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Angular Kinetics

Readings: Chapter 11 [course text]

Hay, Chapter 6 [on reserve] Hall, Chapter 13 & 14 [on reserve] Kreighbaum & Barthels, Modules I & J [on reserve]

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Learning Objectives:

By the end of this lecture, you should be able to: •  For a body or system, calculate and define

center of mass, moment of inertia, and radius of gyration (given appropriate anthropometric information)

•  Use the parallel axis theorem to calculate moment of inertia of a body or system about various axes

•  For a given movement (e.g. running, throwing, diving), describe how these parameters change

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Learning Objectives: •  Use the concepts of moment of inertia, angular

momentum, angular impulse, conservation of angular momentum, work, and conservation of energy to calculate the effects of torques (applied over different time periods and distances) on motion

•  Use the concepts of reaction torques, moment of inertia, conservation of momentum, and angular momentum about multiple axes to explain changes in body position during projectile motion

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Properties of Torques (review from Models and Anthropometry) 1.  Axis of rotation 2.  Force 3.  Moment arm

•  perpendicular distance between force line of action and axis of rotation

4 moment arm

T = F d ┴

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Centre of Mass (review from Models and Anthropometry) Sum of moments (torques) on all sides of a segment’s centre of mass equals zero

i.e. the mass is equally distributed about the centre of mass This is not the same as having equal mass on each side!

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60 N 20 N

0.5 m 1.5 m

Measuring Center of Mass Reaction board method:

In static equilibrium, net F = 0 and net M = 0

i.e. ∑F=0 and ∑M=0

Goal: to measure the vertical height of the center of mass of a person.

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Measuring Center of Mass Reaction board method: Step 1 weigh subject

(mgs). Step 2 – empty board positioned

Scale Reading = R1 Length of Board = 2 m

∑M = 0: -(R1)(2) + (mgb)(b) = 0 -2R1 + bmgb= 0 2R1 = bmgb

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R1

R1

2 m b m mgboard

Reaction board method: Step 3 – person + board

Scale Reading = R2

∑M = 0: -(R2)(2) + (mgb)(b) + (mgs)(s)= 0

mgboard 8

R2

R2

b m mgsubject s m 2 m

Measuring Center of Mass

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Reaction board method: Step 2 But we know 2R1 = bmgb

∑M = 0: -2R2 + bmgbb + smgs = 0 -2R2 + 2R1 = -smgs

mgboard 9

R2

R2

b m mgsubject s m 2 m

Measuring Center of Mass Calculating Center of Mass

We can calculate the locations of the segment center of masses from segment end points and anthropometric information.

If we know the locations of the segment center of masses, we can calculate the total body center of mass.

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Calculating Center of Mass

y (cm)

x (cm)

0

(30, 10) 2 kg

(5, 30) 0.5 kg

(20, 20) 1 kg

Question: What are the coordinates of the system’s center of mass?

1

2

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Calculating Center of Mass Find x-coordinate of system center of mass (xT):

We know that M = F d┴ and ∑M = 0: (mTg)(xT) = (m1g)(x1) + (m2g)(x2) + (m3g)(x3)

All the g’s cancel: mTxT = m1x1 + m2x2 + m3x3

(0.5+1+2)(xT) = (0.5)(5) + (1)(20) + (2)(30) (3.5)xT = 2.5 + 20 + 60 (3.5)xT = 82.5 xT = 82.5/3.5 = 23.6 cm

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4

Stability

Pyramid Quite Stable

Inverted Pyramid Very Unstable

Stability

Pendulum Highly Stable

Ball Neutral Stability

Inverted Pendulum

Abdominals

Quadriceps

Erector Spinae

Gluteus Maximus

Gastrocnemius and Soleus

Anti-Gravity Musculature

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Increased Stability

•  Increased base of support

•  Lower centre of gravity

• Higher mass

•  Position of centre of gravity with respect to base of support (e.g. leaning into a tackle)

Stability

– Which shape is more stable?

To be confident in how to answer that question you would need to ask – stable in which direction?

Ergonomic Chairs

•  The standard is for five legs to increase base of support.

Chairs?

•  Is this a good idea?

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Buckling

•  If the work done on the spine (energy applied) is greater than the work the muscles can do to stiffen the spine, then the spine will buckle.

Moments Created by Mass

The moments created by segment masses can be used to provide resistance during exercise or rehabilitation.

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Angular Kinetic Variables

Angular kinetics is similar to linear kinetics, but uses angular equivalents to many variables.

Torque or moment can be thought of as the angular equivalent of force.

In static situations, segment mass, length, and center of mass location is sufficient for analysis.

In dynamic activities, we also need to know segment moment of inertia because the segments rotate.

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Moment of Inertia

A segment’s moment of inertia (I) represents its resistance to rotation about a given axis.

Moment of inertia is the angular equivalent of mass.

Moment of inertia depends on the distribution of mass about the axis of rotation:

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!

I= imi=1

n

" i2r

The units of I are kg m2.

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Changes to moment of inertia The moment of inertia of a body or system often changes during the course of a movement.

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Leg Position

3.0

2.5

2.0

1.5

1.0 Mom

ent o

f ine

rtia

of

leg

abou

t hip

(k

g m

2 )

Radius of Gyration A segment’s radius of gyration (k) represents the distance at which all of its mass can be said to act.

Radius of gyration is the angular equivalent of center of mass.

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I = mk2

k

Questions: 1. If the leg (shank) of a subject has a mass of 3.6 kg

and is 0.4 m long, what is Ig? 2. Why is the ratio in the table usually higher for a

transverse axis at the distal end than a transverse axis at the proximal end?

Radii of Gyration / segment length (transverse axis)Segment Centre of Mass Proximal Distal

head, neck, trunk 0.503 0.830 0.607upper arm 0.322 0.542 0.645arm 0.303 0.526 0.647hand 0.297 0.587 0.577thigh 0.323 0.540 0.653leg 0.302 0.528 0.643foot 0.475 0.690 0.690

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k L I = mk2

Answers: 1. Value from table (a) = k/L Therefore k = a L.

Ig = mk2 = m(aL)2 = (3.6kg)(0.302 x 0.4m)2 = 0.0525 kg m2

2. Muscle mass is usually distributed closer to proximal joints. This favours speed of movement.

Radii of Gyration / segment length (transverse axis)Segment Centre of Mass Proximal Distal

head, neck, trunk 0.503 0.830 0.607upper arm 0.322 0.542 0.645arm 0.303 0.526 0.647hand 0.297 0.587 0.577thigh 0.323 0.540 0.653leg 0.302 0.528 0.643foot 0.475 0.690 0.690

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I = mk2

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Parallel axis theorem If you know the moment of inertia of a segment or body about an axis through its center of mass, you can calculate its moment of inertia about any parallel axis using the parallel axis theorem:

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I// = Ig + mr2

In this equation: I// is an axis of rotation parallel to Ig Ig is the moment of inertia about the center of mass m is the mass of the segment or body r is the distance between I// and Ig

Ig

r I//

Moment of Inertia Examples

6.5

12-15 10.5-13

3.5

15

4-5

2-2.5 1-1.2

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83

[Hay, 1993 (gymnasts); Hall, 1999]

Angular Momentum Angular momentum (H) is the quantity of angular motion of an object.

Angular momentum is the angular equivalent of linear momentum, and is also a vector:

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H = I ω p = m v linear momentum angular momentum

The units of angular momentum are: (kg m2)(rad/s) = kg m2

s

Conservation of Angular Momentum Conservation of Angular Momentum:

When there is no net external torque acting on a system, then H is constant.

e.g. when gravity is the only external force acting on an object, such as a projectile with no air resistance

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Conservation of Angular Momentum

If momentum remains constant, changing moment of inertia will change the angular velocity.

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H = I ω I = ∑(mi ri

2)

Time (sec)

Moment of inertia

Angular velocity

Tuck position

Conservation of Angular Momentum

Diving: http://www.youtube.com/watch?v=wTcatD-anCw&feature=related

Trampolining: http://www.youtube.com/watch?v=WBv5AJNNL1w&feature=related

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Figure skating: http://www.youtube.com/watch?v=Kl49eYBU88s (spot the fake physics terminology!)

Conservation of angular Momentum Question:

Why don’t bikes fall over?

Answer: Partly due to gyroscopic effects created by conservation of angular momentum. 42

Maximum Linear Velocity

Transfer of momentum via the kinetic chain can maximize linear velocity of distal segments.

Relevant equations for this concept:

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vt = r ω

I = ∑(mi ri2)

τ = I α

H = I ω

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Kinetic Chain

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Kinetic Chain

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Some Linear and Angular Analogues

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Angular variable Angular displacement Δθ Angular velocity ω Angular acceleration α Moment of inertia I Radius of gyration k Torque τ M Angular momentum H

Linear variable Linear displacement Δx Δy Linear velocity v Linear acceleration a Mass m Center of mass Force F Linear momentum ρ

Some Linear and Angular Analogues

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Angular equation

τ = I α H = I ω ΔH = τ t τ t = ΔI Δω KErotational = ½ I ω2 W = τ θ

Linear equation

F = m a ρ = m v Δρ = F t F t = m Δv KElinear = ½ m v2 W = F d

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Some Linear and Angular Analogues

Newton’s First Law An object at rest stays at rest, and an object in motion moves with constant velocity, unless acted on by an external force.

An object at rest stays at rest, and a rotating object moves with constant angular momentum, unless acted on by an external moment/torque.

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Some Linear and Angular Analogues

Newton’s Second Law A change in motion of an object is proportional to the net force acting on the object, and occurs in the same direction as the force.

A change in angular momentum of an object is proportional to the net moment/torque acting on the object, and occurs in the same direction as the moment/torque acts.

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Angular Impulse

The longer you can apply a torque for, the greater the change in angular velocity.

The more torque you can apply, the greater the change in angular velocity.

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ΔH = τ t = ΔI Δω

Work and Energy Similar to linear kinetics, when angular work is performed on a system, it changes the rotational energy of the system.

The angular work done on a system is equal to the change in rotational energy of the system – for our purposes, this usually means a change in rotational kinetic energy.

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KErotational = ½ I ω2

W = ΔE

τ θ = ΔPE + ΔKE = ½ Δ(I ω2)

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Work and Energy If there is no change in rotational potential energy and moment of inertia:

The larger the angle that you can apply a torque over, the greater the change in angular velocity.

The more torque you can apply, the greater the change in angular velocity.

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W = τ θ = ½ Δ(I ω2)

ΔKE ΔE

Work and energy The total work (linear + angular) performed on a system is equal to the total change in energy (linear + angular) of the system.

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W = ΔE

Wlinear + Wrotational = ΔPE + ΔKE

F d + τ θ = ΔPElinear + ΔKElinear + ΔPErotational + ΔKErotational

angular ≡ rotational

Work and Energy A force whose line of action passes through an object’s center of mass causes translation but no rotation:

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d

W = ΔE

If we assume ΔPE=0:

F d = ΔPE + ΔKE

F d = ½ Δ(m v2) F

Work and Energy A force whose line of action does not pass through an object’s center of mass causes translation and rotation:

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d

W = ΔE = ΔElinear + ΔErotational

If we assume ΔPE=0:

F d + τ θ = ΔKElinear + ΔKErotational

F d + τ θ = ½ Δ(m v2) + ½ Δ(I ω2)

θ

F

this is called an eccentric force

Note that the center of mass of the object does not move as far as in the last slide.

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Example: To generate a topspin on a tennis ball, the racket applies an eccentric force to the ball.

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net force of racket on ball

component of force causing rotation (angular movement)

component of force causing translation (linear movement)

center of mass of ball

Fd = mgΔh + ½Δ(mv2)

τθ = ½Δ(Iω2)

racket

Some Linear and Angular Analogues

Newton’s Third Law If one object applies a force on a second object, the second object applies an equal and opposite force on the first.

If one object applies a moment/torque on a second object, the second object applies an equal and opposite moment/torque on the first.

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Reaction Moment

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Conservation of Angular Momentum

Smarter Every Day video #58: http://www.youtube.com/watch?v=RtWbpyjJqrU 64

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Conservation of Angular Momentum

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Conservation of Angular Momentum

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Hang

Hitch

Local and Remote Angular Momentum

The angular momentum of a body segment about its own center of mass is called local angular momentum.

The angular momentum of a body segment about the whole body’s center of mass is called remote momentum.

The total angular momentum of a body segment is composed of both local and remote angular momentum:

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Htotal = Hlocal + Hremote

Local and Remote Angular Momentum

70 [Text, Fig. 11-13]

Htotal = Hlocal + Hremote

= Icm,segωcm,seg + (mr2)ωcm,body

m = mass of segment

r = distance between segment center of mass and body center of mass

this comes from the radius of gyration