Angular Kinematics Angular Kinematics

Chapter 11Chapter 11

Angular MotionAngular Motion

All parts of a body move through the same angle, in theAll parts of a body move through the same angle, in the

same, direction, in the same timesame, direction, in the same time

More prevalent component of general motionMore prevalent component of general motion

Measurement of anglesMeasurement of angles

Joint or relative anglesJoint or relative angles

Angle formed from the long axes of two adjacent body segmentsAngle formed from the long axes of two adjacent body segments

Segmental angles or absolute anglesSegmental angles or absolute angles

Orientation of a single body segment with a fixed line of referenceOrientation of a single body segment with a fixed line of reference

Measurement of Angular VelocityMeasurement of Angular Velocity

ωω = ( = (ΘΘFF – – ΘΘII)/)/∆∆t = t = ∆Θ∆Θ//∆∆tt

Period Hip Knee Period Hip Knee

1-2 s 1-2 s ωω = (170 = (170oo-95-95oo)/1s )/1s ωω = (175 = (175oo-100-100oo)/1s )/1s

= = 75 75 oo/s/s (Ext) = (Ext) = 75 75 oo/s/s (Ext) (Ext)

2-2.5 s 2-2.5 s ωω = (95 = (95oo-170-170oo)/0.5s )/0.5s ωω = (110 = (110oo-175-175oo)/0.5s )/0.5s

= = - 150 - 150 oo/s /s (Flex) = (Flex) = -130 -130 oo/s/s (Flex) (Flex)

2.5-3 s 2.5-3 s ωω = (85 = (85oo-95-95oo)/0.5s )/0.5s ωω = (160 = (160oo-110-110oo)/0.5s)/0.5s

= = -20 -20 oo/s/s (Flex) = (Flex) = 100 100 oo/s/s (Ext) (Ext)

Arc length = r Θ circumference = 2∏r

150o/s = 150/57.3 = 2.6 rad/s

Relationship between linear and angular velocityRelationship between linear and angular velocity

V1

V2

r1 = 0.2 m

= 0.4 m

v (m/s) = r (m) ω (radians/s)

If If ωω = 30 rad/s, what is = 30 rad/s, what is the linear velocity of the the linear velocity of the bat at points 1 and 2?bat at points 1 and 2?

v = r v = r ωω

vv1 1 = 0.2 x 30 = 6 = 0.2 x 30 = 6 m/sm/s

vv2 2 = 0.4 x 30 = = 0.4 x 30 = 12 m/s12 m/s

Angular AccelerationAngular Acceleration

⍺⍺ = (= (ωωFF – – ωωII) / ) / ∆∆t = t = ∆∆ωω / / ∆∆tt

Skater spinning anticlockwise at 198.3º/s comes to a stop in 20s. Skater spinning anticlockwise at 198.3º/s comes to a stop in 20s. What is herWhat is her

angular acceleration?angular acceleration? ⍺⍺ = (= (ωωFF – – ωωII) / ) / ∆∆t = (0 -198.3 º/s) / 20 s = t = (0 -198.3 º/s) / 20 s = - 9.92 º/s- 9.92 º/s22

= - 9.91 / 57.3 rad/s= - 9.91 / 57.3 rad/s2 2 = = -0.17 rad/s-0.17 rad/s22

If the skater’s hand is 0.85 m from the axis of rotation, what is the If the skater’s hand is 0.85 m from the axis of rotation, what is the

tangential acceleration of her hand?tangential acceleration of her hand?

use ause att = r = r ⍺⍺

aatt = r = r ⍺⍺ = 0.85 (- 0.17) = = 0.85 (- 0.17) = -0.14 m/s-0.14 m/s22

“Center-fleeing”

“Center-seeking”

Angular KineticsAngular Kinetics

Angular analogue to mass/inertia Angular analogue to mass/inertia → moment of inertia (→ moment of inertia (I)I)

Resistance to angular acceleration.Resistance to angular acceleration.

nn

I = I = ΣΣ m d m d 22

i = 1i = 1

Determining IDetermining I nn

I = I = ΣΣ m d m d 22

i = 1i = 1

However, determined from However, determined from ΣΣ T = I T = I ⍺⍺

Where Where ΣΣ T = sum of Torques T = sum of Torques I = moment of inertiaI = moment of inertia

⍺⍺ = angular acceleration= angular acceleration

I approximated from cadaver studies: acceleration of aI approximated from cadaver studies: acceleration of a rotating limb measured after applying a known torquerotating limb measured after applying a known torque

Once determined, value characterized by using the formulaOnce determined, value characterized by using the formula

I = mkI = mk2 2 where where I = moment of inertia; m = total mass; k = radius of gyration I = moment of inertia; m = total mass; k = radius of gyration

I = mkI = mk22

Radius of gyration – kRadius of gyration – k

Represents the objects mass distribution with respect to an Represents the objects mass distribution with respect to an axis of rotation. axis of rotation.

It is the distance from the axis of rotation to a point at whichIt is the distance from the axis of rotation to a point at whichthe mass can be theoretically concentrated without altering the mass can be theoretically concentrated without altering the inertial characteristics of the rotating body the inertial characteristics of the rotating body

The length of the radius of gyration changes as the axis ofThe length of the radius of gyration changes as the axis ofrotation changes rotation changes

k for a particular segment can be obtained from anthropometric k for a particular segment can be obtained from anthropometric tables tables

The mass of a particular segment can also be obtained from such The mass of a particular segment can also be obtained from such tablestables

Therefore, net joint torques can be determined for human subjects Therefore, net joint torques can be determined for human subjects by measuring angular accelerations and applying the equation by measuring angular accelerations and applying the equation

ΣΣ T = I T = I ⍺⍺

Whole body moment of inertiaWhole body moment of inertia

Different with respect to different axes of rotationDifferent with respect to different axes of rotation

About which axis is I the smallest? About which axis is I the smallest?

Anteroposterior Mediolateral Mediolateral Longitudinal Longitudinal

12-15 kg.m2 10.5-13.0 kg.m2 4.0-5.0 kg.m2 1.0-1.2 kg.m2 2.0-2.5 kg.m2

Whole body moment of Inertia (I) values

Angular momentumAngular momentum

Quantity of angular motion (vector)Quantity of angular motion (vector)

H = I H = I ωω = (m k = (m k22) ) ωω

Conservation of angular momentumConservation of angular momentum

The total angular momentum of a givenThe total angular momentum of a given

system (e.g., the body) remains constant insystem (e.g., the body) remains constant in

the absence of external torquesthe absence of external torques

If gravity is the only external force, there are no external torques on the body and angular momentum is conserved

H = I ωω

II↑ ↑ ωω↓ ↓

I I ↓↓ ωω↑↑

HA = -IA ωωAA (negative)

HL = IL ωωLL

(positive)(positive)

HHAA = -I = -IAA ωωAA is equal in is equal in magnitude but opposite in magnitude but opposite in direction to Hdirection to HLL = I = ILL ωωLL

IILL (of legs) > I (of legs) > IAA (of arms) (of arms) so arm has > so arm has > ωω

ωωAA > > ωωLL

Conservation of angular momentumConservation of angular momentum

When a body is in the air (angular When a body is in the air (angular momentum conserved), if the angular momentum conserved), if the angular momentum of one body part is increased, momentum of one body part is increased, then all or part of the rest of the body then all or part of the rest of the body must experience a decrease in angular must experience a decrease in angular momentummomentum

While in the air, angular momentum is While in the air, angular momentum is conserved, but it can be transferredconserved, but it can be transferred

Angular impulseAngular impulse

(I (I ωω))BB + + ΣΣT∆t = (I T∆t = (I ωω))AA

A diver produces angular impulse at take-offA diver produces angular impulse at take-offresulting in the angular momentum that he/sheresulting in the angular momentum that he/shepossess in the airpossess in the air

Newton’s Laws of Angular MotionNewton’s Laws of Angular Motion

11stst Law Law

A rotating body will maintain a state of constant angularA rotating body will maintain a state of constant angular

motion unless acted upon by some net external torquemotion unless acted upon by some net external torque

22ndnd Law Law

A net external torque produces angular acceleration of a body A net external torque produces angular acceleration of a body

that is directly proportional to the magnitude of the net that is directly proportional to the magnitude of the net

torque, in the same direction as the net torque, and inversely torque, in the same direction as the net torque, and inversely

proportional to the body’s moment of inertiaproportional to the body’s moment of inertia

ΣΣT = I T = I ⍺⍺ 33rdrd Law Law

For every torque exerted by one body on another, there is anFor every torque exerted by one body on another, there is an

equal and opposite torque exerted by the second body on the equal and opposite torque exerted by the second body on the

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