Transcript Turntables PPT - Physics of Theatre Home

USITT 2007 - Phoenix
Rotational Motion
Verda Beth Martell
Dr. Eric C. Martell
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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.
2
Our starting point
Linear motion starts with Newton’s Second Law:


Fnet  ma

Fnet


Fnet  ma
- 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)


Fnet  ma
3
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)
4
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)
1200 100
lbs 2
lbs
ft
50
lbs
50 lbs * 2 ft = 100 ft.lb
50 lbs at 2 ft = 100 ft.lbs = 1200 in.lbs
5
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
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What is Angular
Acceleration?
Change in angular velocity (w)
tnet=Ia
w
a – Angular acceleration
Measures rate of change of
angular velocity:
w
a
t
Measured in rad/s2
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So, what are we trying to
calculate?
To spec a motor, we’re going to need the drive torque
td – Drive torque
tf – Frictional torques caused by
each ring of casters.
tnet  td-tf1-tf2 -or-
td
tf1 tf2
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.
We’ll start with I…
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Moment of Inertia
Note that the height
of the object does not
matter!
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.
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Moment of Inertia
Many common shapes have known I’s
b
r
r2
r1
a
Solid Cylinders
Rectangles
(turntables, people)
(Walls, Cubes)
I = ½ m r2
I = 1/12 m (a2+b2)
Hollow Cylinders
(Curved Walls)
I = ½ m (r1+r2)2
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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?)
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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 (a2+b2) + md2
a
b
d
d
Columns, people:
I = 1/2 mr2 + md2
12
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
6 ft off-axis: I = 114 slug.ft2
Note: I values are for the block only.
Move 3 ft
off-axis
Move 6 ft
off-axis
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Moment of Inertia
What about people?
Add people as solid cylinders on
the rim of the turntable:
Iactor = ½ mr2 + md2
200 lbs
5’
6’
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
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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
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Finding the Total Moment
of Inertia
Turntable: r = 6 ft
m = 400 lb/32.2 ft/s2 = 12.4 slug
Itt = ½ m r2 = 223.2 slug.ft2
Wall: a = 12.5 ft, b = 2 ft
m = 300 lb/32.2 ft/s2 = 9.3 slug
Iwall = 1/12 m (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
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Remember our goal
We want the Drive Torque
td = Ia+tf1+tf2
We now know how to
find I.
td
tf1 tf2
Next step – how do we
find a?
17
Angular Acceleration
What is a radian? –
Measure of the angle
where s = r.
Remember, a  w/t
Okay, what is w? The change in
angular velocity, or the final angular
velocity (wf) - the initial angular velocity
(w0) (if starting from rest, w0=0).
To find w – start with the distance that
a point travels around the turntable
(say it rotates around once in 30 s).
d = 12 ft
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/t = 6.28 rad/30 sec = .21 rad/s
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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 = w/t = (.21 rad/s – 0 rad/s)/5 s
a = .042 rad/s2
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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
20
Remember our goal
We want the Drive Torque
td = tnet+tf1+tf2
We now know how to
find tnet=Ia
td
tf1 tf2
Now we need to find the
torque caused by friction.
21
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
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:
Weight on ring [from turntable]
= (Total weight)*(# of casters in ring)
F
Ring 2
Ring 1
q
r
(Total # of casters)
How does this work?
22
Finding the Frictional
Torques
Weight on each ring
Ring 1 – 4 casters
Ring 2 – 8 casters
Total = 12 casters
F
Ring 2
Ring 1
q
Weight of turntable = 400 lb
Weight on ring 1 = 400 lb * (4/12)
= 133.3 lb
r
Weight on ring 2 = 400 lb * (8/12) = 266.7 lb
What about objects on turntable?
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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
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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
Friction in ring 1 = 283.3 lb * 0.05
= 14.2 lb
Friction in ring 2 = 716.7 lb * 0.05
= 35.8 lb
F
Ring 2
Ring 1
q
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
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Calculating Drive Torque
Let’s put the pieces together
td = tnet+tf1+tf2
tnet = 23.4 ft.lb
tf1 = 28.4 ft.lb
tf2 = 196.9 ft.lb
td
tf1 tf2
td = 248.7 ft.lb
Now, how big of a motor do
we need to accomplish this?
26
Calculating Drive Force
t = r F sin(q)
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.
td
tf1 tf2
If the force is applied on the edge:
Fd = 248.7 ft.lb/6 ft = 41.5 lb
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Calculating HP
HP = (Fd * v)/550
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.
td
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
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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.
r2
r1
Hollow Cylinders
(Curved Walls)
I = ½ m (r1+r2)2
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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.
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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
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Research supported by grants
from:
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