Lecture 32: Dynamics of Rigid Bodies

Transcription

1 Lecture 3: Dynamics of Rigid Bodies Our next topic is the study of a special ind of system of particles a rigid body The defining characteristic of such a system is that the distance between any two particles in the system is constant Any solid obect is a reasonable approximation of a rigid body But not a perfect one, since all real obects are made up of atoms that can vibrate with respect to each other The unique type of motion that rigid bodies undergo is rotation, so that will be the focus of our studies

2 What We Already Know This isn t the first time you ve seen rotations In fact, we ve been using some of our nowledge throughout the semester Hoops rolling down planes, and the lie Relations lie the following should be familiar: L = Iω, T = Iω The quantity I is the moment of inertia We ll have plenty more to say about it! What we ll find is that everything we now is true but it s rather limited i.e., it only deals with rotations about a single plane, about an axis of symmetry or parallel to one

3 Rotational Energy-The General Case Let s remove the restrictions and consider the energy of a rotating body Choose our origin to be the point that the body is rotating about This must be a fixed distance from the center of mass, so the entire inetic energy is that due to motion with respect to our origin: We can always write: T That s ust the definition of ω for a particle rot = v = r m v

4 If the entire body is rotating about the origin, though, every particle must have the same angular velocity, so: T rot = m r To interpret this, we need to apply a vector identity: Therefore, A B = ABsinθ = A B cos θ A B = A B Trot = m ω r In terms of the Cartesian components of ω and r this is: T rot = m ω i x, ωi x, i ω x, i i r

5 Now write ω i in a non-obvious, but ultimately helpful, way: ω = ω δ ; ω = ω ω δ = ω ω δ i i i i i i i Putting this into the expression for rotational inetic energy gives: T = m ω ω δ x ω ω x x = ω iω m δ i x, x, i x, i, rot i i, i, i, i, i, This term depends on properties of the obect not how it s moving

6 The Inertia Tensor We can simplify the expression for energy by defining: Then m δ x x x I i,, i, i T rot = ω ω I Note that if the obect is rotating about the i axis, the expression becomes: i, This loos more lie what we remember from our introductory course! i, i i Trot = ω ω δ I = I ω i i i ii i

7 But in earlier courses, you probably got the impression that the moment of inertia for an obect was described by three numbers: I x, I y, I z Note, though, that no one tried to write this as a vector Now it seems we need nine numbers to express the same quantity For convenience, can write them as a matrix: I I I I 3 = I I I 3 I I I The Inertia Tensor So, who s been lying to you? No one, really, though you weren t told the whole story before. More on that later

8 One step towards reconciling the nine vs. three numbers dilemma is that the nine is really six I = m δ x x x i i,, i, = m δ x x x = I So the only unique numbers are i,,, i i I, I, I3, I, I3, I33 If the rigid body is a continuous distribution of matter rather than a collection of discrete points, the I i are found by: Ii = ρ r δ i x xi x dv V

9 Angular Momentum The general definition of angular momentum is: = L r p The value depends on our choice of origin. One can choose any point, but two choices mae things easy:. If a point in the body is fixed in some inertial reference frame, choose that point as the origin. If there is no fixed point, use the center of mass as the origin In other words, try to choose an origin about which the body is rotating! For such an origin, we now that v = r, so that: L = r m v = m r r

10 Here again we find that a vector identity is useful: A B A = A B A A B Using this, L = m r r r or, in terms of Cartesian components: L = m ω x x x ω i i,, i, = m ω δ x x x ω i,, i, = ω m δ x x x = ω I i,, i, i The inertia tensor appears again!

11 In tensor notation, we write this as: L = I This implies something quite interesting: Despite all you ve heard about L = Iω, the L and ω vectors are not parallel in general Example: dumbbell rotating about non-symmetric axis: m L z ω L TOT m L Tae CM as the origin The ω vector is directed upwards Le L = L + L = r p + r p z Note also that L is not constant direction changes as dumbbell rotates. External torque must be applied to cause this motion.