Rigid bodies - general theory
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1 Rigid bodies - general theory Kinetic Energy: based on FW-26 Consider a system on N particles with all their relative separations fixed: it has 3 translational and 3 rotational degrees of freedom. Motion with One Arbitrary Fixed Point: continuum limit: it is convenient to define the inertia tensor: = 0 in a it is constant and depends only on the mass distribution the kinetic energy has a simple form: both I and ω depend on the choice of the coordinate system, but T does not! Angular momentum: General motion with No Fixed Point: with the origin at the center of mass a parallel copy of the with the same origin as the body fixed frame - the center-of-mass frame (in general it is not an ) continuum limit: using the inertia tensor it can be written as: and it is related to the kinetic energy: I, ω and L depend on the choice of the coordinate system! 0 for rigid bodies
2 Kinetic Energy and Angular Momentum: with the origin at the center of mass a parallel copy of the with the same origin as the body fixed frame - the center-of-mass frame (in general it is not an ) Review: Inertia tensor based on FW-26 We have to learn how to evaluate the inertia tensor in a body-fixed frame. First, let s define the inertia tensor in a with the origin at the center of mass: internal motion about the CM Then the inertia tensor in a with the origin displaced by a is given by: center-of-mass relations Formulas for T and L are identical to the case of a motion with one fixed point! parallel-axis theorem! Example (a uniform disk of radius a): Principal axes (a coordinate system in which the inertia tensor is diagonal): Diagonalizing a real symmetric matrix: all the formulas simplify in such a coordinate system moment of inertia about a perpendicular axis through the edge of the disk (would be tedious to calculate directly) moment of inertia about a perpendicular axis through the center of the disk (easy to calculate in cylindrical polar coordinates) non-trivial solution only if: similar to solving for normal modes, v I and m 1 we get 3eigenvalues, 3 eigenvectors, and we can form the modal matrix: T new orthonormal basis - principal axes
3 Modal matrix diagonalizes the inertia tensor: Angular Momentum in the new coordinate system: we can define a new set of angular velocities: T projections of ω along the principal axes Principal moments of inertia: Kinetic Energy becomes: projections of r along principal axes Euler s Equations Applications - Compound Pendulum Describing rotational motion: based on FW-27 valid in any but also in the cm frame: in terms of body-fixed observables: Euler s equations: a rigid body constrained to rotate about a fixed axis: stationary both in the inertial frame and in we can choose the coordinate system so that: one degree of freedom ignoring friction, the only torque comes from gravity: projecting onto the principal body axes simple formulas, but limited use, since both angular velocities and torques are evaluated in a time dependent (principal axes); used mostly for torque-free motion or partially constrained motion (examples follow) in both inertial and body frame perp. distance from the 3rd axis
4 equation of motion of an arbitrary compound pendulum: oscillation frequency about the axis through Q: for small oscillations, the motion is simple harmonic: length of a simple pendulum with the same frequency center of percussion Let s rewrite it in terms of the moment of inertia about the axis going through the center of mass: radius of gyration equivalent radial distance of a point mass M leading to the same moment of inertia radius of gyration about the axis through Q invert the pendulum and suspend it from the parallel axis going through P The oscillations frequencies and periods about the axis through Q and P are identical: is used to measure g! Response of a baseball bat to a transverse force: motion of the center of mass (N-2nd): torque equation: reaction force at the point of support the only direction R can move due to constraints applied force (we neglect gravity) Rolling and Sliding Billiard Ball a ball is struck at the center (h=0) with a horizontal force. When does it start rolling? initial conditions:.. x = 0, x = v, ϕ = 0, ϕ = 0 force equation: solution: 0 torque equation: Pure rolling without sliding occurs only for: if the force is applied at the center of percussion the reaction force at the point of support vanishes! if applied beyond that point, the reaction force has the same direction as applied force a = 5 2 µgt at this time:
5 Rolling and Sliding Billiard Ball a ball is struck at h above the center with a horizontal force. When does it start rolling? the same equations of motion: initial conditions: initial angular velocity Pure rolling without sliding occurs only for: rolls immediately, no sliding (friction) slides, then rolls rolls too fast?! independent of the impulse! 147
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