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Holt Physics Chapter 7

Holt Physics Chapter 7. Rotational Motion. Measuring Rotational Motion. Spinning objects have rotational motion Axis of rotation is the line about which rotation occurs A point that rotates around an axis undergoes circular motion. s=arc length, r=radius

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Holt Physics Chapter 7

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  1. Holt Physics Chapter 7 Rotational Motion

  2. Measuring Rotational Motion • Spinning objects have rotationalmotion • Axis of rotation is the line about which rotation occurs • A point that rotates around an axis undergoes circular motion

  3. s=arc length, r=radius • =angle of rotation (measured in degrees or radians) corresponding to s s  r

  4. (rad) = s/r • If s = r, then  = 1 radian • If s = 2r (circumference), • then  (rad) = 2r/r= 2 radians • Therefore 360 degrees = 2 radians, or 6.28 • See figure 7-3, page 245

  5. s  r

  6. Equation to convert between radians and degrees •  (rad) = [/180º] X  (deg) • Sign convention: • clockwise is negative, counterclockwise is positive!

  7. Angular Displacement () describes how much an object has rotated •  = s/r

  8. s  r

  9. Book Example 1: While riding on a carousel that is rotating clockwise, a child travels through an arc length of 11.5m. If the child’s angular displacement is 165 degrees, what is the radius of the carousel? •  = -165 degrees but…we need radians! • rad= [/180º] Xdeg = (/180º)(-165º) = -2.88 radians • s = -11.5 m r=? •  = s/r so, r=s / • r = -11.5/-2.88 = 3.99m

  10. Example 2. A child riding on a merry-go-round sits at a distance of 0.345 m from the center. Calculate his angular displacement in radians, as he travels in a clock-wise direction through an arc length of 1.41 m.

  11. Given: r = 0.345 m Ds = -1.41 m Find: Dq Solution: The negative sign in our answer indicates direction.

  12. Homework • Practice 7A, page 247

  13. Angular Speed (ave) is equal to the change in angular displacement(in radians!) per unit time. • ave= / t , in radians/sec • Remember, One revolution = _______ radians, 2p

  14. so the number of revolutions =/2rad

  15. s  r

  16. Book Example 3: A child at an ice cream parlor spins on a stool. The child turns counterclockwise with an average angular speed of 4.0 rad/s. In what time interval will the child’s feet have an angular displacement of 8.0  radians. • =8 rad • ave= 4.0 rad/s t=? • ave= / t, so t=/ ave • t= 8 rad / 4.0 rad/s = 2.0  s • = 6.3 sec

  17. Example 4. A man ties a ball to the end of a string and swings it around his head. The ball’s average angular speed is 4.50 rad/s. A) In what time interval will the ball move through an angular displacement of 22.7 rad? B) How many complete revolutions does this represent?

  18. Given: wavg = 4.50 rad/s Dq = 22.7 rad Find: A) Dt B) # of revolutions Solution:

  19. Angular Acceleration (ave) is the change in angular speed () per unit time(s). • ave= /t = (2-1)/(t2-t1) • Unit is rad/s2

  20. Book Example 1: A car’s tire rotates at an initial angular speed of 21.5 rad/s. The driver accelerates, and after 3.5 sec the tire’s angular speed is 28.0 rad/s. What is the tire’s average angular acceleration during the 3.5 s time interval? • 1=21.5 rad/s 2=28.0 rad/s • t=3.5 s ave= ? • ave= 2-1/t = • ave =(28.0rad/s-21.5rad/s)/3.5s • ave=1.9 rad/s2

  21. Example 2. The angular velocity of a rotating tire increases from 0.96 rev/s to 1.434 rev/s with an average angular acceleration of 6.0 rad/s2 . Find the time required for given angular acceleration. First change revolutions/sec to radians/sec by multiplying by 2. (0.96rev/s)X 2 = 6.0 rad/sec (1.434rev/s)X 2 = 9.01 rad/sec

  22. w1 = 6.0 rad/s w2=9.01 rad/s aavg=6.0rad/s2 Given: t = ? Find: Solution: 0.50s t = 9.01 rad/s – 6.0 rad/s = 6.0 rad/s2

  23. Homework Practice 7B (1,2, and 4) page 248 and 7C (1 and 3), pg 250

  24. Angular Kinematics • All points on a rotating rigid object have the same angular acceleration and angular speed. • Angular and linear quantities correspond if angular speed () is instantaneous and angular acceleration () is constant. • See table 7-2, pg. 251 or your formula sheet

  25. Equations for Motion with Constant Acceleration Linear Rotational

  26. Book Example 1 (page 251)The wheel on an upside-down bicycle moves through 11.0 rad in 2.0 sec. What is the wheel’s angular acceleration if its initial angular speed is 2.0 rad/s? • =11.0 rad t=2.0 s • i= 2.00rad/s  = ? • Use one of the angular kinematic equations from Table 7-2 to solve for  … which one? • it  ½ (t)2

  27. Solve for  •  =2(-it)/(t)2 •  =2[11.0rad-(2.00rad/s)(2.0s)]/(2.0s)2 •  = 3.5 rad/s2

  28. Example 2. A CDinitially at rest begins to spin, accelerating with a constant angular acceleration about its axis through its center, achieving an angular velocity of 3.98 rev/s when its angular displacement is 15.0 rad. what is the value of the CD’s angular acceleration?

  29. wi = 0.0 rad/s wf = 3.98 rev/s X 2 = 25.0 rad/s Dq = 15.0 rad Given: Find: a Original Formula: Thewi = 0.0 rad/s, so

  30. wi = 0.0 rad/s wf = 25.0 rad/s Dq = 15.0 rad Given: Find: a Formula: Now, we isolate a

  31. wi = 0.0 rad/s wf = 25.0 rad/s Dq = 15.0 rad Given: Find: a Working Formula:

  32. wi = 0.0 rad/s wf = 25.0 rad/s Dq = 15.0 rad Given: Find: a Solution:

  33. Homework • Problems 7D, 1-5 on page 252

  34. Tangential Speed • Although any point on a fixed object has the same angular speed, their tangential speeds vary based upon their distance from the center or axis of rotation. • Tangential speed (vt)– instantaneous linear speed of an object directed along the tangent to the objects circular path. (see fig. 7-6, pg. 253)

  35. Formula • vt = r • Make sure  is in radians/s • r is in meters • vt is in m/s

  36. Book Example 1 (pg. 254): The radius of a CD in a computer is 0.0600m. If a microbe riding on the disc’s rim had a tangential speed of 1.88m/s, what is the disc’s angular speed? • r=0.0600m vt=1.88m/s • =? • Use the tangential speed equation (vt = r )to solve for angular speed. • =vt/r • =1.88m/s/0.0600m • =31.3 rad/s

  37. Example 2. A CD with a radius of 20.0 cm rotates at a constant angular speed of 1.91 rev/s. Find the tangential speed of a point on the rim of the disk. Given: r =20.0cm= 0.200 mw = 2X1.91rev/s= 12.0 rad/s nt Find: Solution:

  38. Tangential Acceleration • Tangential Acceleration (at)- instantaneous linear acceleration of an object directed along the tangent to the objects circular path. • Formula • at= r • Make sure that at is in m/s2 • r is in meters •  is in radians/s2

  39. Example 1(at): A spinning ride at a carnival has an angular acceleration of 0.50rad/s2. How far from the center is a rider who has a tangential acceleration of 3.3 m/s2? • =0.50rad/s2 at= 3.3 m/s2 • r=? • Use the tangential acceleration equation (at= r) and rearrange to solve for r. • r=at/ • r = (3.3 m/s2 )/(0.50rad/s2) • r=6.6m

  40. Example 2. The angular velocity of a rotating CD with a radius 20.0 cm increases from 2.0 rad/s to 6.0 rad/s in 0.50 s. What Is the tangential acceleration of a point on the rim of the disk during this time interval? Hint: Start by finding the angular acceleration

  41. Given: r = 0.200 m wi = 2.0 rad/s wf= 6.0 rad/s Dt = 0.50 s a Find: Solution:

  42. Given: r = 0.200 m a = 8.0 rad/s2Dt = 0.50 s at Find: Solution:

  43. Problems 7E (1,2 and 4), page 255 and 7F (all) pg 257

  44. Centripetal Acceleration • Centripetal acceleration is the acceleration directed toward the center of a circular path. • Caused by the change in direction • Formula ac = vt2/r • And also……ac = r2

  45. Example 1 pg. 258: A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05 m/s2. What is its tangential speed? • r=48.2m ac = 8.05 m/s2 vt=? • Use the first centripetal acceleration equation (ac = vt2/r) to solve for vt. • vt2= acXr • vt= acXr • vt= (8.05m/s2X48.2m) • vt= 19.7m/s

  46. Example 2. An object with a mass of 0.345 kg, Moving in a circular path with a radius of 0.25 m, Experiences a centripetal acceleration of 8.0 m/s2. Find the object’s angular speed.

  47. Example 2. An object with a mass of 0.345 kg, Moving in a circular path with a radius of 0.25 m, Experiences a centripetal acceleration of 8.0 m/s2. Find the object’s angular speed. Given: m = 0.345 kg r = 0.25 m ac = 8.0 m/s2 w Find: Original Formula:

  48. Example 2. An object with a mass of 0.345 kg, Moving in a circular path with a radius of 0.25 m, Experiences a centripetal acceleration of 8.0 m/s2. Find the object’s angular speed. Given: m = 0.345 kg r = 0.25 m ac = 8.0 m/s2 w Find: Solution:

  49. Using the Pythagorean Theorem • Tangential and centripetal acceleration are perpendicular, so you can use Pythagorean theorem and trig to find total acceleration and direction, if asked for. • See figure 7-9, on page 259

  50. at ac atotal

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