1. Summer School 2007B. Rossetto1 8. Dynamics of a rigid body  Theorems r CM : location of the center of mass referred to an inertial frame /Oxyz v CM : velocity of the center of mass /Oxyz P, L : total momentum, angular momentum of the system /Oxyz P/ CM, L/ CM : momentum, angular momentum of the system referred to the center of mass F ext,  ext : external force, torque, M: total mass of the system

2. Summer School 2007B. Rossetto2 8. Dynamics of a rigid body  Translation and rotation Variable: position r Variable: angle  Linear velocity v=dr/dt Angular velocity =d/dt Linear acceleration  =dv/dt Angular acceleration =d/dt Fundamental law dp/dt=F Fundamental law dL/dt  Force perpendicar to momentum F=xp Torque perpendicular to angular momentum =xL Kinetic energy E k =½mv 2 Kinetic energy E k =½  Power P=F.v Power P=. TranslationRotation (Bold letter are vectors)

3. Summer School 2007B. Rossetto3 8. Moment of inertia  Theorem Proof.  Definition G O

4. Summer School 2007B. Rossetto4 8. Moment of inertia  r r sin  x y z r d r sin d r  Homogeneous sphere/z’Oz Contribution of the element dm, length: r sin dweidth: r d height : dr,distance: r sin r sin  0 0   Radius: R, mass: M, density : r

5. Summer School 2007B. Rossetto5 8. Dynamics of a rigid body  Variable mass system Example: rocket vertical motion. If -dm is the positive value of the mass of expelled gases, v and v’ the rocket and gas exhaust velocities relative to earth axes, the total momentum at t+dt is: (m+dm)(v+dv)+(-dm)v’ =mv+mdv-v 0 dm, with v 0 =v’-v Momentum conservation requires that it is equal to momentum at t: mv+mdv-v 0 dm=mv From the second law: and then:

6. Summer School 2007B. Rossetto6 8. Dynamics of a rigid body  Gyroscopic precession O G Fundamental theorem Torque of the weight: and The axis of rotation of the gyroscope, given by the direction of the angular momentum, turns aroud a vertical axis, parallel to weight Exercice. Find the precession velocity.. Variation of the angular momentum