Rotation

1. y x Rotation • Imagine watching a spinning bicycle wheel: • How would you describe the position of a point (a reflector, for example) on this wheel with time? • You could keep track of the (x,y) coordinates of this point: (x,y)

2. y x Angular Position q • Instead, it’s more convenient to describe the angleqthat the reflector point makes with thex axis on a fixed coordinate system: • Defining a positive rotation direction defines the sign (+ or –) of coordinateq: • We use radians (rad) as the unit to measure angleq (angular position) + qnegative + y y qpositive q q x x q = 1 rad when it subtends arc length s = r

3. Angular Velocity • Thusq(in radians) = s / r or s = rq • To convert between radians and degrees, use a conversion factor like: 180° = p radians • To determine the number of radians in 32°:  (x rad)(180°) = (p rad)(32°)  x = 0.5585 rad • Rotational motion can be described in analogy to linear motion: • We can thus define an average angular velocitywaveto be: At time t = t1: At time t = t2: q1 q2 x x (depends on angular displacementq2 - q1)

4. Angular Velocity • We can also define an instantaneous angular velocitywsimilar to the linear case: • The units of angular velocity are radians/sec • 1 rev/s = 2p rad/s so 1 rev/min = 1 rpm = 2p/60 rad/s • The direction ofwis given by the Right–Hand Rule: • Curl fingers of right hand in direction of the rotation • Thumb points in direction ofw • Thuswpoints parallel to axis of rotation (not in direction of rotation) • Note that all points on a rotating object move in a circle centered about the axis of rotation

5. Angular Acceleration • The average time rate of change ofwis called the average angular accelerationaave: • The angular acceleration at a particular instant in time is the instantaneous angular accelerationa: • When the rotation axis is fixed,alies along that axis (likew), but could point in the same or opposite direction asw (i.e. parallel or antiparallel to w)

6. Qualitative Relationships Betweenwanda • The qualitative behavior ofwandacan be summarized by the following table: • These were the same relationships between linear velocity and acceleration in Chapter 2! (faster rotation in + dir.) Same sign (faster rotation in – dir.) (slower rotation in – dir.) Opposite sign (slower rotation in + dir.)

7. Constant Angular Acceleration w  Constant angular acceleration means constant rate of increase forw • For special case of constant angular acceleration, thenwvs.t graph becomes a straight line: • In analogy to motion with constant linear acceleration: • Notice the similarity between linear(x, v, a) and angular (q, w, a) quantities and relationships t

8. v r q x Axis of rotation Relating Linear and Angular Kinematics • Consider the reflector on the rotating bike wheel again: • The angleq(in radians)and arc lengthssubtended byqare related bys = rq • Thus linear displacement Ds and angular displacement Dq are related by: (think of string unwinding from a wheel) • Called the “non-slip” condition • A similar relationship holds between the linear (or tangential) speedvand the angular speedw: • The farther a point is from the rotation axis, the greater its linear speed for a given value ofw

9. Relating Linear and Angular Kinematics • This relationship holds true for every point in a rigid body • Rigid body = object that has perfectly definite and unchanging shape and size • All points thus have the same angular velocityw • Taking this a step further, we see another similar relationship between tangential acceleration and angular acceleration: • In general, a tangential acceleration changes the linear speed of a point on a rotating object (i.e. the magnitude of the velocity)

10. Example Problem #7.12 A 45.0-cm diameter disk rotates with a constant angular acceleration of 2.50 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x–axis at this time. At t = 2.30 s, find (a) the angular speed of the wheel, (b) the linear velocity and the tangential acceleration of P, and (c) the position of P (in degrees, with respect to the positive x–axis). • Solution (details given in class): • 5.75 rad/s • 1.29 m/s, 0.562 m/s2 • 76.2° CCW from the +x–axis

11. CQ1: Interactive Conceptual Example: Which animation shows the correct path of the ball once the string breaks? • Animation A • Animation B • Animation C • Animation D • Animation E PHYSLET Exercise 7.1.2, Prentice Hall (2001)

12. Uniform Circular Motion • Circular motion of a puck on a frictionless horizontal table with constant speed: Top view: tangent to the circle at each pt.  Changing direction of means an acceleration directed toward center of circle R

13.  Magnitude of = arad = v2 / R(“centripetal” acceleration) Uniform Circular Motion R  Or, sincev = Rw: arad = R2w2 / R = Rw2 • Source of arad: centripetal force • From Newton’s 2nd law: F = Fnet = marad = mv2 / R

14. Circular Motion in a Vertical Plane mg • Consider the circular motion of an object swung in a vertical circular path: • At the top: Fc = Fnet = –mg – T • At the bottom: Fc = Fnet = T – mg T T mg

15. Fp Fp (top) (bottom) mg mg Analysis of Swinging Pail of Water Top: +y Bottom: vt vb Free–body diagrams of water: Force exerted on water by pail at top: SFy = may = m(–vt2 / r)  – Fp – mg = m(–vt2 / r)  Fp=m (vt2 / r) – mg

16. Analysis of Swinging Pail of Water Minimum value ofvtfor water to remain in pail:  Minimum force pail can exert on water is zero, so setFp = 0 (arad = –g) and solve for minimum speedvt,min: 0 - mg = m(–vt,min2 / r)  vt,min2 = rg vt,min = (rg)1/2 Force exerted on water by pail at bottom: SFy = may = m(vb2 / r)  Fp – mg = m(vb2 / r)  Fp=m(vb2 / r) + mg *Remember that centripetal force is not a special external force acting on a body – it is just the name of the net external force directed toward the center of the circular motion

17. 1 2 H – 2R y y2 = 2R H y1 = 0 Design of a Loop–the–Loop Roller Coaster Suppose we wish to design the following Loop–the–Loop roller coaster: R What is the minimum height H necessary for the roller coaster car to remain on the track at the top of the loop? (Car falls under the influence of gravity only.) • Conservation of mechanical energy: ½ mv12 + mgy1 = ½ mv22 + mgy2 • Assume that roller coaster starts from rest at top of hill. Then we have:mgH = ½ mv22 + mg(2R) • v22 = 2mg(H – 2R) / m = 2g(H – 2R)

18. Design of a Loop–the–Loop Roller Coaster • For car to make it safely over the loop:acar = arad g(remember water in bucket) • arad = v22 / Rg • 2g(H – 2R) / R g • H  R / 2 + 2R = 5R / 2 • H  5R / 2  So, for example, ifR = 22 cm thenHneeds to be at least55 cm.

19. Example Problem #7.75 In a popular amusement park ride, a rotating cylinder of radius 3.00 m is set in rotation at an angular speed of 5.00 rad/s (see figure at right). The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider’s clothing and the wall is needed to keep the rider from slipping? Solution (details given in class): ms = 0.131

20. Example Problem #7.31 A 40.0-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 18.0 m. (a) What is the centripetal acceleration of the child? (b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? (c) What force does the seat exert on the child at the highest point of the ride? (d) What force does the seat exert on the child when the child is halfway between the top and bottom? 9.0 m (Original figure from University Physics, Young and Freedman, 11th Ed., Pearson Addison Wesley) Solution (details given in class): (a) 1.58 m/s2 (b) 455 N upward (c) 329 N upward (d) 397 N directed inward at 80.8° above the horizontal

21. CQ2: Interactive Example Problem:Racecar Safety What is the maximum linear speed of the car before it starts skidding off the track? • 20 m/s • 30 m/s • 50 m/s • 100 m/s • 400 m/s (ActivPhysics Online Problem #4.5, Pearson/Addison Wesley)

22. Gravitation • We are already familiar with concept of gravitational attraction from concept of weight (gravitational attractive force exerted by Earth): • Newton generalized this force to that acting between any 2 bodies in his Law of Gravitation (1687): • In Newton’s words:I thereby compared the force requisite to keep the Moon in her orb with the force of gravity at the surface of the Earth, and found them answer pretty nearly. • Magnitude of gravitational force: • G = constant = 6.67  10–11 Nm2 / kg2 • Very small value  gravity is a weak (the weakest!) force • Gravity very important for large masses, however W = mg

23. m1 Fg Gravitation r Demo Fg m2 • Direction of the gravitational force: • Along line joining 2 particles (attractive force) • m1exerts force of equal magnitude, opposite direction asm2exerts onm1(Newton’s 3rd Law) • Newton also showed (using integral calculus) that: • True of points outside the matter distribution • Inside a sphere, it’s a different story – for a solid sphere of uniform density,Fgis proportional tor m1 m2 Fg Fg Fg Fg m1 m2 = r r (point masses concentrated at centers) (spherically symmetric masses)

24. Gravitation |Fg| R • Graph of|Fg| vs. rfor a solid sphere of uniform density: • When 1 of the masses is the Earth (having massmE):Fg = GmEm / r2 where r = dist. frommto Earth’s center • At Earth’s surface:Fg = GmEm / RE2(RE = radius of Earth) • This is just the weight of mass m at the surface:W = Fg = GmEm / RE2(again, measured at Earth’s surface) ~ 1/r2 ~ r r

25. Weight • But, we also defined weight asW = mg, so evidently: g = GmE/ RE2 = (6.67  10–11 Nm2/kg2)(5.97  1024 kg) / (6.38  106 m)2 = 9.8 m/s2(in agreement with the value we have been using thus far) • We can now see how both the weight and the gravitational acceleration decrease with distance from Earth: • W = GmEm / r2 • a = Fg / m = GmE/ r2 • So why does an astronaut in a spacecraft orbiting Earth experience a feeling of weightlessness?

26. CQ3: If M is the mass of the earth, m is the mass of the moon, and d is the distance between their centers, which of the following gives the instantaneous velocity of the moon as it orbits the earth? (The universal gravitational constant is given by G.)

27. CQ4: If the radius of the orbit of a satellite orbiting the earth is reduced by a factor of 2, the gravitational force on the earth will: • decrease by a factor of 2. • remain the same. • increase by a factor of 2. • increase by a factor of 4.

28. Example Problem #7.41 • A satellite moves in a circular orbit around Earth at a speed of 5000 m/s. Determine • the satellite’s altitude above the surface of Earth and • the period of the satellite’s orbit. Solution (details given in class): (a) 9.57  106 m (b) 5.57 h