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```					       8 Rotational Dynamics
• describe/predict rotational behavior:
• motion, energy, momentum.
• with concepts of: rotational inertia & torque

• Homework:
• 30, 33, 39, 60, 65, 83, 89, 111.

1
Translation and Rotation
•   Translation: up, down, left, right, in, out
•   Rotation:
•   clockwise (cw),
•   counterclockwise (ccw)
•   //

2
Rotational Dynamics:
Newton’s 2nd Law for Rotation

net rotational action

rotational inertia

3
Mass-Distribution Effect
Larger       Larger          Larger
radius      Speed          Effort

Rotational
Inertia       ~   MR2

4
Rotational Inertia ( I )
I  m r  m r 
1 1
2
2 2
2
kg(m)2

Example

I  (4kg)(3m) 2  (5kg)( 2m) 2
 36  20  56 kg  m    2

5

I  m r  m r 
1 1
2
2 2
2

Example: Four identical masses. Axis
passes symmetrically through center
of mass.

I  m1a  m2a  m3a  m4a
2          2           2           2

I  4ma        2

6

I  m r  m r 
1 1
2
2 2
2

Example: Four identical masses. Axis
passes through one end.

I  m02  m02  m(2a)2  m(2a)2
I  4ma  4ma  8ma
2          2          2

7
You Try It

Calculate the Rotational
Inertia of this system:

8
Calculated Rot. Inertias, p.273
• rotational inertias of solid objects can be
calculated
• need to know: a, b, c, d, e, g, h, j, k.
• omit: f, i

9
Rotational Kinetic Energy
• Rotational K = ½(I)w2.
• Example: Constant Power Source has
100kg, 20cm radius, solid disk rotating at
• I = ½MR2 = ½(100kg)(0.2m)2 = 2kgm2.

• Rot K = ½ (2kgm2)(7000/s)2 = 49 MJ

10
Torque       (t)        [N·m]
F                      F

                   

t  F
 = lever-arm               11
Torque Concept
• a small force can cause a large torque
• a large force can create no torque

12
Rotationally Effective Application of Forces

13
Rotationally Ineffective Application of Forces

14
• torque = Fsin(f) (r)

15
1. A pair of forces with equal magnitudes and
opposite directions is acts as shown. Calculate
the torque on the wrench.

16
Example
Rod is 3m Long.

  (3m) cos40        40

t  F  (5N )(2.3m)  11.5Nm

17
18
Newton’s 2nd Law (Rotation)

t Net  I

FNet  ma

19
Concept Review
• Torque: rotational action
• Rotational Inertia: resistance to change in
rotational motion.
• Torque ~ force x lever-arm

20
Equilibrium
• Translational: net-force = 0
• Rotational: net-torque = 0

21
The drawing shows a person
whose weight is 584N.
Calculate the net force with
which the floor pushes on each
end of his body.

22
Rotational Work-Energy Theorem
• (Work)rot = tDq.
• Example: torque of 50 Nm is applied for
one revolution.
• rotational work = (50Nm)(2prad) = 314 J
• (Rotational Work)net = DKrot.
• Krot = ½Iw2.

23
Angular Momentum (L)
• analog of translational momentum
• L = Iw [kgm2/s]

• Example: Disk R = 1m, M = 2kg, w = 10/s
• I = ½MR2 = ½(1)(1)2 = 0.5
• L = Iw = (0.5kgm2)(10/s) = 5kgm2/s

24
Conservation of Angular
Momentum
• For an isolated system
• (Iw)before = (Iw)after
• Example: Stationary disk M,R is dropped
on rotating disk M, R, wi.
• (Iw)before = (Iw)after
• (½MR2)(wi) = (½MR2 + ½MR2)(wf)
• (wf) = ½ (wi)
25
8 Summary
•   All TRANSLATIONAL quantities;
•   speed, velocity, acceleration,
•   force, inertia, energy, and momentum,
•   have ROTATIONAL analogs.

26
Problem 33
•   Pivot at left joint, Fj = ?, but torque = 0.
•   ccw (Fm)sin15(18) = mg(26) = cw
•   ccw (Fm)sin15(18) = (3)g(26) = cw
•   (Fm) = (3)g(26)/sin15(18) = 160N
•   Note: any point of arm can be considered
the pivot (since arm is at rest)

27
#39
•   Left force = mg = 30g, Right = 25g
•   mg = 30g + 25g m = 55kg
•   ccw mg(xcg) = cw 30g(1.6)
•   (55)g(xcg) = 30g(1.6)
•   (55)(xcg) = 30(1.6)
•   Xcg = (30/55)(1.6)

28
#60, z-axis
• Each mass has r2 = 1.52 + 2.52.
• I = sum mr2 = (2+3+1+4)(1.52 + 2.52)

29
#65
• First with no frictional torque, then with
frictional torque as specified in problem.
• M = 0.2kg, R = 0.15m, m1 = 0.4, m2 = 0.8

30
#83
• Pulley M, R. what torque causes it to
reach ang. Speed. 25/s in 3rev?
• Alpha: use v-squared analog eqn.
• Torque = I = (½MR2)()

31
#89, uniform sphere part
• Rolling at v = 5m/s, M = 2kg, R = 0.1m
• K-total = ½mv2 + ½Iw2.
• = ½(2)(5x5) + ½[(2/5)(2)(0.1x0.1)](5/0.1)2.
• = 25 + 10 = 35J
• Roll w/o slipping, no heat created, mech
energy is conserved, goes all to Mgh.
• 35 = Mgh h = 35/Mg = 35(19.6) = 1.79m

32
#111
•   Ice skater, approximate isolated system
•   Therefore:
•   (Iw)before = (Iw)after
•   (100)(wi) = (92.5)(wf)
•   (wf) = (100/92.5)(wi)
•   K-rot increases by this factor squared
times new rot. Inertia x ½.

33
Example: Thin rod formulas.

34
K = Ktranslation + Krotation

35
Angular Momentum
•   Symbol: L      Unit: kg·m2/s
•   L = mvr = m(rw)r = mr2w = Iw.
•   v is perpendicular to axis
•   r is perpendicular distance from axis to line
containing v.

36
Angular Momentum
•   Symbol: L      Unit: kg·m2/s
•   L = mvr = m(rw)r = mr2w = Iw.
•   v is perpendicular to axis
•   r is perpendicular distance from axis to line
containing v.

37
13) Consider a bus designed to obtain its motive power from a large
rotating flywheel (1400. kg of diameter 1.5 m) that is periodically
brought up to its maximum speed of 3600. rpm by an electric motor
at the terminal. If the bus requires an average power of 12. kilowatts,
how long will it operate between recharges?
Diff: 2 Var: 1 Page Ref: Sec. 8.4

38
6) A 82.0 kg painter stands on a long horizontal board 1.55 m from one
end. The 15.5 kg board is 5.50 m long. The board is supported at each
end.
(a) What is the total force provided by both supports?
(b) With what force does the support, closest to the painter, push
upward?

39
28) A 4.0 kg mass is hung from a string which is wrapped around a
cylindrical pulley (a cylindrical shell). If the mass accelerates
downward at 4.90 m/s2, what is the mass of the pulley?
A) 10.0 kg
B) 4.0 kg
C) 8.0 kg
D) 2.0 kg
E) 6.0 kg

40
19) A solid disk with diameter 2.00 meters and mass 4.0 kg freely rotates about a
vertical axis at 36. rpm. A 0.50 kg hunk of bubblegum is dropped onto the disk and
sticks to the disk at a distance d = 80. cm from the axis of rotation.
(a) What was the moment of inertia before the gum fell?
(b) What was the moment of inertia after the gum stuck?
(c) What is the angular velocity after the gum fell onto the disk?

(a) 2.0 kg-m2
(b) 2.3 kg-m2                                                                  41
(c) 31. rpm
42
3. The drawing shows the top view of
two doors. The doors are uniform and
identical. The mass of each door is M
and width as shown below is L. How do
their rotational accelerations compare?

43
A Ring, a Solid-Disk, and a Solid-Sphere are released from
rest from the top of an incline. Each has the same mass and
radius. Which will reach the bottom first?

44
5. The device shown below is spinning with rotational rate wi when the
movable rods are out. Each moveable rod has length L and mass M. The central
rod is length 2L and mass 2M.
Calculate the factor by which the angular velocity
is increased by pulling up the arms as shown.

45
How large is the height h in the drawing? The ball’s initial speed is
4.0 m/s. Does it depend on M and R?

solid ball
rolls without
slipping

46
Rotational Review   (angles in radians)

s  rq                             Dq
wavg    
Dt
vt  rw
Dw
 avg   
at  r                             Dt

a c  rw                    
a  a c  at
2

+ 4 kinematic equations

47
Graph showing mt2 is more constant than mt

Rotation Disk-Hanging Mass

2500

2000
mt and mt2/10

1500
mt
mt2
1000

500

0
0          200        400         600
mass (grams)
48
Angular Momentum Calculation

L = Iw

Example: Solid Disk M = 2kg R = 25cm
Spins about its center-of-mass at 35 rev/s

L  Iw  1 (2kg)(.25m) 2 )(35rev / s)(2prad / rev) )
2

 13.7kgm2 / s

49
4. A one-meter-stick has a mass of 480grams.
a) Calculate its rotational inertia about an axis perpendicular to the
stick and through one of its ends.
b) Calculate its rotational inertia about an axis perpendicular to the
stick and through its center-of-mass.
c) Calculate its angular momentum if spinning on axis (b) at a rate of

50
Conservation of Angular
Momentum
• Example: 50 grams of putty shot at 3m/s
at end of 200 gram thin 80cm long rod free
to rotate about its center.
• Li = mvr = (0.050kg)(3m/s)(0.4m)
• Lf = Iw = {(1/12)(0.200kg)(0.8m)2 +
(0.050kg)(0.4m)2}(w)
• final rotational speed of rod&putty =
51
Dq
wavg    
Dt

Dw
 avg   
Dt

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55
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