Rigid Body: an Introduction
For the dynamical problem of the motion of a rigid body (one that does not bend or vibrate), we need six coordinates: three for the position (usually the position of the center of mass), and three for the orientation. We have two vector equations (and hence six scalar equations) to use:
SF = MR’’ = dp/dt where p = MV, and V = R’
Sti = dL/dt where ti = ri ´ F, and L = S(ri ´ mivi)
where the t may depend on both the position and orientation; and where, if the body is free, the torque and angular momentum are measured about the center of mass; and if the body is constrained to be at point Q, the torque and angular momentum are measured about that point, Q.
For the orientation, we have no simple set of coordinates that are symmetric. The set (θx, θy, and θz) does not specify a unique orientation because the final orientation depends on which of the angles you rotate through first. For instance, if you take an object and rotate it first about the x axis by 90o degrees, and then about the y axis by 90o degrees, the final orientation is in a different orientation than if you rotate first about the y axis by 90o and then about the x by 90o.
Rotation about an axis
In the case of a rotation about a specified axis, we need only one coordinate to specify the orientation. In this case, it is usually easiest to use cylindrical coordinates (s, θ, z), and use the z axis as the specified axis of rotation. [Note that here I use s instead of r as I did earlier, since now r will be used for density.] Please keep in mind that the s coordinate is the distance from the z-axis, not the distance from the origin. The r vector is from the origin. In this case, then
L = S (ri ´ mi vi) = Lzz ,
where the component of vi that is perpendicular to the radial and to the z directions is vqi = siq’, so
Lz = Smivqisi = Smisi2θ’ = (Smisi2) θ’ = Izθ’ ,
Iz = (Smisi2) = òòòr s2 dV ,
since θ’ = dθ/dt is the same for all mi, where r is the mass density, and dV = ds (sdθ) dz in cylindrical form.
It is sometimes useful to define a radius of gyration, kz, as Mkz2 = Iz , so kz = [Iz/M]½ .
The torque equation in this case is:
Stz = dLz/dt = Izθ’’ .
We can also define a potential energy, V(θ), based on the torque just like we defined a V(r) based on the force: Work = ∫ F· dl = ∫ Fq sdq = ∫ t(θ) dθ :
V(θ) = - θi∫θf t(θ) dθ and the inverse: tz(θ) = -dV(θ)/dθ .
The kinetic energy in the rotations in this case is:
KE = S½mivθi2 = ½Smi(siθi’)2 = ½(Smi si2) θ’2 = ½Izθ’2 .
Note also that Power = rate of using energy, so P = dW/dt = d[ ∫ F· dl ]/dt = F· v and P = dW/dt = d[ ∫ t · dq ]/dt = t · w .
Homework Problem #29: Find the steady state motion of a propeller with moment of inertia, I, that has an applied torque, ta = to[1 + a*cos(wat)] and a frictional torque due to air resistance of tf = -bw where to, a, wa, and b are constants.
Homework Problem #30: The balance wheel of a watch consists of a ring of mass, M, and radius, R, with spokes of negligible mass. The hairspring exerts a restoring torque, ths = -kq. Find q(t) if initially q(t=0) = qo and w(t=0) = 0.