Lecture 17
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1) Linear and angular kinetics relate external forces/torques to inertia, displacement/angular displacement, velocity/angular velocity, and acceleration/angular acceleration respectively. 2) Moment of inertia represents an object's resistance to changes in angular motion and depends on the object's mass distribution and the axis of rotation. 3) Angular momentum is the product of an object's moment of inertia and angular velocity, and is conserved if no external torque is applied to the system.
Lecture 17
- 1. Linear vs. Angular Kinetics • Every concept in linear kinetics has an analogue in Angular Inertia & Momentum angular kinetics. Linear Kinetics Angular Kinetics Objectives: Relationship between: Relationship between: • Define moment of inertia, angular momentum • External forces (F) • External torques (T) • Understand factors that determine moment of inertia • Inertia (mass) • Inertia • Understand and apply the principle of • Displacement (∆p) • Angular displacement (θ) conservation of angular momentum • Velocity (v) • Angular velocity (ω) • Understand how momentum can be transferred between segments or axes • Acceleration (a) • Angular acceleration (α) Inertia Moment of Inertia • Concept relating to the difficulty with which an • For a particle: mi object’s motion is altered Ii = mi ri ² • Inertia = tendency to resist acceleration • In linear kinetics, inertia is represented by mass • For an object: ri Greater mass Greater force to produce a given acceleration I = Σ Ii = Σ mi ri² axis of • In angular kinetics, inertia is represented by the where: rotation mass moment of inertia (I) of an object – mi : mass of particle i – ri : distance of particle i from axis of rotation Greater torque to produce a Greater I • SI Units: kg·m² given angular acceleration 1
- 2. Example Problem #1 Factors Affecting Moment of Inertia Compute the mass moment of inertia of the object • Since: I = Σ mi ri² y below about the axis of rotation shown. I increases with: x' x' – Greater mass C = 1 kg – Greater distance from center 0 .2 m m axis of rotation of mass 15 A = 1 kg 0. x • Therefore, I depends on: x 0.1 m B = 1 kg – Mass of the object 0. – Shape of the object 15 axis of 0.2 m m rotation – Location of the axis of rotation D = 2 kg (distance from center of mass) – Direction of the axis of rotation y Body Position & Moment of Inertia Computing Moment of Inertia • Body moment of inertia depends on body position • In biomechanics, determine I using: and axis of rotation – Formulas for geometric solids (e.g. a cylinder) – Measurements using pendulum techniques or medical imaging – Radius of gyration (k): I = m k² • Could lump all of an object’s mass at a distance k from the axis and I would be the same I = 10.5 – I = 4.0 – I = 1.0 – I = 2.0 – • k has been measured for different body segments 13 kg·m2 5.0 kg·m2 1.2 kg·m2 2.5 kg·m2 2
- 3. Example Problem #2 Angular Momentum A table of anthropometic data gives the following • The quantity of angular motion information: • Measured as the product of an object’s moment of inertia (I) and its angular velocity (ω; in rad/s): Leg Length 24.6% of body height Leg + Foot Mass 6.1% of body mass H= Iω Foot + Leg Radius of This equation applies to two cases: 73.5% of leg length Gyration * – rotation about a fixed, stationary axis * About the mediolateral axis of the knee – rotation about an axis through the center of mass For a man with a body mass of 75 kg and height of (which can be moving) 1.75 m, find the moment of inertia of the leg + foot • SI Units: kg·m2/s about the knee during flexion-extension Multi-segment Angular Momentum Angular Momentum: General Case • Angular momentum of a multi-segment object is • Angular momentum of an object about a point the sum of angular momentum of its parts = (ang. momentum of the object • For special case of segments rotating about the about its center of mass) same fixed axis: ω1 + (ang. momentum of the center d H = Σ Hi = Σ (Ii ωi) Segment1 of mass about the point) where: Axis of H = Icm ω + m v d 90° – Ii : moment of inertia of Rotation segment i about the axis of rotation ω2 • Used to compute angular – ωi : angular velocity of momentum of the whole ω v Segment 2 segment i about the axis of body about its center of mass rotation 3
- 4. Conservation of Angular Momentum Example Problem #3 • If the net external torque applied to a system is A figure skater is spinning at 1 rev/s in the position on zero, the total angular momentum of the system the left. Her moment of inertia is 5 kg m 2 (about either a fixed axis of rotation or the While spinning, she changes to the position on the system’s center of mass) remains constant right. Her moment of inertia in the new position is: (a) 3 kg m2, (b) 5 kg m2, or (c) 7 kg m 2 ? If Σ Texternal = 0, then: What is her angular velocity in the new position? ω ω H = I ω = a constant • Practical implications: – If I of a body increases, ω decreases – If I of a body decreases, ω increases – In both cases, direction of ω stays the same Example Problem #4 Transfer of Angular Momentum A diver performs a backward 1 ½ somersault, as • Applies to bodies made up of multiple segments shown below. • When momentum is conserved: Sketch his angular momentum, moment of inertia, changing the angular momentum of one segment and angular velocity about his center of mass as a changes the angular momentum of the other function of time. segments • For 2 segments rotating about ωarm a fixed axis: I1inital ω1initial + I2initial ω2initial ωbody = I1final ω1final + I2final ω2final 4
- 5. Example Problem #5 Transfer of Momentum Between Axes A skateboarder attempts a frontside 180. • How does a diver initiate a twisting somersault from Case 1: starts twisting in midair with body straight a somersault? Case 2: starts twisting in midair with body piked • Total angular momentum is constant, so by swinging the arms to rotate the long axis of the body, These are diagrammed below. What is the upper momentum is redirected into twisting and rotation body angular velocity in each case? Ixωsomersault Hinitial = 0 Hinitial = 0 Iupper = 1 kg m2 Iupper = 3 kg m2 H = I ωsomersault H Ilower = 1 kg m2 Ilower = 1.5 kg m2 Iyωtwist ωlower = 6 rad/s ωlower = 6 rad/s 5