1. Chapter 11
Rotation

2. Uniform Circular Motion
Recall: v is tangent to the circle a v R
The magnitude of the centripetal acceleration is,
pointing towards the center of the circle of radius R
Show clip 2d_circular_motion.mov
Period (the time to make one revolution):
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3. Polar and Cartesian Coordinates
Cartesian to Polar: y q x
Polar to Cartesian:
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4. Radians
If we take a line of length r and stretch it along the edge of the circle, the angle swept out is 1 radian r q
p radians = 180°
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5. Linear Motion Along CircleDs 
s is the arclength along the circle
s = r
Only true for q in radians
Linear speed along circle: v = ds/dt
= r dq/dt
Define angular velocity: w = dq/dt, then:
v = rw
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6. Linear Motion Along Circle
v = rw
If an extended object is rotating, we have different v at different r
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7. Uniform Circular Motion Easier in Polar Coordinates
angular position
radians 
angular velocity
radians/second
 = v/r where v is the linear speed around the circle
Similar to before
(t) = t + 0
x(t) = vxt + x0
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8. Example: uniform circular motion
Angular Acceleration
Plays same role in rotational motion as acceleration in linear motion
Example: uniform circular motion
but
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9. Angular Acceleration r Can break linear acceleration into two parts:
Tangential:
atan
Radial:
arad r
Then:
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10. Vectors? Analogous to 1D kinematics x, v, a  q, w, a
Vectors point in direction of right-hand rule
Clockwise is into page
Counter-clockwise is out of page
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11. Rotational Motion Analogous to motion in a straight line position x 
Translation
Rotation
position x 
velocity
v = dx/dt
 = d/dt
acceleration
a = dv/dt
 = d/dt
However, linear acceleration is never zero for rotational motion.
There is always a radial acceleration
(otherwise it wouldn’t be rotational motion!)
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12. Rotational Kinematics
We will be primarily dealing with “1-D” rotations,
so the vector direction will be indicated by the sign and we can drop the vector notation for simplicity
If a is constant:
As with linear kinematics, we can use these two equations to solve most constant a problems!
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13. Constant Angular Acceleration
We can derive variants of these equations which are also useful:
Equation
Missing Quantity
wf = w0 + a t Dq
Dq = w0 t + ½ a t2 wf
wf 2 = w a Dq t
Dq = ½ (w0 + wf) t a
Dq = vf t – ½ a t2 w0
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14. Fake Gravity Spin ship to simulate Earth-like gravity
Rotation: half turn in about 13 s
  =  rad / 13 s = 0.24 rad/s
Inner radius: about 18 ladder rungs
 r = 18 ft = 5.5 m
Centripetal acceleration:
ac = v2/r = 2 r = (0.24 s-1)2  5.5 m = 0.32 m/s2
ac = 3.3% g !
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15. Example: Hampster on a Wheel
A hamster is running in place on a spinning wheel with R =0.5 m,
w0 = -1 rad/s
a = rad/s2
At what time will the wheel have w = 2 rad/s?
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16. Example: (continued) w0
How far will the wheel have rotated from its position at t = 0?
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17. Example: (continued) w0 What is his displacement at that time?
What is his linear acceleration?
Radial:
Tangential:
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18. Rotational Kinetic Energy (Point Particle)w r
K = ½mv2
K = ½m(wr)2
K = ½ (mr2) w2 I
I = mr2 is the rotational inertia or moment of inertia for a point particle
K = ½ I w2
I is the rotational equivalent of mass
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19. Rotational Inertia of Point Mass
For a single particle I = mr2
(all mass at same r)
A single particle
The same particle farther out
1 m
4 kg
2 m
4 kg
I = (4 kg)(1 m)2
= 4 kg m2
I = (4 kg)(2 m)2
= 16 kg m2
Four times the rotational “mass”
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20. Rigid Bodies
A rigid body is one where all the particles maintain their relative position
State College
Sydney
As the body rotates
Each particle moves
Relative positions don’t change
Cities on the earth are always moving
But they don’t get closer together
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21. Kinetic Energy (Rigid Body)
Divide rigid body into bits:
m1, m2, m3, …  ri vi
For a rigid body: vi = riw (same w for every part)
Where I = Smiri2 is “rotational inertia” of body
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22. Rotational Inertia: Extended Objects
For an extended object, we break it into differential pieces and sum them up: z dm y x
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23. Rotational Inertia aka “Moment of Inertia”
Plays the role of mass for rotational motion
I = Smiri2 = r2dm
ri is the distance of mi to the axis of rotation
For a single particle I = mr2
For a rigid body, I depends on how the mass is distributed in an object relative to the axis of rotation
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24. Example:
Four masses are arranged in a square as shown. Find I for each of the given axes.
m1 = 10 kg
m2 = 20 kg m1 m2 1
d=4m m2 m1 3 2
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25. Example: 1 m1 m2
d1= 8
4 m
The moment of inertia around an axis going through the center is:
m1 = 10 kg
m2 = 20 kg
I1 = 2m1d12 + 2m2d12
= 2(10kg)(8 m2)+2(20 kg)(8 m2)
= 480 kg m2
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26. Example: The moment of inertia around axis 2 is: I2= 2m1d22 + 2m2d22
= 2(10 kg)(4 m2)+2(20 kg)(4 m2)
= 240 kg m2
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27. Example The moment of inertia around axis 3 is: I3= 2m1d2 + 2m2×0
= 2(10 kg)(4 m)2 + 0
= 320 kg m2
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28. Rotational Inertia: Common Objectsℓ ℓ R R a b
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More in the textbook (pg 227)

29. Rotational Inertia Example
A rod with blocks: Mrod = Mblock = M
7 times larger!!!
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30. Example: Solid Cylinder
How do we get: I = mR2/2? h
dV = dx dy dz = (rdθ) dr dz
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31. Parallel-Axis Theorem
I about CM is “easy” to calculate
I about other axis of rotation is not! D
Bar rotating about CoM
Parallel Axis Theorem:
Bar rotating about one end
I = ICM + MD2
I about CM
Object of Mass M
distance D
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32. Faster Than a Falling Rock
The rod starts at rest and falls over
How fast is it going when it hits the table?
Conservation of Energy: Ki+Ui = Kf+Uf
Initial: Ui = mg(L/2) and Ki = 0
Final: Uf = and Kf = ½Iw2
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33. Faster Than a Falling Rock
I = mL2/3
Ei = Ef
A little algebra:
Moment of inertia for the bar:
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34. Faster Than a Falling Rock
Center of mass:
End of the rod:
The end of the rod is falling faster than a dropped stone!
Dropped Rock:
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35. Torque on a Point Particleat
How does an applied tangential force lead to rotational acceleration? r Ft
Ft = mat = m(ra)
Want to replace m with I = mr2
r Ft = mr2a = Ia
Define torque t = r Ft (simplified definition)
t = Ia rotational equivalent of F = ma
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36. Torque: Rotational “Force”
Door on a hinge:
How do you open? F3 F1 F2
Define a new vector quantity: Torque
r = vector from origin to point where force is applied
Usually take origin
to be on axis of rotation
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37. “Cross” or “Vector” Product
Recall:
Magnitude only…
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38. Right Hand Rule
Point your fingers in the direction of the first vector
Curl them in the direction of the second vector
Your thumb now points in the direction of the cross product
Sanity check your answers!
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39. Torque: Key Points ttot and a direction both given by right-hand-rule
1) t is a vector with unit Nm  J
2) t is defined around a certain origin
3) tnet = Ia (just like Fnet = ma)
ttot and a direction both given by right-hand-rule
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40. Torque: How to Think of It
r sin θ θ q r
r sinq is the component of r perpendicular to F
“Moment arm”
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41. Torque on a Door: ExampleF4
Top View
q=30° F3 F1 F2
A 3N force is applied to different places on 50 cm door.
Find the magnitude of the torques?
1 = 0 (r=0)
2 = 0.5m  3N = 1.5 Nm out of screen
3 = 0 (θ=180°)
4 = 0.5m  3N  sin(150°) = 0.75 Nm into screen
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42. A Falling Stuntman Stuntmen sometimes need to fall large distances
without getting hurt! But it has to look like they fall
Cylinder (R)
The rope is around the spindle
Spindle (r) M m
Reduced gravity!
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43. Torque and Work Recall: 2(θf – θi) = f2 - i2
Rearrange and multiply by I
I(θf – θi ) = ½ If2 - ½ Ii2
tot Dθf = DK = W
Can also write:
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44. Rotational Analogy to Linear Motion
Position  x
Velocity
 = d/dt
v = dx/dt
Acceleration
 = d/dt
a = dv/dt
Translation
Kinetic Energy
K = ½Iw2
K=½mv2
Mass
I = Smiri2 m
F = ma
 = I
Newton’s 2nd Law
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