1 / 106

Chapter 12Rotational Motion

Chapter 12Rotational Motion. 1) 1 radian = angle subtended by an arc (l) whose length is equal to the radius (r) 2) q = l r 360 0 = 2 p radians Radians are dimensionless. l. r. q.

oistin
Download Presentation

Chapter 12Rotational Motion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 12Rotational Motion 1) 1 radian = angle subtended by an arc (l) whose length is equal to the radius (r) 2) q = l r • 3600 = 2p radians • Radians are dimensionless l r q

  2. A bird can only see objects that subtend an angle of 3 X 10-4 rad. How many degrees is that? 3 X 10-4 rad 360o = 0.017o 2p rad

  3. How small an object can the bird distinguish flying at a height of 100 m? q = l r l = q r l = (3 X 10-4 rad)(100m) l = 0.03 m = 3 cm q r l (approx.)

  4. How at what height would the bird be able to just distinguish a rabbit that is 30 cm long (and tasty)? (ANS: 1000 m) q r l (approx.)

  5. The Mighty Thor swings his hammer at 400 rev/min. Express this in radians/s. 400 rev 1 min 2p rad 1 min 60 s 1 rev = 13.3p rad/s or 41.9 rad/s

  6. Formula Review v = rw at = ra ar = v2 or ar = w2r r

  7. Converting between Angular and Linear Quantities atan Linear = Radius X Angular v = rw atan = ra Note the use of atan to differentiate from centripetal acceleration, ac or ar: ar

  8. Angular Kinematics v=vo + at w=wo + at x = vot + ½ at2 q = wot + ½ at2 v2 = vo2 + 2ax w2 = wo2 + 2aq

  9. A DVD (Encino Man, Director’s Cut), rotates from rest to 31.4 rad/s in 0.892 s. • Calculate the angular acceleration. (35.2 rad/s2) • How many revolutions did it make? (2.23 revolutions)

  10. A car engine idles at 500 rpm. When the light turns green, it accelerates to 2500 rpm in 3.0 s. • Convert the angular velocities to rad/s • Calculate the angular acceleration • Calculate the number of revolutions the wheel undergoes.

  11. A bicycle slows from vo = 8.4 m/s to rest over a distance of 115 m. The diameter of each wheel is 68.0 cm. • Calculate the angular velocity of the wheels before braking starts. (24.7 rad/s) • How many revolutions did each wheel undergo?( 53.8 rev) • What was the angular acceleration? (-0.903 rad/s2)

  12. Center of mass • one point on a particle that follows the same path. • Point at which the force of gravity can be considered to act (uniform gravity field, Center of gravity)

  13. General Motion • Translational Motion • all points of an object follow the same path • Sliding a book across a table • Rotational Motion • General Motion – combination of translational and rotation motion

  14. Translational Translational and Rotational

  15. xcm = 1 ∫ mixi = m1x1 + m2x2 + …. M m1 + m2 xcm = 1 ∫ x dm M

  16. A 500 g ball and a 2.0 kg ball are connected by a massless 50 cm rod. • Calculate the center of mass (0.10 m) • Calculate the linear speed of each ball if they rotate at 40 rpm. (0.42 m/s, 1.68 m/s)

  17. Where is the center of mass for the Earth-Moon system. Assume the center of the Earth is the origin. Some values you need are: mEarth = 5.97 X 1024 kg mMoon = 7.35 X 1022 kg Earth-Moon distance = 3.84 X 108 m (4.45 X 106 m)

  18. Using calculus, find the center of mass of a thin uniform rod of mass M and length L. Use this result to rind the tangential acceleration of a 1.60 m long rod rotates about its center with an angular acceleration of 6.0 rad/s2. xcm = 1 ∫ x dm M dm = M dx (substitute and integrate from 0 to L) L

  19. Translational and Rotational Speed • Does a rotating helicopter blade have kinetic energy before the helicopter takes off? • How about afterwards? • Does all of the energy of the fuel go into moving the helicopter?

  20. Translational and Rotational Speed Translational Speed (v) • speed of the center of a wheel with respect to the ground • Can also be called linear speed • Use regular KE = ½ mv2 Rotational speed (w) • angular speed of the wheel • Use KE = ½ Iw2

  21. Deriving the Rotational KE KE = S½ mv2 v = rw KE = S ½ m(rw)2 KE = S ½ mr2w2 I = S mr2 KE = ½ Iw2(I is moment of inertia)

  22. Law of Conservation of Mechanical Energy Emech= KEt + KEr + PEt

  23. Calculate the moment of inertia of the following “widget.” Assume the connecting rods are massless. (2.14 X 10-3 kg m2) • Calculate the angular velocity in rpm if the system has 100 mJ of rotational kinetic energy (92 rpm)

  24. Rotational KE: Example 1 What will be the translational speed of a log (100 kg, radius = 0.25 m, I= ½ mr2) as it rolls down a 4 m ramp from rest? 4 m

  25. (KEt + KEr + PEt)i = (KEt + KEr + PEt)f (0 + 0 + mgy)i = ( ½ mv2 + ½ Iw2 + 0)f mgy = ½ mv2 + ½ Iw2 2mgy = mv2 + Iw2 (multiplied both sides by 2) v = rw so w = v/r 2mgy = mv2 + Iv2 r2 I = 1/2 mr2 2mgy = mv2 + 1mr2v2 2r2

  26. 2mgy = mv2 + 1mr2v2 2r2 2gy = v2 + 1v2 2 2gy = 2v2 + 1v2 2 2 2gy = 3v2 2 v = 4gy = 4(9.8m/s2)(4.00 m) = 7.23 m/s 3 3

  27. Now we can calculate the angular speed v = rw so w = v/r • = 7.23 m/s = 2.9 rad/s 0.25 m 2.9 rad 1 rev = 0.46 rev/s s 2p rad

  28. Moment of Inertia (I) • Measure of Rotational Inertia • An objects resistance to a change in angular velocity • Would it be harder to push a child on a playground merry-go-round or a carousel?

  29. I = moment of inertia • I = mr2 • More properly I = Smr2 = m1r12 + m2r22 +…. St= Ia Would it be harder (require more torque) to twirl a barbell in the middle (pt. M) or the end (Pt. E) E M

  30. Moment of Inertia: Example 1 Calculate the moment of inertia (I) for the barbell when rotated about point M. We will assume the barbell is 1.0 m long, and that each weight is a point mass of 45.4 kg. I = Smr2 = (45.4 kg)(0.50 m)2 + (45.4 kg)(0.50 m)2 I = 22.7 kg-m2 M

  31. Moment of Inertia: Example 2 Now calculate I assuming Mr. Fredericks twirls the barbells from point E. I = Smr2 = (45.4 kg)(0 m)2 + (45.4 kg)(1 m)2 I = 45.4 kg-m2 E

  32. Calculate I for Bouncing Boy (75 kg, radius = 1.2 m). Use the formulas from the book. I = 2/5 MR2 I = (2)(75 kg)(1.2 m)2 5 I = 43.2 kg-m2

  33. What will be the translational speed of Bouncing Boy (75 kg, radius = 1.2 m) as he rolls down a 3 m ramp from rest? (6.50 m/s) 3 m

  34. What would his speed be if he just slid down the ramp? (KEt + KEr + PEt)i = (KEt + KEr + PEt)f (0 + 0 + mgy)i = ( ½ mv2 + 0 + 0)f mgy = ½ mv2 gy = ½ v2 v2 = 2gy v = \/2gy = (2 X 9.8 m/s X 3.00 m)1/2 v = 7.7 m/s Why is this larger than if he rolls?

  35. Calculate whether a 5.0 kg sphere, a 5.0 kg hoop, or a 5.0 kg cylinder will reach the bottom of a 1.0 m tall ramp first.

  36. A 1.0 m long, 200 g rod is hinged at one end and allowed to fall. (I = 1/3ML2) . Note: use the height of the center of mass for potential energy height. • Derive the formula for the linear speed of the tip of the bar at the bottom on the fall. (v=(3gL)1/2) • Calculate the speed of the tip of the rod as it hits the wall. (5.4 m/s)

  37. Calculus and Moment of Inertia. I = ∫ r2 dm Always need to substitute for dm in terms of r

  38. Calculate the moment of inertia of a thin rod of length L and mass M that pivots at one end. I = ∫ x2 dm dm = M dx L I = ∫ x2M dx L

  39. Calculate the moment of inertia of a thin disk of radius R and Mass M. I = ∫ r2 dm dm = M dA A dA = 2prdr

  40. Calculate the moment of inertia of a thin rod of length L and mass Mthat pivots through the middle. (Hint: integrate from -1/2L to 1/2L)

  41. Parallel Axis Theorem I = Icm + Md2 Determine the moment of inertia of a thin rod of mass M and length L one-third of the length from one end.

  42. Determine the formula for the moment of inertia of a solid sphere about an axis on the edge of the sphere. (7/5Mr2) Calculate the moment of inertia if the mass is 12.0 kg and the radius is 0.75 m. (9.45 kg m2)

  43. Torque – tendency of a force to rotate a body about some axis (the force is always perpendicular to the lever arm) • = Frsinq • = Ia r pivot F

  44. Torque Sign Conventions Counter-clockwise Torque is positive Clockwise Torque is negative

  45. A wrench is 20.0 cm long and a 200.0 N force is applied perpendicularly to the end. Calculate the torque. t = Fr t = (200.0 N)(0.20 m) t = 40.0 m-N 20.0 cm 200.0 N

  46. Suppose that same 200.0 N force is now applied at a 60o angle as shown. Calculate the Torque. (34.6 m-N) 20.0 cm 200.0 N 60o

  47. The biceps muscle exerts a 700 N vertical force. Calculate the torque about the elbow. • = Fr = (700 N)(0.050 m) • = 35 m-N

  48. Two wheels, of radii r1 = 30 cm and r2 =50 cm are connected as shown. • Calculate the net torque on this compound wheel when two 50 N force act as shown. (-6.5 m-N) 50 N 30o r2 Note that Fx will pull the wheel r1 50 N

  49. Out in space, two rockets are docked, connected by a 90 m tube. One has a mass of 100,000 kg, the other 200,000 kg. They both fire their rockets with 50,000 N of thrust in opposite directions. • Calculate the center of mass of the system (60 m) • Calculate the moment of Inertia about the center of mass (use mr2 for the rockets) (5.4 x 108) • Calculate the net torque. (4.5 X 106 Nm) • Calculate the angular acceleration. (8.33 X 10-3rad/s2)

More Related