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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.

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Chapter 12Rotational Motion

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  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)

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