1. Review: Uniform Circular Motion In Part 2 we looked at uniform circular motion: r = constant (def. of circular motion)  =  t (  changes with time at uniform rate where  = constant) v r = 0 (radius doesn’t change) v  =  r (where  t) a r =  2 r (v changes direction toward center) a  = 0 (angular speed doesn’t change)

2. Angular Acceleration Here we generalize circular motion to include the case where the angular speed can change. We define angular acceleration as:  t. [Recall arclength = s, and  =s/r, so s=r , v  =r  ] Since a  =  v  /  t =  r)/  t = r(  t) =  r. We still have all the equations we had before, except a  is no longer zero but is instead  r.

3. NON-uniform Circular Motion Rectangular viewpoint Circular motion is defined by: r = constant. We still have d  /dt = , but  is no longer held constant. This means that d  /dt =  rather than zero. Converting polar to rectangular, we have: x = r cos(  ) = r cos[  (t)] y = r sin(  ) = r sin[  (t)].

4. Non-uniform Circular Motion Rectangular viewpoint x = r cos[  (t)] y = r sin[  (t)] Now we use: v x = dx/dt and v y = dy/dt: v x = dx(t)/dt = dx  d  * d  /dt = r (- sin(  ) ) *  = -  r sin(  ) v y = dy(t)/dt = dy  d  * d  /dt = r (+ cos(  ) ) *  =  r cos(  )

5. Non-uniform Circular Motion Rectangular viewpoint x = r cos(  ) y = r sin(  ) v x = -  r sin(  )This v is the same as for v y =  r cos(  )uniform circular motion Now we use: a x = dv x /dt and a y = dv y /dt. Since  is not constant, we need to use the product rule. a x = -  r sin(  ) + -    r cos(  ) a y = +  r cos(  ) + -    r sin(  ).

6. Acceleration v x = -  r sin(  ) v y =  r cos(  ) Now we use: a x = dv x /dt and a y = dv y /dt. Since  is not constant, we need to use the product rule : a x = dv x /dt = d[-  r sin(  ) ]/dt = r d[-  /dt sin(  ) + -  r d[sin(  )]/dt = -  r sin(  ) + -    r cos(  ). In a similar way, we get: a y = +  r cos(  ) + -    r sin(  ).

7. Non-uniform Circular Motion Back to Polar To convert back to polar for position, we use the inverse transformation equations: x = r cos(  ) y = r sin(  ) r = [x 2 + y 2 ] 1/2 = [r 2 cos 2 (  ) + r 2 sin 2 (  )] 1/2 = r[cos 2 (  ) + sin 2 (  )] 1/2 = r  = inv tan[y/x] = inv tan[ r sin(  ) / r cos(  )] = inv tan[tan(  )] = .

8. Non-uniform Circular Motion Back to Polar To convert back to polar for velocity, we use the inverse transformation equations: v x = -  r sin(  )v y =  r cos(  ) v = [v x 2 + v y 2 ] 1/2 = [  2 r 2 sin 2 (  ) +  2 r 2 cos 2 (  )] 1/2 =  r[sin 2 (  ) + cos 2 (  )] 1/2 =  r  v = inv tan[v y /v x ] = inv tan[  r cos(  ) / -  r sin(  )] Note that cos(  ) = sin(  +90 o ) and –sin(  ) = cos(  +90 o ) so  v =  . Note that the direction of the velocity,  v, is perpendicular to the direction of the position (the radius), which means the velocity is tangent to the circle.

9. Non-uniform Circular Motion Back to Polar  v =   Note that the direction of the velocity,  v, is perpendicular to the direction of the position (the radius), which means the velocity is tangent to the circle.

10. Non-uniform Circular Motion Back to Polar To convert acceleration: a x = -  r sin(  ) + -    r cos(  ) a y = +  r cos(  ) + -    r sin(  ). a = [a x 2 +a y 2 ] 1/2 = [  2 r 2 sin 2 (  ) + 2  2 r 2 sin(  )cos(  ) +  4 r 2 cos 2 (  ) +  2 r 2 cos 2 (  ) - 2  2 r 2 sin(  )cos(  ) +  4 r 2 sin 2 (  )] 1/2 = [(  r) 2 + (   r) 2 ] 1/2 = [a t 2 +a r 2 ] 1/2. We can identify (  2 r) as the radial acceleration - due to the turning of the velocity, and we now identify (  r) as the tangential acceleration - due to the changing speed of the velocity.

11. Non-uniform Circular Motion Back to Polar a x = -  r sin(  ) + -    r cos(  ) a y = +  r cos(  ) + -    r sin(  ). a = [a t 2 +a r 2 ] 1/2 where a t =  r and a r =  2 r. If we look at just the  r terms, then the direction of these terms is:  at = inv tan[ cos(  ) / -sin(  ) ] =  +/- 90 o (just like for  v which is tangent to the circle), And if we look just at the  2 r terms, then the direction of these terms is:  ar = inv tan[ -sin(  ) / -cos(  ) ]  +/- 180 o (which is towards the center).

12. Constant Angular Acceleration In the special case of constant angular acceleration (  = constant), we have equations analogous to those we had for constant (regular) acceleration: d  /dt =  = constant, becomes  o   d  = t=0  t  dt, or    t like v x = v xo + a x t. d  /dt =  becomes  o   d  = t=0  t  dt = t=0  t (  o  t)  dt, or     t + (1/2)  t 2 like x = x o +v xo t+(1/2)a x t 2.

13. Constant Angular Acceleration If a car rolls on wheels that do not slip, we can relate the motion of the car to the spinning of the wheels: s = distance of car = arclength of wheels, v = speed of car = v  of wheels, a = acceleration of car = a  of wheels;  = angle of wheel,  = angular speed of wheel, and  = angular acceleration of wheel.

14. Constant Angular Acceleration For circular motion: s = r ,r = constant (circular) v = r  v  v r =0 a = r  = a  a r =  2 r (due to turning) For constant acceleration: s = s o + v o t + (1/2)at 2 v = v o + at     t + (1/2)  t 2    t. This gives many relations so that, by knowing just a few things, we can solve for many other things about the motion.

15. Example Constant Angular Acceleration Consider a car that accelerates from rest to a speed of 20 m/s (45 mph) in 10 seconds on wheels (tires) of outside diameter 66 cm. Assume that the tires roll nicely and do not slip on the road. Through what angle (θ) do the tires rotate during the 10 seconds? What is the final angular speed (ω) at the 10 second mark? What is the angular acceleration (α) of the wheels during the 10 seconds?

16. Example We can use the equations for constant angular acceleration, or we can use the relations between distance and angle. x o = 0 m v o = 0 m/s t = 10 seconds x = ? v = 20 m/s θ o = 0 rad ω o = 0 rad/sec θ = ? ω = ? α = ? d = 66 cm, r = d/2 = 33 cm a = ? t = 0

17. Example Since the car is rolling (and not slipping), we recognize that x = s = rθ, and that v = v   = r  Using the equations for constant angular acceleration along with r  v  :     t + (1/2)  t 2    t v = 20 m/s, t = 10 sec.,    0  rad, r =.33 m, v o = 0 m/s, so   = 0 rad/sec.

18. Example (1)r  v  becomes:.33 m *  20 m/s     t + (1/2)  t 2 becomes  = 0 + 0 + (1/2)*  *(10 sec) 2    t becomes:  = 0 +  *(10 sec) From Eq. 1:  = (20 m/s) /.33 m = 60.6 rad/s From Eq. 3:  =  /t = (60.6 rad/s) / 10 sec = 6.06 rad/s 2. From Eq. 2:  = (1/2)*  *(10 sec) 2 = (1/2)*  rad/s 2 )*(10 sec) 2 = 303 rad * (1 cycle/2  rad) = 48.2 revolutions

19. Example – another way Instead of using the constant angular acceleration formulas, we could solve the regular motion and use the relations between regular quantities and angular quantities: s = r ,r = constant (circular) v = r  v  v r = 0 a = r  = a  a r =  2 r (due to turning)

20. Example another way s = r ,r = constant (circular) v = r  v  v r = 0 a = r  = a  a r =  2 r (due to turning) s = s o + v o t + (1/2)at 2 v = v o + at 20 m/s = 0 + a*(10 sec), so a = (20 m/s) / (10 s) = 2 m/s 2, so a =  r, or  = a/r = (2 m/s 2 ) / (.33 m) = 6.06 rad/s 2. s = 0 + 0 + (1/2)*a*(10 sec) 2 = (1/2)*(6.06 rad/s 2 )*(10 sec) 2 = 303 radians. v =  r, or  = v/r = (20 m/s) / (.33 m) = 60.6 rad/s.

21. Rotational Kinematics The computer homework program, Rotational Kinematics, Volume 2 #7, contains explanations and problems involving Rotational Kinematics using the equations in the previous slide.

22. Rotational Force (Torque) Forces cause change in the motion, but in Part 2 we only considered motion that changes the position of the object. What about changing the “spin” or rotation of an object? To get a nice introduction to the idea of torque, see the computer homework program on Introduction to torque (Vol.2 #5), and continue with the next slide.

23. Torque There are two important quantities in torque: Force and where you apply the force (called radius).  = r x F = r F sin(  rF ) To get a large torque, we need to use a large radius, a large force, and apply the force perpendicular to the radius!

24. Statics and Torque Just as  F = 0 when the object is static, so also  = 0 when the object is not spinning (or spinning at a constant rate). In static cases, there is no obvious center to measure the radius from, so we are free to choose any point. However, some points may be simpler to use than others.

25. Your elbow Let’s consider as an example of torque how your muscles, bones and joints work. Consider holding up a ball of weight 5 lb. How does this work? (We will choose to measure distances from the elbow.) First we draw a diagram: biceps triceps weight = elbow r w rbrb

26. Your elbow In terms of forces and distances, the diagram looks like this: Estimate the distance from your elbow joint F c F b to your biceps connect point, r b ; and to your hand, r w. r c =0 r b r w W

27. Your elbow If the ball weights 5 lb, how much force does your biceps pull up with? How much force of contact does your upper arm push down with on your lower arm at the elbow? What is the basic principle to use? Statics:  F = 0 and .

28. Your elbow From  F = 0 we have: -F c + F b - W = 0 And from  and measuring from the elbow gives: F c *r c + F b *r b - W*r w = 0. We have two equations and we have two unknowns (F c and F b ).

29. Your elbow We can use the torque equation first, since r c =0 eliminates one of the unknowns, F c. F c *r c + F b *r b - W*r w = 0 or F b = W*r w /r b. Then we can use the force equation to find F c :- F c + F b - W = 0, or F c = F b - W.

30. Your elbow By putting in reasonable values for r b and r w, you can see that the biceps have to exert a large force to hold up a relatively light weight! What advantage does this give? Note how far the biceps have to contract in order to move the weight! This is the advantage of the elbow set- up! In practice, we use clubs and rackets to make this difference even greater!

31. Car going around a turn: Sliding off the road or flipping over For a car going around a turn, if the car goes too fast the car will either slide off the road or flip over. What determines which will occur, and how fast can the car go around the turn without either happening? Since we now introduce “flipping over”, we have to consider rotational motion, and hence consider torque. This means we need to consider not only what the forces are, but where they act.

32. Sliding or Flipping We’ll consider a car making a right turn on an unbanked (horizontal) turn. We have the weight of the car (W=mg), which acts at the center of gravity. We have the contact force, which acts on each of the left and right wheels at the road: F cL and F cR. We have the friction force, which also acts on each of the left and right wheels at the road: F fL and F fR, where F f   F c We also have an acceleration to the right: a c = v 2 /r.

33. a c = v 2 /r Sliding or Flipping h The weight, W=mg, acts at the center of gravity, which is at a height, h, above the road. The contact forces, F c act at the tires which are on the road. The friction forces, F f, also act at the road. The tires are a distance, w, apart. w

34. Sliding or Flipping h  F x = F fL + F fR = m v 2 /r  F y = F cL + F cR - mg = 0 Let’s choose to consider the center of gravity as the point for rotations. The distances for the torques need to be distances that are  to the direction of the forces. We will also choose clockwise as positive for torques. w a c = v 2 /r

35. w Sliding or Flipping h  F x = F fL + F fR = m v 2 /r  F y = F cL + F cR - mg = 0  = -F fL *h - F fR *h + F cL *(w/2) - F cR *(w/2) + mg*0 = 0. Since the F cL is the only force causing a + torque, it must increase and F cR must decrease if the  = 0. This can happen until F cR = 0 and F cL = mg. When this happens, F fR = 0. a c = v 2 /r

36. Sliding or Flipping  F x = F fL + 0 = m v 2 /r, or F fL  m v 2 /r  F y = F cL + 0 - mg = 0, or F cL = mg  = -F fL *h - 0*h + F cL *(w/2) - 0*(w/2) + mg*0 = 0, or F fL *h = F cL *(w/2). If friction is maximum (limit for sliding) we have: F fL  m v 2 /r with F f =  F c =  mg, so  mg = m v 2 /r, or v max = [  gr] 1/2. If flipping is to be prevented we have F fL = mv 2 /r, F cL =mg, or mv 2 /r * h = mg*(w/2), or v max = [grw/2h] 1/2.

37. Sliding or Flipping v max = [  gr] 1/2 applies when sliding is a concern; v max = [grw/2h] 1/2 applies when flipping over is a concern. Which maximum speed is lower - that tells us the true maximum speed to continue around the turn: If  < w/2h, then sliding is the main concern; if  > w/2h, then flipping is the main concern. For big trucks, h is large, and unless the street is very slick, flipping over will be bigger; for sports cars, h is small and sliding is more of a concern.

38. Statics and Torque The computer homework program, Statics and Torque, Volume 2 #6, contain explanations and problems dealing with this topic.

39. Torque and Rotations Forces (when not balanced) cause changes in motion. Torques (when not balanced) cause changes in rotational motion. Forces are related to acceleration by Newton’s Second Law:  F = ma. How are  and  related?

40. Rotational Dynamics Consider:  F = ma and  = r F sin(  rF ). If we multiply both sides of  F = ma by r sin(  rF ), we get:  = m a r sin(  rF ) ; but a sin(  ra ) = a   r, so  = mr 2 . If the mass is distributed at various radii, then the mr 2 becomes  m i r i 2. We define the moment of inertia be be I =  m i r i 2 so that we have  = I .

41. Moment of Inertia Note that the moment of inertia relates the torque to the angular acceleration just like the mass relates the force to the regular acceleration. Note that the moment of inertia depends on the mass of the object and also on the shape of the object.

42. Moment of Inertia Objects that have most of their mass in the center have smaller moments of inertia than objects that have most of their mass on the outside. Using the calculus, we can replace the summation of small mass pieces with an integral: I =  m i r i 2 becomes I =  r 2 dm. Each small dm has its own radius, or distance from the axis of rotation, so r is really a function of which dm we are using: r(m).

43. Moment of Inertia I =  r 2 dm It is awkward to express r(m). We can, however, use density:  = m ass /V olume = dm/dV to convert dm into  dV. Then the integral is really a Volume integral (triple integral). I =  r 2  dV Using this, we can derive nice formulas for the moment of inertias of certain shapes: I ring = MR 2 I sphere = (2/5)MR 2 ; I cylinder = (1/2)MR 2.

44. Sphere versus Cylinder The blue object in the figures is the side view of a cylinder with the same radius and volume (mass) as the orange object (a sphere). [The top view would like the same: circles.] Note that most of the mass is in the same place, but the top and bottom of the sphere must be moved away to provide the corners for the cylinder. Hence, the Moment of Inertia of the cylinder is slightly larger than that of the sphere: I sphere = (2/5)MR 2 ; I cylinder = (1/2)MR 2. axis R

45. Rotational Energy Now that we have the equivalent of Newton’s Second Law for rotations:  = I , let’s see if we can get a rotational energy relation. Recall that W ork =  F  ds. If we exert a force to spin an object, the ds becomes a small arc length that is perpendicular to the radius. Recall the definition of an angle in radians:  = s/r, so that ds = r d . But F times r is torque, so we get: W ork =   d .

46. Rotational Power W ork =   d  The definition of power is: P = dW ork /dt, and so for rotations we have: P = dW/dt = d[   d  dt =  This formula for rotational power is similar to that for regular power: P = F  v P = .

47. Rotational Kinetic Energy If something is spinning, it can still not be moving so that it’s regular kinetic energy is zero. However, parts of the object are moving, so there is kinetic energy associated with its spinning. We use three relations we have already developed: KE = ½mv 2, v  =  r, and I =  m i r i 2 KE total =  ½m i v i 2 =  ½m i (  r i ) 2 = ½  2  m i r i 2 = ½I  2. This energy is used in the Conservation of Energy relation just as any other energy term is used.

48. Angular Momentum Another important tool in solving problems is Conservation of Momentum. Is there a similar tool for rotations? We already have: p = mv and  F = dp/dt. If we multiply both sides of Newton’s Second law by the radius we have (with v =  r):  = r  F = r  dp/dt = d(r  p)/dt = dL/dt (where we have used the fact that dr/dt  p = v  mv = 0) where we have defined L = r  p, where L is called the Angular Momentum.

49. Conservation of Angular Momentum  = dL/dt Just like  F = dp/dt leads to Conservation of momentum if no external forces are present, so  = dL/dt leads to Conservation of Angular momentum if no external torques are present. Note: p = mv, and L = r  p = r  mv = r m v  = r m  r = mr 2  = I .

50. Angular Momentum In class we will observe several demonstrations about Angular Momentum and talk about its uses in controlling rockets and spinning athletes (divers and gymnasts in particular). We will also show how angular momentum relates to riding bicycles, and how this relates to the spinning earth!

51. Review of Rotational Equations Basically we replace F with , m with I, v with , a with  and  p with L (where L is the angular momentum):  F = ma  = I  W ork = =  F  dsW ork =   d  P ower = F  vP ower =  KE = (1/2)mv 2 KE rotation = (1/2)I  2 p = mvL = I   F =  p/  t  =  L/  t.

52. Rotational Motion The computer homework program, Volume 2 #8 on Rotational Dynamics gives an overview and several problems involving rotational motion.

53. Example: rolling ball How fast will a ball be going at the base of a ramp if it is released at the top of the ramp, where the length of the ramp is 2 meters and the high end is 50 cm above the floor (base of the ramp)? We assume the ball rolls without slipping. Does the mass or radius of the ball affect the speed? If a cylinder of the same mass and radius were also released, would the ball or cylinder win the race down the ramp?

54. Rolling ball Ball: v i = 0,  i =0 mass = m radiius = r h i = 50 cm s = 2 meters v f = ?  f = ?

55. Rolling ball We could use Newton’s Laws of Motion:  F = ma, and  = I  along with a=  r and the equations for constant acceleration, or we could use Conservation of Energy (KE regular + KE rotational + PE gravity ) initial = (KE regular + KE rotational + PE gravity ) initial + E lost along with v=  r.

56. Rolling ball Since we’re not interested in direction or time, we’ll use Conservation of Energy. We know that for a ball (sphere), I = (2/5)mr 2. We know that KE regular = (1/2)mv 2. We know that KE rotational = (1/2)I  2. We know that PE gravity = mgh. We know that without slipping, even though friction causes the ball to roll, there is no E lost because the ball does not slide.

57. Rolling ball Conservation of Energy (KE regular + KE rotational + PE gravity ) initial = (KE regular + KE rotational + PE gravity ) initial + E lost 0 + 0 + mgh = (1/2)mv 2 + (1/2)I  2 + 0 + 0. Substituting I=(2/5)mr 2 and  =v/r gives: mgh = (1/2)mv 2 + (1/2)[(2/5)mr 2 ][v 2 /r 2 ], or mgh = (1/2)mv 2 + (1/5)mv 2 = (7/10)mv 2.

58. Rolling ball mgh = (7/10)mv 2 which becomes when solving for v: v = [(10/7)gh] 1/2 = [(10/7)*9.8m/s 2 *.50m] = 2.645 m/s. Note: both the mass and the radius cancel out of the solution, and so the mass and radius don’t matter.

59. Rolling ball Note: the angle of the ramp doesn’t make any difference either, since only the height and not the length is involved. The steeper the ramp, the faster the ball will reach its final speed, but the final speed will not be affected!

60. Sphere versus Cylinder Would a cylinder or a sphere win a race down the ramp? What difference does the shape make in our equation? The difference comes in the KE for rotations in the moment of inertia, I. For a sphere (ball), I = (2/5)mr 2, while for a cylinder it would be I = (1/2)mr 2. This ultimately changes the (10/7) factor into (4/3), the value from 2.65 m/s to 2.55 m/s.

61. Sphere versus Cylinder We can understand the slightly lower speed for the cylinder by considering that the moment of inertia of the cylinder is larger. This will make the kinetic energy larger for the same speed. Hence there will be less kinetic energy for the regular speed if we have the same potential energy to begin with!