There is no time to cover: Chapter 10 Please Read it!!! - PowerPoint PPT Presentation

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There is no time to cover: Chapter 10 Please Read it!!!

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  1. I am sorry!! There is no time to cover: • Chapter 10 Please Read it!!!

  2. Chapter 10. Rotation of a Rigid Object About Fixed Axis

  3. 10.1 Angular Position, Velocity and Acceleration • Topic of Chapter: Bodies rotating • First, rotating, without translating. • Then, rotating AND translating together. • Assumption:Rigid Object

  4. Rigid Object • A rigid object is one that is non-deformable • The relative locations of all particles making up the object remain constant • All real objects are deformable to some extent, but the rigid object model is very useful in many situations where the deformation is negligible • Rigid Object motion≡Translational motion of center of mass (everything we’ve done up to now) + Rotational motion about axis through center of mass (this chapter!). Can treat two parts of motion separately.

  5. Angular Position • Axis of rotation is the center of the disc • Choose a fixed reference line • Point P is at a fixed distance rfrom the origin • Point P will rotate about the origin in a circle of radius r • Every particle on the disc undergoes circular motion about the origin, O

  6. Angular Position, 2 • Polar coordinates: • Convenient to use to represent the position of P (or any other point) • P is located at (r, q) where ris the distance from the origin to P and qis the measured counterclockwise from the reference line • Motion: • As the particle moves, the only coordinate that changes is q • As the particle moves through q, it moves though an arc length s. • The arc length and r are related: s = qr(10.1a)

  7. Radian • This can also be expressed as (10.1b) • q • is a pure number, ratio of 2 lengths (dimensionless) • commonly is given the artificial unit, radian • Solving problems in this chapter, need calculators inRADIAN MODE!!!! • One radian is the angle subtended by an arc length equal to the radius of the arc • When srq1 Radian

  8. Conversions • Comparing degrees and radians • Converting from degrees to radians • Converting from radians to degrees

  9. Example 10.1Birds of Pray- in Radians • Bird’s eye can distinguish objects that subtended an angle no smaller than 3x10- 4 rad • (a) 310-4 rad =  º? = (310-4 rad)(180º/πrad)  =0.017º • (b)r = 100 m,  = ?  = r = (100)  (310-4)  = 0.03 m = 3 cm MUST be in radians in part (b)!

  10. Angular Position, final • We can associate the angle q with the entire rigid object as well as with an individual particle • Remember every particle on the object rotates through the same angle • The angular position of the rigid object is the angle qbetween the reference line on the object and the fixed reference line in space • The fixed reference line in space is often the x-axis

  11. Angular Displacement • The angular displacement of the rigid object is defined as the angle the object rotates in a time interval t • This is the angle that the reference line of length r sweeps out

  12. Average Angular Speed • The average angular speed, (Greek omega), of a rotating rigid object is the ratio of the angular displacement() to the time interval (t) (10.2)

  13. Instantaneous Angular Speed • The instantaneous angular speed is defined as the limit of the average speed as the time interval approaches zero (10.3)

  14. Angular Speed, final • Units of angular speed are radians/sec • rad/s or s-1 since radians have no dimensions • Angular speed will be positive if θ is increasing (counterclockwise) • Angular speed will be negative if θ is decreasing (clockwise)

  15. Average Angular Acceleration • The average angular acceleration, (Greek alpha) of a rotating object is defined as the ratio of the change in the angular speed () to the time it takes for the object to undergo the change: (10.4)

  16. Instantaneous Angular Acceleration • The instantaneous angular acceleration is defined as the limit of the average angular acceleration as the time goes to 0 (10.5)

  17. Angular Acceleration, final • Units of angular acceleration are rad/s² or s-2 since radians have no dimensions • Angular acceleration will be positive if an object rotating counterclockwise is speeding up • Angular acceleration will also be positive if an object rotating clockwise is slowing down

  18. Angular Motion, General Notes • When a rigid object rotates about a fixed axis in a given time interval, every portion on the object rotates through the same angle in a given time interval and has the same angular speed and the same angular acceleration • So q, w, a all characterize the motion of the entire rigid objectas well as the individual particles in the object

  19. Directions, details • Strictly speaking, the speed and acceleration (w, a ) are the magnitudes of the velocity and acceleration vectors • The directions are actually given by the right-hand rule • The direction of a follows from its definition a = d/dt • Same aswif increases • Opposite toif decreases

  20. Hints for Problem-Solving • Similar to the techniques used in linear motion problems • Constantangularacceleration techniques ≡ Constantlinearacceleration techniques • There are some differences to keep in mind • For rotational motion, define a rotational axis • The choice is arbitrary • Once you make the choice, it must be maintained • The object keeps returning to its original orientation, so you can find the number of revolutions made by the body

  21. 10. 2 Rotational Kinematics: Rotational Motion with Constant Angular Acceleration • Under constant angular acceleration, we can describe the motion of the rigid object using a set of kinematic equations • These are similar to the kinematic equations for linear motion • The rotational equations have the same mathematical form as the linear equations

  22. Rotational Kinematics, final • We’ve seen analogies between quantities: LINEAR and ANGULAR Displacement Angular Displacement x ≈  Velocity Angular Velocity v ≈  Acceleration Angular Acceleration a ≈ 

  23. Rotational Equations • For αconstant,we can use the equations from Chapter 2 with this replacements!! (10.6) (10.7) (10.8) (10.9)

  24. Comparison Between Rotational and Linear Equations These are ONLY VALID if all angular quantities are in radian units!!

  25. Displacements Speeds (10.10) Accelerations (10.11) Every point on the rotating object has the same angular motion Every point on the rotating object does not have the same linear motion 10.3 Angular and Linear Quantities

  26. Speed Comparison • The linear velocity is always tangent to the circular path • called the tangential velocity • The magnitude is defined by the tangential speed (10.10)

  27. Acceleration Comparison • The tangential acceleration is the derivative of the tangential velocity (10.11)

  28. Speed and Acceleration Remarks • All points on the rigid object will have the same angular speed, but not the same tangential speed • All points on the rigid object will have the same angular acceleration, but not the same tangential acceleration • The tangential quantities depend on r, and r is not the same for all points on the object

  29. Centripetal Acceleration • An object traveling in a circle, even though it moves with a constant speed, will have an acceleration • Therefore, each point on a rotating rigid object will experience a centripetal acceleration (10.12)

  30. Resultant Acceleration • The tangential component of the acceleration is due to changing speed • The centripetal component of the acceleration is due to changing direction • Total acceleration can be found from these components (10.13)

  31. Translational-RotationalAnalogues & Connections ANALOGUES TranslationRotation Displacementx  Velocityv  Accelerationa  CONNECTIONS s = r v = rat = r  ac = v2/r = 2 r

  32. Rotational Motion Example • For a compact disc player to read a CD, the angular speedmust vary to keep the tangential speed constant (vt = wr ) • At the inner sections,the angular speed isfasterthan at the outer sections

  33. Example 10.2Bicycle • A bicycle slows down uniformly from vi = 8.4 m/s. To rest (f = 0) over a distance of 115 m. Diameter of wheel = 0.69 m ( r = 0.34 m). • FIND: • (a) i : v = r i =vi /r = 8.4 m/s/0.34 m  i =24.7 rad/s 

  34. Example 10.2Bicycle, final • (b)Total revolutions of the wheels before come to rest:  s/r = 115 m/0.34 m = 338.2 rad/2πrad/rev 53.8 rev • (c).= (f2 – i2)/2 = (0 – 24.72)/2(53.8)rad/s2 =0.902 rad/s2 • (d). Time to stop: t = (f – i)/= (0 – 24.7)/0.902 s  t =27.4 s

  35. 10.4 Rotational Kinetic Energy • An object rotating about some axis with an angular speed, ω, has rotational kinetic energy even though it may not have any translational kinetic energy • Each particle has a kinetic energy of Ki = ½ mivi2 • Since the tangential velocity depends on the distance, r, from the axis of rotation, we can substitute vi = wi r

  36. Rotational Kinetic Energy, cont • The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles (10.14) (10.16) • I≡ Σmi ri2 (10.15)MOMENT OF INERTIA

  37. Rotational Kinetic Energy, final • There is an analogy between the kinetic energies associated with linear motion (K = ½ mv 2) and the kinetic energy associated with rotational motion (KR= ½ Iw2) • Rotational kinetic energyis not a new type of energy, the form is different because it is applied to a rotating object • The units of rotational kinetic energy are Joules (J)

  38. Example 10.3Four Rotating Objects • Find Iy and KR for this system. • Radii of the spheres <<a or b. • Iy≡ Σmi ri2  Iy= m(0) + m(0) + Ma2 +Ma2 Iy= 2Ma2 • KR = ½Iyw2 = ½ (2Ma2)w2 • KR = Ma2w2

  39. Example 10.3Four Rotating Objects, final • Find Iz and KR for this system. • Radii of the spheres <<a or b. • Iz≡ Σmi ri2  Iz= mb2 + mb2 + Ma2 +Ma2 Iz= 2mb2 +2Ma2 • KR = ½Izw2  KR = ½ (2mb2 +2Ma2)w2 KR = (2mb2 +Ma2)w2

  40. 10.5 Moment of Inertia • The definition of moment of inertia is • The dimensions of moment of inertia are ML2 and its SI units are kg.m2 • We can calculate the moment of inertia of an object more easily by assuming it is divided into many small volume elements, each of mass Dmi

  41. Moment of Inertia, cont • We can rewrite the expression for Iin terms of Dm (10.17) • Since r = m/V (Volumetric Mass Density). • For the small volume segment r = dm/dV or dm = r dV, thenIwill be: (10.17a) • If ris constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known

  42. Notes on Various Densities • Volumetric Mass Density: mass per unit volume: r≡m/V • For the small volume segment r = dm/dV or dm = rdV • Surface Mass Density: mass per unit area: ≡m/A • For the small area segment  = dm/dA or dm = dA • of a sheet of uniform thickness, t : s≡m/A= mt/At= (m/V )t = rt  s≡rt • Linear Mass Density: mass per unit length: l≡m /L • For the small length segment l = dm/dLor dm = ldL • lof a rod of uniform cross-sectional area A: l = m /L = mA /AL = (m/V )A =rA  l = rA

  43. Example 10.4Moment of Inertia of a Uniform Thin Hoop • Since this is a thin hoop, all mass elements are the same distance (R) from the center

  44. Example 10.5Moment of Inertia of a Uniform Rigid Rod • The shaded area has a mass • dm = l dx • Then the I thru its CM is:

  45. Example 10.6Moment of Inertia of a Uniform Solid Cylinder • Divide the cylinder into concentric shells with radius r, thickness dr and length L • Then for I

  46. Moments of Inertia of Various Rigid Objects

  47. Parallel-Axis Theorem • In the previous examples, the axis of rotation coincided with the axis of symmetry of the object • For an arbitrary axis, the parallel-axis theorem often simplifies calculations • The theorem states: I = ICM + MD 2(10.18) • Iis about any axis parallel to the axis through the center of mass of the object • ICM is about the axis through the center of mass • D is the distance from the CM axis to the arbitrary axis

  48. Parallel-Axis Theorem Example • The axis of rotation goes through O • The axis through CM is shown • The moment of inertia about the axis through Owould be: IO = ICM + MD 2

  49. Example 10.7 Moment of Inertia for a Rod Rotating Around One End • Iof the rod about its CM is • D is equal to ½L • Therefore:

  50. Torque • Torque, t, is the tendency of a force to rotate an object about some axis • Torque is a vector • t = r F sin f = F d • F is the force • f is the angle the force makes with the horizontal • d is the moment arm (or lever arm)