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Classical Mechanics Rigid Body Dynamics

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Classical Mechanics Rigid Body Dynamics Classical Mechanics Rigid Body Dynamics Dipan Kumar Ghosh UM-DAE Centre for Excellence in Basic Sciences Kalina, Mumbai 400098 November 14, 2016 1 Introduction A rigid body is defined as a collection of particles such that the relative distance between any pair of particles in the body remains fixed. The position of the rigid body gets fixed if we specify the positions of three non-colinear points in the body with respect to a fixed (external to the body) coordinate system OXYZ. Such a space-fixed coordinate system will also be called as the inertial coordinate system. Since three particles have a total of 9 degrees of freedom, the rigid body itself has six degrees of freedom because the distance between three possible pair of particles are fixed by three holonomic constraints. (Adding another particle would introduce another three degrees of freedom but it will bring in three additional constraints as well!). It is convenient to define a second set of axes which is fixed with the body, i.e. it moves or rotates with the body. We denote this set of axes by Cxyz where C is the origin of this \body-fixed” system of coordinates. Frequently (but not necessarily) C is chosen to be the centre of mass of the body. The six degrees of freedom of the rigid body is specified by the position vector R~ of the origin C of the body fixed system (with respect to the origin O of the inertial frame) and three independent angles that Cxyz makes with OXYZ (e.g. we can fix Cx with resoect to OX two angles - polar and azimuthal; Cy can then be chosen by one more angle; Cz gets fixed since it is perpendicular to both Cx and Cy). 1 c D. K. Ghosh, IIT Bombay 2 Z z r’ y C r x R O Y X Let us consider the situation where Cxyz axes rotate about the OZ axis where the points O and C coincide. Our interest will be to find the relationship between dynamical quantities in the rotating frame of reference with those in the fixed frame.(Note that what follows in the next section is true for a rotating frame which and is not just for the rigid body). 2 Velocity and Acceleration of a particle as seen from a Rotating Frame Let us consider what happens to a vector when it is rotated about a fixed axis by an angle δ' with an angular velocity Ω.~ r r’ r Under rotation let the vector ~r ! ~r0 = ~r + δr~ Clearly, δr~ is perpendicular to the plane containing Ω~ and ~r and its magnitude is δ' j δr~ j= 2r sin( ) ≈ rδ' 2 Thus, if we fix the direction of δr~ by the cross product, we have, δr~ = δ'~ × ~r c D. K. Ghosh, IIT Bombay 3 where δ'~ = δ'Ω.~ Note that while the unit vectors of the Cxyz system ^i; ^j; k^ do not change in the body fixed frame but they change in the space fixed frame. For instance, δ^i = δ'~ × ^i Thus ! d^i d~' = × ^i = Ω~ × ^i dt dt fixed The subscript “fixed” on the l.h.s. is actually superfluous because in the rotating frame the unit vectors would not change. What we want to do is to relate the derivatives of a vector A~ in the fixed frame with the derivative of the same quantity with respect to the rotating frame. Note that a vector is the same in both the frames though its representation may differ from frame to frame. Consider a vector A~ as seen from a rotating frame. Its derivative in the rotating frame can be wrtten as ! dA~ dA dA dA = x^i + y ^j + z k^ dt dt dt dt rot The derivative, as seen from the fixed frame is the sum of this with a term which arises ^ because the unit vectors themselves change with time. Denoting ^i; ^j; k byx ^i with i = 1; 2; 3, we have, ~ ! dA X dAi X dx^i = x^ + A dt dt i i dt fixed i i ! dA~ X = + A (Ω~ × x^ ) dt i i rot i ! dA~ X = + Ω~ × A x^ dt i i rot i ! dA~ = + Ω~ × A~ (1) dt rot Note that Ω~ itself is frame independent. 2.1 Velocities and Acceleration in Rotating Frame Note that if O and C do not coincide, the position vectors in the two frames would be different ~r0 = R~ + ~r c D. K. Ghosh, IIT Bombay 4 where ~r0 is the position vector of a point P from the origin of fixed frame while ~r is the position vector in the rotating frame. Thus ! dR~ d~r ~v = + f dt dt f f d~r = V~ + dt f d~r = V~ + + Ω~ × ~r dt rot ~ ~ = V + ~vr + Ω × ~r (2) Here V~ is the velocity of C with respect to the fixed frame. A further differentiation will give relationship between the acceleration as seen from the two frames. ! ! dV~ d~v dΩ~ d~r ~a = + r + × ~r + Ω~ × f dt dt dt dt f f f f ~ ~ ~_ ~ ~ = A + [~ar + Ω × ~vr] + Ω × ~r + Ω × (~vr + Ω × ~r) ~ ~ ~_ ~ ~ = A + ~ar + 2Ω × ~vr + Ω × ~r + Ω × (Ω × ~r) (3) Thus the acceleration of a point P as observed in the rotating frame is related to acceleration of P as seen from the fixed frame by ~ ~ ~_ ~ ~ ~ar = (~af − A) − 2Ω × ~vr − Ω × ~r − Ω × (Ω × ~r) (4) We know that Newton's laws are valid in inertial reference frame. In such a frame the cause of acceleration is a force whose physical origin can always be traced to a real source. Thus in the inertial reference frame, the acceleration of the particle at P is ~af which must arise from a real force whose value is m~af . Likewise the acceleration of C must be traced to a real force mA~. Hence it is understandable that as seen from the rotating frame the acceleration of the particle is given by the first term of (4). The remaining terms on the right hand side of the above equation obviously have no force apparently associated with them and must be caused by the nature of the rotating frame. If we multiply these terms with the mass of the particle at P, we would get quantities having the dimensions of force. Such forces without apparent physical origin which seem to arise in rotating (in general in a non-accelerating frame) are called pseudo forces. Existence of such forces are postulated only to make Newton's laws valid in the non-inertial frames as well. The last term in (4) −Ω~ × (Ω~ × ~r) is the familiar centrifugal force. The second term ~ −2Ω × ~vr is known as the Coriolis Force which arises if there is a lateral velocity of the particle in the rotating frame. We will drop the third term by assuming that the angular velocity is constant. c D. K. Ghosh, IIT Bombay 5 2.2 Examples: Example 1: Effective gravity on the earth's surface Consider a person standing on the surface of the earth at a latitude λ. The centrifugal force is −mΩ~ ×(Ω~ ×~r). Ω~ ×~r has a magnitude R! cos λ and is directed tangentially along the latitude circle which has a radius of R cos λ. centrifugal mg The centrifugal term is directed away from the rotation axis of the earth and has a magnitude mRΩ2 cos λ. The force of gravity is directed towards the centre of the earth. Because of the centrifugal correction, the effective gravity is slightly towards south in the northern hemisphere. By taking the component of the centrifugal acceleration along the true gravitational force direction, it can be seen that the acceleration due to gravity gets reduced by an amount m!2R cos2 λ. At the equator the centrifugal term is directly opposite to the direction of m~g and it reduces the acceleration due to gravity by about 0.3%. Note that at any other latitude there is a horizontal component of the centrifugal force with the magnitude m!2R cos λ sin λ. As a result of this a plumb-line at these places will not point towards true vertical but will deviate from it towards south in the northern hemisphere and towards north in the southern hemisphere. Example 2: Fc v A person attempts to walk on the surface of a rotating turn table moving from its edge towards the centre. The turn table is rotating anticlockwise with an angular speed ~ ! when viewed from above. He will experience a Coriolis force Fc = −2m~! × ~v, directed towards his right. In order to be able to walk straight, he will have to counter this with c D. K. Ghosh, IIT Bombay 6 a frictional force of equal magnitude to his left. How does this force arise? dL~ The torque acting on the person is . where L~ is the angular momentum of the person dt having a magnitude mr2!, ! being the angular velocity of the person as viewed from the laboratory frame which is equal to the angular speed of the turn table. Since the person d~r is moving towards the centre, = −~v. Thus dt dL d! = −2mr!v + mr2 dt dt = −2mr!v where we have assumed ! to be constant. Thus the torque acting on the person is r times (−2m!v), the last expression being the tangential friction that is applied by the turn table on the man. In case the man chooses not to apply frictional force, in the rotating frame he will experience a tangential acceleration 2!v because of the Coriolis force.
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