acceleration of a particle moving along a trajectory. • a’= a (dvo/ dt = 0)

  21. y(t) motion governed by 1) a = -gy 2) vy = v0y – gDt 3) y = y0 + v0y – g Dt2/2 Reference frame on the moving cart. x motion: x = vxt Reference frame on the ground. Net motion: R = x(t) i + y(t) j(vector) Central concept for problem solving: “x” and “y” components of motion treated independently. • Example: Man on cart tosses a ball straight up in the air. • You can view the trajectory from two reference frames:

  22. No Net Force, No acceleration…a demo exercise • In this demonstration we have a ball tied to a string undergoing horizontal UCM (i.e. the ball has only radial acceleration) 1 Assuming you are looking from above, draw the orbit with the tangential velocity and the radial acceleration vectors sketched out. 2 Suddenly the string brakes. 3 Now sketch the trajectory with the velocity and acceleration vectors drawn again.

  23. Chaps. 5, 6 & 7What causes motion?(What is special about acceleration?)What are forces ?What kinds of forces are there ?How are forces and changes in motion related ?

  24. Newton’s First Law andIRFs An object subject to no external forces moves with constant velocity if viewed from aninertial reference frame (IRF). If no net force acting on an object, there is no acceleration. • The above statement can be used to define inertial reference frames.

  25. IRFs • An IRF is a reference frame that is not accelerating (or rotating) with respect to the “fixed stars”. • If one IRF exists, infinitely many exist since they are related by any arbitrary constant velocity vector! • In many cases (i.e., Chapters 5, 6 & 7) the surface of the Earth may be viewed as an IRF

  26. Newton’s Second Law The acceleration of an object is directly proportional to the net force acting upon it. The constant of proportionality is the mass. • This is a vector expression: Fx, Fy, Fz • Units The metric unit of force is kg m/s2 = Newtons (N) The English unit of force is Pounds (lb)

  27. Lecture 6 Assignment: Read rest of chapter 6

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