Rigid Body Dynamics

# Rigid Body Dynamics - Rigid Body Dynamics chapter 10...

This preview shows pages 1–17. Sign up to view the full content.

Rigid Body Dynamics chapter 10 continues around and around we go …

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Rigid Body Rotation Rotation Axis { } 2 2 2 2 2 2 2 2 2 1 2 1 ) ( 2 1 2 1 ) ( 2 1 ) ( ϖ I r m KE r m r m KE r v v m KE i i i total i i i i i i i i i i = = = = = = Moment of Inertia: 2 2 = i i i r m I
CORRESPONDENCE: I m v I mv ϖ 2 2 2 1 2 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Rotational Rotational Kinetic Energy We there have an analogy between the kinetic energies associated with linear motion ( K = ½ mv 2 ) and the kinetic energy associated with rotational motion ( K R = ½ I ϖ 2 ) Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object The units of rotational kinetic energy are also Joules (J)
Important Concept: Moment of Inertia The definition of moment of inertia is The dimensions of moment of inertia are ML 2 and its SI units are kg . m 2 We can calculate the moment of inertia of an object more easily by assuming it is divided into many small volume elements, each of mass m i 2 i i i I r m =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Moment of Inertia, cont We can rewrite the expression for I in terms of m With the small volume segment assumption, If ρ is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known lim 2 2 0 i m i i i I r m r dm = = 2 I r dV =
Question – WHAT IS THE MOMENT OF INERTIA OF THIS OBJECT??

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Let’s Look at the possibilities c d
Two balls with masses M and m are connected by a rigid rod of length L and negligible mass as in Figure P10.22. For an axis perpendicular to the rod, show that the system has the minimum moment of inertia when the axis passes through the center of mass. Show that this moment of inertia is I = L 2, where = mM /( m + M ).

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Remember the Various Densities Volumetric Mass Density –> mass per unit volume: ρ = m / V Face Mass Density –> mass per unit thickness of a sheet of uniform thickness, t : σ = t Linear Mass Density –> mass per unit length of a rod of uniform cross-sectional area: λ = m / L = ρΑ
Moment of Inertia of a Uniform Thin Hoop Since this is a thin hoop, all mass elements are the same distance from the center 2 2 2 I r dm R dm I MR = = =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Moment of Inertia of a Uniform Rigid Rod The shaded area has a mass dm = λ dx Then the moment of inertia is / 2 2 2 / 2 2 1 12 L L M I r dm x dx L I ML - = = =
Moment of Inertia of a Uniform Solid Cylinder Divide the cylinder into concentric shells with radius r , thickness dr and length L Then for I ( 29 2 2 2 2 1 2 z I r dm r Lr dr I MR πρ = = = dV=L(2 π rdr)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Moments of Inertia of Various Rigid Objects
Parallel-Axis Theorem In the previous examples, the axis of rotation coincided with the axis of symmetry of the object For an arbitrary axis, the parallel-axis theorem often simplifies calculations The theorem states I = I CM + MD 2 I </