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USITT 2007 - Phoenix Rotational Motion Verda Beth Martell Dr. Eric C. Martell

USITT 2007 - Phoenix Rotational Motion Verda Beth Martell Dr. Eric C. Martell. What we’re going to talk about. How big of a motor do you need to rotate a turntable with stuff on it? Builds on last year’s talk about linear motion.

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USITT 2007 - Phoenix Rotational Motion Verda Beth Martell Dr. Eric C. Martell

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  1. USITT 2007 - PhoenixRotational Motion Verda Beth MartellDr. Eric C. Martell

  2. What we’re going to talk about • How big of a motor do you need to rotate a turntable with stuff on it? • Builds on last year’s talk about linear motion. • Differences between rotational and linear motion – both conceptually and logistically. • Build a working model of a turntable system.

  3. Our starting point Linear motion starts with Newton’s Second Law: - the sum of the forces acting on the object (lbs) • m - the mass of the object (slugs) • a - the acceleration of the object (ft/s2)

  4. For rotating objects Rotational analogue to Newton’s Second Law: tnet=Ia • tnet - the sum of the torques acting on the object (ft.lb) • I - the moment of inertia for the object (slug.ft2) • a - the angular acceleration of the object (rad/s2)

  5. What is Torque? Torque – a force acting about an axis. t = rF(sinq) F = force (lbs) r = radius from axis to force (ft) q = angle between r and F (will be 90o for turntable drives) 50 lbs * 2 ft = 100 ft.lb 1200 lbs 50 lbs 100 lbs 50 lbs 2 ft 50 lbs 50 lbs at 2 ft = 100 ft.lbs = 1200 in.lbs

  6. What is the Moment of Inertia? tnet=Ia I – Moment of Inertia Like mass only for rotation Measure of how hard it is to get object to rotate faster or slower Measured in slug.ft2

  7. What is Angular Acceleration? Change in angular velocity (w) tnet=Ia a – Angular acceleration Measures rate of change of angular velocity: Measured in rad/s2 w

  8. So, what are we trying to calculate? To spec a motor, we’re going to need the drive torque td td – Drive torque tf – Frictional torques caused by each ring of casters. tnet = td-tf1-tf2 -or- td = tnet+tf1+tf2=Ia+tf1+tf2 So, to determine the drive torque, we need to find the moment of inertia, the angular acceleration, and the frictional torques. tf1 tf2 We’ll start with I…

  9. Moment of Inertia Note that the height of the object does not matter! The Moment of Inertiais the distribution of the mass about the rotational axis. Calculated by splitting the object into very small parts, multiplying the mass of each part by the square of the distance to the axis of rotation, and then adding the results together.

  10. Moment of Inertia Many common shapes have known I’s r1 b r2 r a Hollow Cylinders (Curved Walls) I = ½ m (r1+r2)2 Solid Cylinders (turntables, people) I = ½ m r2 Rectangles (Walls, Cubes) I = 1/12 m (a2+b2)

  11. Moment of Inertia – object on turntable If we had a wall (centered on the axis of rotation) on the turntable: Irectangle= 1/12 m (a2+b2) Use the rectangle formula for solid walls, acting blocks, furniture, etc. These equations are for objects that are centered on the axis of rotation. (What if they’re not?)

  12. Moment of Inertia What if the object is not centered? a The parallel axis theorem can be used for any object that is off center b d Furniture, blocks, walls: I = 1/12 m (a2+b2)+ md2 Furniture, blocks, walls: I = 1/12 m (a2+b2) d Columns, people: I = 1/2 mr2+ md2

  13. Moment of Inertia How big a difference this makes… A square box, 2 ft on a side, weighing 100 lb is placed on-axis: m = 100 lb/32.2 ft/s2 = 3.11 slug I = 1/12*3.11 slug*((2 ft)2 + (2 ft)2) = 2.07 slug.ft2 3 ft off-axis: I = 2.07 slug.ft2 + 3.11 slug*(3 ft)2 I = 30.1 slug.ft2 Move 3 ft off-axis Move 6 ft off-axis 6 ft off-axis: I = 114 slug.ft2 Note: I values are for the block only.

  14. Moment of Inertia What about people? 200 lbs Add people as solid cylinders on the rim of the turntable: Iactor = ½ mr2 + md2 6’ 5’ Let’s work it out… mactor = 200 lbs/32.2 ft/s2 =6.2 slugs Iactor = ½ mr2 + md2 = ½ (6.2 slugs)(1ft)2 + (6.2 slugs)(5 ft)2 = 158.1 slug.ft2

  15. Moment of Inertia The total moment of inertia is the moment of inertia of the turntable plus the moments of inertia of everything on it: Itot = Iactor + Iblock + Iwall + Itt

  16. Finding the Total Moment of Inertia Turntable: r = 6 ft m = 400 lb/32.2 ft/s2 = 12.4 slug Itt = ½m r2 = 223.2 slug.ft2 Wall: a = 12.5 ft, b = 2 ft m = 300 lb/32.2 ft/s2 = 9.3 slug Iwall = 1/12m (a2 + b2) = 124.2 slug.ft2 Box: a = 2 ft, b = 1 ft, d = 4 ft m = 100 lb/32.2 ft/s2 = 3.1 slug Ibox = 1/12 m (a2 + b2) + m d2= 50.9 slug.ft2 Actor: r = 1 ft, d = 5 ft m = 200 lb/32.2 ft/s2 = 6.2 slug Iactor = ½ m r2 + m d2= 158.1 slug.ft2 Itotal = Itt + Iwall + Ibox + Iactor Itotal = 556.4 slug.ft2

  17. Remember our goal We want the Drive Torque td td = Ia+tf1+tf2 We now know how to find I. Next step – how do we find a? tf1 tf2

  18. Angular Acceleration What is a radian? – Measure of the angle where s = r. Remember, a = Dw/Dt Okay, what is Dw? The change in angular velocity, or the final angular velocity (wf) - the initial angular velocity (w0) (if starting from rest, w0=0). d = 12 ft To find w – start with the distance that a point travels around the turntable (say it rotates around once in 30 s). s = pd = 37.68 ft Then we find the angular distance (in radians): q = s/r = 37.68 ft/6 ft = 6.28 rad Finally, w = q/Dt = 6.28 rad/30 sec = .21 rad/s

  19. Angular Acceleration Now to find a Final angular velocity wf= .21 rad/s Initial angular velocity w0 = 0 rad/s d = 12 ft How much time do you have to get up to speed, starting from rest? Say 5 seconds. a = Dw/Dt = (.21 rad/s – 0 rad/s)/5 s a = .042 rad/s2

  20. Net Torque tnet = Ia Total moment of inertia Itotal = 556.4 slug.ft2 Angular acceleration a = 0.042 rad/s2 tnet = 556. 4 slug.ft2 * 0.042 rad/s2 tnet=23.4 ft.lb

  21. Remember our goal We want the Drive Torque td td = tnet+tf1+tf2 We now know how to find tnet=Ia Now we need to find the torque caused by friction. tf1 tf2

  22. Finding the Frictional Torques Consider an empty turntable supported by two rings of casters: There’s a Frictional Force resisting the motion. F = mr*weight on casters in each ring F The weight on the casters in each ring can be calculated a number of ways – simple model – weight on each caster is the same, so: Ring 2 q Weight on ring [from turntable] = (Total weight)*(# of casters in ring) Ring 1 r (Total # of casters) How does this work?

  23. Finding the Frictional Torques Weight on each ring F Ring 1 – 4 casters Ring 2 – 8 casters Total = 12 casters Ring 2 q Ring 1 Weight of turntable = 400 lb r Weight on ring 1 = 400 lb * (4/12) = 133.3 lb Weight on ring 2 = 400 lb * (8/12) = 266.7 lb What about objects on turntable?

  24. Finding the Frictional Torques What about objects on turntable? Actor on edge – weight on outside ring Box near edge – weight on outside ring Wall – weight on both rings Weight on ring 1 = 133.3 lb + ½*weight of wall 133.3 1b + ½*(300 lb) = 283.3 lb Weight on ring 2 = 266.7 lb +½*weight of wall + weight of actor + weight of box = 266.7 lb + ½*(300 lb) + 100 lb + 200 lb = 716.7 lb

  25. Finding the Frictional Torques Ring 1: r1 = 2 ft Ring 2: r2 = 5.5 ft q = 90o Frictional force = mr*weight on ring Coefficient of rolling friction: mr = 0.05 F Friction in ring 1 = 283.3 lb * 0.05 = 14.2 lb Ring 2 q Friction in ring 2 = 716.7 lb * 0.05 = 35.8 lb Ring 1 r Remember: t = r F sin(q) tf1 = 2 ft * 14.2 lb = 28.4 ft.lb tf2 = 5.5 ft * 35.8 lb = 196.9 ft.lb

  26. Calculating Drive Torque Let’s put the pieces together td td = tnet+tf1+tf2 tnet = 23.4 ft.lb tf1 = 28.4 ft.lb tf2 = 196.9 ft.lb td = 248.7 ft.lb tf1 tf2 Now, how big of a motor do we need to accomplish this?

  27. Calculating Drive Force t = r F sin(q) td We just found the drive torque: td = 248.7 ft.lb To find the drive force, we work backwards – Fd = td/r, so we need to know where the drive force is being applied. tf1 tf2 If the force is applied on the edge: Fd = 248.7 ft.lb/6 ft = 41.5 lb

  28. Calculating HP HP = (Fd * v)/550 td In addition to the drive force we just calculated, we also need to know the linear velocity of a point on the turntable where the force is being applied. tf1 tf2 Linear velocity = angular velocity * radius (v = w*r) v = 0.21 rad/s * 6 ft = 1.22 ft/s HP = (41.5 lb * 1.22 ft/s)/550 = 0.09 HP

  29. Things to keep in mind Our model assumes that each item is homogenous despite the fact that they are not. If your turntable has a heavy outer ring (like a steel reinforced facing or multilayered facing), you may want to figure the I value of the ring separately from the turntable itself. r1 r2 Hollow Cylinders (Curved Walls) I = ½ m (r1+r2)2

  30. Things to keep in mind Our model assumes that each item is homogenous despite the fact that they are not. Same thing goes for archways – break the archway into smaller shapes and calculate each same separately.

  31. More Resources Physics of Theatre Website http://www2.kcpa.uiuc.edu/kcpatd/physics/index.htm (You can Google “Physics of Theatre”) • Sample lectures – including this one (next week) • Excel spreadsheets to help with calculation

  32. Research supported by grants from:

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