Lecture 6

1. Lecture 6 Goals • Discuss uniform and non-uniform circular motion • 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.

2. Uniform Circular Motion (UCM) has only radial acceleration Dq v UCM changes only in the direction of 1. Particle doesn’t speed up or slow down! 2. Velocity is always tangential, acceleration perpendicular !

3. Again vT ar r Uniform circular motion involves only changes in the direction of the velocity vector Acceleration is perpendicular to the trajectory at any point, acceleration is only in the radial direction. Centripetal Acceleration ar = vT2/r = w2r Circular motion involves continuous radial acceleration

4. Uniform Circular Motion (UCM) is common so we have specialized terms s vT r q New definitions 1. Period (T): The time required to do one full revolution, 360° or 2p radians 2. Frequency (f): f≡ 1/T, number of cycles per unit time Angular velocity or speed in UCM w = Dq / Dt = 2p / T = 2p f (radians per unit time)

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 !!! but a neutron star surface is at 1012 m/s2 bb5

6. 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

7. 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 )

8. 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

9. 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

10. Sketch of q vs. time q = qo + wDt q = 20prad + (2p x2/5) 5 rad q = 24 rad 30p q = qo + wDt q = 0 + 4p 5rad 20p q (radians) 10p 0 10 5 time (seconds)

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

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

13. Non-uniform CM: What if w is linearly increasing … • Then angular velocity is no longer constant so dw/dt ≠ 0 • Define tangential acceleration as aT ≡ dvT/dt = r dw/dt • Define angular acceleration a ≡ dw / dt • Let a be constant & integrating: • = w0 + aDt Integrating again: q = q0 + w0Dt + ½ aDt2 • Multiply by r rq= r q0 + r w0Dt + ½ r aDt2 s= s0 + vTDt + ½ aTDt2 • Many analogies to linear motion but it isn’t one-to-one

14. 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).

15. 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 =0.48 p m/s2 • 7 What is the magnitude and direction of the acceleration after 10 seconds?

16. 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

17. 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!

18. 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 ?

19. 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.

20. 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.

21. Lecture 6 Assignment: Read Chapter 5 & 6.1-6.4