1. 1 USITT 2007 - Phoenix Rotational Motion Verda Beth Martell Dr. Eric C. Martell

2. 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. 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/s 2 )

4. 4 For rotating objects Rotational analogue to Newton’s Second Law:  net =I    net - the sum of the torques acting on the object (ft. lb)  I - the moment of inertia for the object (slug. ft 2 )   - the angular acceleration of the object (rad/s 2 )

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

6. 6 What is the Moment of Inertia?  net =I  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. ft 2

7. 7 What is Angular Acceleration? Change in angular velocity (  )  net =I   – Angular acceleration Measures rate of change of angular velocity: Measured in rad/s 2 

8. 8 So, what are we trying to calculate? dd  f1  f2  d – Drive torque  f – Frictional torques caused by each ring of casters.  net  d -  f1 -  f2 -or-  d  net +  f1 +  f2 =I  +  f1 +  f2 So, to determine the drive torque, we need to find the moment of inertia, the angular acceleration, and the frictional torques. To spec a motor, we’re going to need the drive torque We’ll start with I…

9. 9 Moment of Inertia The Moment of Inertia is 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. Note that the height of the object does not matter!

10. 10 Moment of Inertia Many common shapes have known I’s Solid Cylinders (turntables, people) I = ½ m r 2 Rectangles (Walls, Cubes) I = 1 / 12 m (a 2 +b 2 ) a b Hollow Cylinders (Curved Walls) I = ½ m (r 1 +r 2 ) 2 r1r1 r2r2 r

11. 11 Moment of Inertia – object on turntable If we had a wall (centered on the axis of rotation) on the turntable: I rectangle = 1/12 m (a 2 +b 2 ) 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. 12 Moment of Inertia What if the object is not centered? The parallel axis theorem can be used for any object that is off center Furniture, blocks, walls: I = 1/12 m (a 2 +b 2 ) d a b Columns, people: I = 1/2 mr 2 + md 2 d Furniture, blocks, walls: I = 1/12 m (a 2 +b 2 ) + md 2

13. 13 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/s 2 = 3.11 slug I = 1/12*3.11 slug*((2 ft) 2 + (2 ft) 2 ) = 2.07 slug. ft 2 Move 3 ft off-axis 3 ft off-axis: I = 2.07 slug. ft 2 + 3.11 slug*(3 ft) 2 I = 30.1 slug. ft 2 Move 6 ft off-axis 6 ft off-axis: I = 114 slug. ft 2 Note: I values are for the block only. Moment of Inertia

14. 14 What about people? Add people as solid cylinders on the rim of the turntable: I actor = ½ mr 2 + md 2 Let’s work it out… m actor = 200 lbs/32.2 ft/s 2 = 6.2 slugs I actor = ½ mr 2 + md 2 = ½ (6.2 slugs)(1ft) 2 + (6.2 slugs)(5 ft) 2 = 158.1 slug. ft 2 5’ 6’ 200 lbs Moment of Inertia

15. 15 I tot = I actor + I block + I wall + I tt The total moment of inertia is the moment of inertia of the turntable plus the moments of inertia of everything on it: Moment of Inertia

16. 16 Turntable: r = 6 ft m = 400 lb/32.2 ft/s 2 = 12.4 slug I tt = ½ m r 2 = 223.2 slug. ft 2 Wall: a = 12.5 ft, b = 2 ft m = 300 lb/32.2 ft/s 2 = 9.3 slug I wall = 1/12 m (a 2 + b 2 ) = 124.2 slug. ft 2 Box: a = 2 ft, b = 1 ft, d = 4 ft m = 100 lb/32.2 ft/s 2 = 3.1 slug I box = 1/12 m (a 2 + b 2 ) + m d 2 = 50.9 slug. ft 2 Actor: r = 1 ft, d = 5 ft m = 200 lb/32.2 ft/s 2 = 6.2 slug I actor = ½ m r 2 + m d 2 = 158.1 slug. ft 2 I total = I tt + I wall + I box + I actor I total = 556.4 slug. ft 2 Finding the Total Moment of Inertia

17. 17 Remember our goal We want the Drive Torque dd  f1  f2  d = I  +  f1 +  f2 We now know how to find I. Next step – how do we find  ?

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

19. 19 Angular Acceleration Now to find  d = 12 ft How much time do you have to get up to speed, starting from rest? Say 5 seconds. Final angular velocity  f =.21 rad/s Initial angular velocity  0 = 0 rad/s  =  /  t = (.21 rad/s – 0 rad/s)/5 s  =.042 rad/s 2

20. 20 Net Torque  net = I  Total moment of inertia I total = 556.4 slug. ft 2 Angular acceleration  = 0.042 rad/s 2  net = 556. 4 slug. ft 2 * 0.042 rad/s 2  net = 23.4 ft. lb

21. 21 Remember our goal We want the Drive Torque dd  f1  f2  d =  net +  f1 +  f2 We now know how to find  net =I  Now we need to find the torque caused by friction.

22. 22 Finding the Frictional Torques Ring 2 Ring 1 Consider an empty turntable supported by two rings of casters: There’s a Frictional Force resisting the motion. F =  r *weight on casters in each ring 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: F r  Weight on ring [from turntable] = (Total weight)*(# of casters in ring) (Total # of casters) How does this work?

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

24. 24 Finding the Frictional Torques 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 What about objects on turntable? 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. 25 Frictional force =  r *weight on ring Finding the Frictional Torques Coefficient of rolling friction:  r = 0.05 Friction in ring 1 = 283.3 lb * 0.05 = 14.2 lb Friction in ring 2 = 716.7 lb * 0.05 = 35.8 lb  f1 = 2 ft * 14.2 lb = 28.4 ft. lb  f2 = 5.5 ft * 35.8 lb = 196.9 ft. lb Remember:  = r F sin(  ) Ring 1: r 1 = 2 ft Ring 2: r 2 = 5.5 ft  = 90 o Ring 2 Ring 1 F r 

26. 26 Calculating Drive Torque Let’s put the pieces together dd  f1  f2  d =  net +  f1 +  f2  net = 23.4 ft. lb  f1 = 28.4 ft. lb  f2 = 196.9 ft. lb  d = 248.7 ft. lb Now, how big of a motor do we need to accomplish this?

27. 27 Calculating Drive Force dd  f1  f2 We just found the drive torque:  d = 248.7 ft. lb To find the drive force, we work backwards – F d =  d /r, so we need to know where the drive force is being applied.  = r F sin(  ) If the force is applied on the edge: F d = 248.7 ft. lb/6 ft = 41.5 lb

28. 28 HP = (F d * v)/550 Calculating HP dd  f1  f2 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. Linear velocity = angular velocity * radius (v =  *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. 29 Our model assumes that each item is homogenous despite the fact that they are not. Things to keep in mind 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. Hollow Cylinders (Curved Walls) I = ½ m (r 1 +r 2 ) 2 r1r1 r2r2

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

31. 31 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 More Resources

32. 32 Research supported by grants from: