1. Angular Mechanics - Torque and moment of inertia Contents:
Review
Linear and angular Qtys
Tangential Relationships
Angular Kinematics
Torque
Example | Whiteboard
Moment of Inertia D D
Torque and Moment of inertia

2. Angular Mechanics - Angular Quantities Linear:
(m) s
(m/s) u
(m/s) v
(m/s/s) a
(s) t
(N) F
(kg) m
Angular:
 - Angle (Radians)
o - Initial angular velocity (Rad/s)
 - Final angular velocity (Rad/s)
 - Angular acceleration (Rad/s/s)
t - Uh, time (s)
 - Torque
I - Moment of inertia
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3. Angular Mechanics - Tangential Relationships Linear:
(m) s
(m/s) v
(m/s/s) a
Tangential: (at the edge of the wheel)
= r - Displacement
= r - Velocity
= r - Acceleration*
*Not in data packet
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4. Angular Mechanics - Angular kinematics
Linear:
s/t = v
v/t = a
u + at = v
ut + 1/2at2 = s
u2 + 2as = v2
(u + v)t/2 = s
ma = F
Angular:
 = /t
 = /t*
 = o + t
 = ot + 1/2t2
2 = o2 + 2
 = (o + )t/2*
 = I
*Not in data packet
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5. Angular Mechanics - Torque
Torque A twisting force that can cause an angular acceleration.
So how do you increase torque?
Push farther out: F F
Push with more force: or
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6.  = rxF F r Angular Mechanics - Torque
Torque A twisting force that can cause an angular acceleration. F
 = rxF r
If r = .5 m, and F = 80 N, then  = (.5m)(80N)
 = 40 Nm. Torque is also in foot pounds
(Torque Wrenches)
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7. F  = rxF = rFsin Fsin r Angular Mechanics - Torque 
Torque A twisting force that can cause an angular acceleration.  F
 = rxF = rFsin
Fsin r
Only the perpendicular component gives rise to torque. (That’s why it’s sin)
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8.  = 56o Example: What’s the torque here? F = 16 N r = 24 cm
Ok – This is as tricky as it can be:
use  = rFsin, r = .24 m, F = 16 N
But  is not 56o. You want to use the angle between r and F which is 90 – 56 = 34o
Finally,  = (.24 m)(16 N)sin(34o) = 2.1 Nm
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9. Whiteboards:
Torque
1 | 2 | 3
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10. What is the torque when you have 25 N of force perpendicular 75 cm from the center of rotation?
 = rFsin
= (.75 m)(25 N)sin(90o) = Nm
= 19 Nm W
19 Nm

11. F =(52Nm)/((.34m) sin(65o))=168.75 N F = 169 N
If you want 52.0 Nm of torque, what force must you exert at an angle of 65.0o to the end of a .340 m long wrench?
 = rFsin
52 Nm = (.34 m)(F) sin(65o)
F =(52Nm)/((.34m) sin(65o))= N
F = 169 N W
169 N

12. 200. ft lbs = (r)(145 lbs) sin(90o)
The axle nut of a Volkswagen requires 200. foot pounds to loosen. How far out (in feet) do you stand on the horizontal wrench if you weigh 145 lbs?
 = rFsin
200. ft lbs = (r)(145 lbs) sin(90o)
r = (200. ft lbs)/(145 lbs) = feet
r = 1.38 feet (1’ 41/2”) W
1.38 feet

13. Angular Mechanics – Moment of inertia
Moment of Inertia - Inertial resistance to angular acceleration.
Question - If the blue masses were identical, would both systems respond identically to the same torque applied at the center?
Demo
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14. Angular Mechanics – Moment of inertia
F = ma - We can’t just use “m” for “I”
 = I (The position of “m” matters!)
F = ma
rF = mar (Mult. by r)
rF = m(r)r (a = r)
= (mr2) = I ( = rF)
So I = mr2 for a point mass, r from the center. r F
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15. Angular Mechanics – Moment of inertia
What about a cylinder rotating about its central axis?
Some parts are
far from the axis
Some parts are
close to the axis
In this case,
I = 1/2mr2
(You need calculus to derive it)
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16. In general, you look them up in your book.
Demos:
TP, Solids
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17. mr2 - Hoop (or point mass)
Three main ones:
½mr2 - Cylinder (solid)
mr2 - Hoop (or point mass)
2/5mr2 – Sphere (solid)
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18. Example: Three 40. kg children are sitting 1
Example: Three 40. kg children are sitting m from the center of a merry-go-round that is a uniform cylinder with a mass of 240 kg and a radius of 1.5 m. What is its total moment of inertia?
The total moment of inertia will just be the total of the parts:
Children – use mr2 (assume they are points)
MGR – use 1/2mr2 (solid cylinder)
I = 3((40kg)(1.2m)2) + ½(240kg )(1.5m)2
I = kgm2 = 440 kgm2
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19. Whiteboards:
Moment of Inertia
1 | 2 | 3
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20. What is the moment of inertia of a 3
What is the moment of inertia of a 3.5 kg point mass that is 45 cm from the center of rotation?
I = mr2
I = (3.5kg)(.45m)2 = kgm2
I = .71 kgm2 W
.71 kgm2

21. A uniform cylinder has a radius of 1
A uniform cylinder has a radius of m and a moment of inertia of kgm2. What is its mass?
I = ½mr2
572.3 kgm2 = ½m(1.125 m)2
m = 2(572.3 kgm2)/(1.125 m)2
m = = kg W
904.4 kg

22. A sphere has a mass of 45. 2 grams, and a moment of inertia of 5
A sphere has a mass of 45.2 grams, and a moment of inertia of x 10-6 kgm2. What is its radius? (2 Hints)
m = kg
I = 2/5mr2
5I/(2m) = r2
r = (5I/(2m))1/2
r = (5(5.537x10-6kgm2)/(2(.0452kg)))1/2
r = m W
.0175 m

23. F = ma  = I Angular Mechanics – Torque and moment of inertia
Now let’s put it all together, we can calculate torque and moment of inertia, so let’s relate them:
The angular equivalent of F = ma is:
F = ma
 = I
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24. Example: A string with a tension of 2. 1 N is wrapped around a 5
Example: A string with a tension of 2.1 N is wrapped around a 5.2 kg uniform cylinder with a radius of 12 cm. What is the angular acceleration of the cylinder?
I = ½mr2 = ½(5.2kg)(.12m)2 = kgm2
 = rF = (2.1N)(.12m) = .252 Nm
And finally,
 = I,
 = /I = (.252 Nm)/( kgm2) = 6.7 s-2
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25. Torque and Moment of Inertia
Whiteboards:
Torque and Moment of Inertia
1 | 2 | 3 | 4 | 5
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26. What torque is needed to accelerate a 23
What torque is needed to accelerate a 23.8 kgm2 wheel at a rate of 388 rad/s/s?
 = I
 = (23.8 kgm2)(388 rad/s/s)
I = Nm = 9230 Nm W
9230 Nm

27. I = (6.34 Nm)/(23.1 rad/s/s) = .274 kgm2
An object has an angular acceleration of 23.1 rad/s/s when you apply 6.34 Nm of torque. What is the object’s moment of inertia?
 = I
(6.34 Nm) = I(23.1 rad/s/s)
I = (6.34 Nm)/(23.1 rad/s/s) = .274 kgm2 W
.274 kgm2

28. If a drill exerts 2. 5 Nm of torque on a. 075 m radius, 1
If a drill exerts 2.5 Nm of torque on a .075 m radius, 1.75 kg grinding disk, what is the resulting angular acceleration? (1 hint)
 = I, I = ½mr2
I = ½mr2 = ½(1.75 kg)(.075 m)2 = kgm2
 = I, 2.5 Nm = ( kgm2)
 = /I = (2.5 Nm)/( kgm2) = 510 s-2 W
510 s-2

29. What torque would accelerate an object with a moment of inertia of 9
What torque would accelerate an object with a moment of inertia of 9.3 kgm2 from 2.3 rad/s to 7.8 rad/s in .12 seconds? (1 hint)
 = I,  = o + t
7.8 rad/s = 2.3 rad/s + (.12 s)
 = (7.8 rad/s rad/s)/(.12 s) = s-2
 = I = (9.3 kgm2)(45.833s-2) = Nm
 = 430 Nm W
430 Nm

30.  = rF, I = 1/2mr2 ,  = o+t,  = I  = 1.35 s-1/5.0 s = .27 s-2
A merry go round is a uniform solid cylinder of radius 2.0 m. You exert 30. N of force on it tangentially for 5.0 s and it speeds up to 1.35 rad/s. What’s its mass? (1 hint)
 = rF, I = 1/2mr2 ,  = o+t,  = I
 = 1.35 s-1/5.0 s = .27 s-2
 = rF = 30.N(2.0m) = 60 Nm
 = I, I=/ (60Nm)/(.27s-2)=222.2kgm2
I = 1/2mr2, m = 2I/r2
m = 2(222.2kgm2)/(2.0m)2 = kg
m = 110 kg W
110 kg