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Equation of motion, rolling disk

  1. Mar 30, 2013 #1
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    I'm trying to get the position vector of the offset mass relative to the inertial frame of reference. Do I account for the fact that the disk has rolled a little bit with respect to the inertial frame of reference? With that, I have
    r,vector = d n1 + R n2 + R/2 er = R*theta n1 + R n2 + R/2 er,
    where d = R*theta because the disk is assumed to roll without slip.
    Or is it just r,vector = R n2 + R/2 er?
     
  2. jcsd
  3. Mar 30, 2013 #2
    So, it's all a matter of where you define your origin to be. As drawn, [itex]\vec{r}[/itex] is with respect to the center of the disk. If the center of the disk is the origin, then [itex]\vec{r} = \frac{R}{2} \hat{e}_r[/itex].

    If you say [itex]\vec{r} = R \hat{n}_2 + \frac{R}{2} \hat{e}_r[/itex], then you have defined your origin as the point of contact between the disk and the incline. Since the disk is accelerating, that origin is the origin of a non-inertial reference frame. Therefore, i don't think that's what you want since you say "relative to the inertial frame of reference".

    I'm assuming that you mean the frame of reference of the incline (wherein the incline does not move), assuming, of course, that the incline is fixed in place by some means. Then, your first answer, [itex]\vec{r} = R \theta \hat{n}_1 + R \hat{n}_2 + \frac{R}{2} \hat{e}_r[/itex], is correct. Here, you have set your origin to be the point of contact between the incline and the disk when [itex]\theta = 0[/itex].

    I'm guessing that you're already aware of this, but just be careful of the fact that while [itex]\hat{n}_1[/itex] and [itex]\hat{n}_2[/itex] are constant vectors, [itex]\hat{e}_e[/itex] and [itex]\hat{e}_{\theta}[/itex] are not; they change as the disk rolls down.
     
  4. Mar 31, 2013 #3
    It doesn't specify in the question whether the N frame is inertial or not, but I do need to derive the equation of motion, and the equation of motion is found by F = ma for an inertial frame of reference, right?

    Am I right in thinking that technically, an inertial frame of reference is one in which the unit base vectors are constant, such as n1 and n2 in this case? And thus, Newton's second law cannot be applied to the E = {er, e,theta, ek} frame?

    I'm just having a little trouble understanding setting up the frames, but once I have this down, the problem of deriving the acceleration vector to apply to F = ma should be straightforward.
     
    Last edited: Mar 31, 2013
  5. Apr 1, 2013 #4
    Well, the unit vectors of a frame can always be made constant with respect to an observer in that frame so their constancy alone can't determine whether or not the frame is inertial. I won't bore you with inertial reference frames since that's probably on wikipedia or something and the problem is not really about that.

    You are correct to say that you can't just apply [itex]F=ma[/itex] to the [itex]\{\hat{e}_r, \hat{e}_{\theta} \}[/itex] frame. You would have to take fictitious forces into account, which should be doable in principle, but not worth the trouble. However, you should be able to express these vectors in terms of [itex]\{ \hat{n}_1, \hat{n}_2 \}[/itex] since they span the same plane (the plane of the drawing). This relationship will be a function of [itex]\theta[/itex]. For example, when [itex]\theta = 0[/itex], you should have [itex]\hat{e}_{\theta} = \hat{n}_1[/itex] and [itex]\hat{e}_r = \hat{n}_2[/itex]. Once you've done that, then you'll have the position of the point mass just in terms of the constant vectors: [itex]\vec{r} = f_1 ( \theta ) \hat{n}_1 + f_2 ( \theta ) \hat{n}_2[/itex]. Of course, [itex]\theta[/itex] is a function of time, so [itex]\vec{a} = \ddot{\vec{r}} \neq 0[/itex].

    By the way, if I were given free rein to solve the problem as I wished, I would use the Lagrangian formalism. Once you have the position as above, you can calculate the kinetic energy, [itex]T[/itex], and potential energy, [itex]V[/itex], as usual, write the Lagrangian [itex]L=T-V[/itex], and derive the Lagrangian equation of motion: [itex]\bigl( \frac{d}{dt} \frac{\partial}{\partial \dot{\theta}} - \frac{\partial}{\partial \theta} \bigr) L = 0[/itex]. You can even keep the fact that the disk is on the incline and rolling without slipping as a separate constraint and use a constrained Lagrangian formalism. If you already know these things, I'm sorry for boring you; if you don't, I'm sorry for bombarding you with useless information.
     
  6. Apr 1, 2013 #5
    Thanks for bombarding me with useless information :), it's nice to know there are other ways to do this, although I do not know of the Lagrangian techniques.

    Although F = ma cannot directly be applied to the E-frame, what fictitious forces must be taken into account to solve the problem? I know I can just rotate the E-frame into the inertial N-frame by a rotation matrix (or in this case, use er = sintheta n1 + costheta n2, etheta = costheta n1 - sintheta n2), and then use F = ma, but I'm curious as to what fictitious forces you would include to apply F = ma to the non-inertial E-frame.

    I'm guessing the answer to my next question is no, but if the N-frame was moving (pure translation, no rotation) with the disc, so that its origin is always at the point of contact between the disc and the incline, would F = ma be applicable?
     
  7. Apr 1, 2013 #6
    All the possible fictitious forces pertain to this problem, which is partly what makes going to the E-frame difficult. There is a linear fictitious force parallel to the incline just because the disk as a whole is accelerating down the incline. There are also the three rotational fictitious forces, Coriolis, Centrifugal, and Euler. In this case, I believe that Centrifugal points away from the center of the disk, Euler points up the incline, and Coriolis has both components, although I may be wrong here.

    Roughly speaking, you know that some such fictitious forces must exist by the following argument. In the N frame, the resultant force on the point mass has at least a component pointing radially towards the center of the disk and another pointing down the incline. After all, it is rotating around the center of the disk and the whole disk is going down the incline. If you just naively transform these two components to the rotating frame, E, the result would certainly not be zero. But, the point mass is supposed to be at rest in the frame E! So the over-all fictitious force in the E-frame must have a component pointing away from the center of the disk and another pointing up the incline.

    If it isn't already obvious, generally speaking, fictitious forces are best avoided when possible. Of course, sometimes you really can't; for example, the Coriolis force is the reason why hurricanes rotate in the counter-clockwise direction in the northern hemisphere and clockwise in the southern hemisphere (as viewed from above).

    The answer to your last question is "no", as you guessed. The disk is accelerating down the incline and so the point of contact between the disk and the incline is also accelerating down the incline... at the same rate. However, in this case, the only fictitious force is the linear one. In this frame, the disk does not move linearly, it just rotates and as it does the incline "slides" underneath it faster and faster. Linearly accelerated frames are pretty much as easy to deal with as inertial ones.
     
  8. Apr 3, 2013 #7
    One more question: Can F = ma be applied to the inertial acceleration vector after expressing it in E-frame components? Would this vector still be considered "inertial?"
    I would think so, because it is still the same vector just expressed in a different frame, but I'd like to make sure.

    Another thing I'm wondering:
    Is the equation of motion still the same regardless of what inertial frame I'm using? (Like for example, I arbitrarily place the origin of the N-frame somewhere else in space). I derived the EOM for several different inertial frames and seem to be getting the same result each time, but I'd like to know if this is generally the case.
     
  9. Apr 3, 2013 #8
    In Newtonian mechanics, [itex]F=ma[/itex] will hold. Actually, that's tautological because Newtonian mechanics is defined by Newton's laws of motion and [itex]F=ma[/itex] is Newton's second law.

    Anyway, often the tricky part is to figure out what [itex]F[/itex] is. In the E-frame, or any other non-inertial frame, you just have to be vary careful and include into [itex]F[/itex] all of the real as well as the fictitious forces.

    A word on terminology: I wouldn't call a vector inertial. A reference frame can be inertial. As you say, a vector exists independently of the reference frame you choose to use to describe it.

    The answer to your last question is "yes". The reason is that while it is possible for inertial frames to be moving relative to one another, they cannot be accelerating relative to one another. More mathematically, if [itex]\vec{r}_1[/itex] and [itex]\vec{r}_2[/itex] are two expressions for the same vector just in two different inertial reference frames, [itex]S_1[/itex] and [itex]S_2[/itex], then there must exist some constant velocity vector, [itex]\vec{v}[/itex], such that [itex]\vec{r}_2 = \vec{r}_1 + \vec{v}t[/itex], where [itex]t[/itex] is time. Note that time is universal in Newtonian mechanics (Galilean relativity) as opposed to relativistic mechanics (Einsteinian relativity). The velocities associated with the two different expressions of the vector are simply shifted by the constant velocity: [itex]\vec{v}_2 = \dot{\vec{r}}_2 = \dot{\vec{r}}_1 + \vec{v} + \dot{\vec{v}} t = \vec{v}_1 + \vec{v} + \vec{0}[/itex]. Therefore, the accelerations are the same: [itex]\vec{a}_2 = \dot{\vec{v}}_2 = \dot{\vec{v}}_1 + \dot{\vec{v}} = \vec{a}_1 + \vec{0}[/itex].

    You mentioned that you get the same equation of motion regardless of exactly where in the N-frame you chose to place the origin. That's the same as shifting the origin by a constant. In the language of the previous paragraph, that's [itex]\vec{v}=0[/itex]. The previous paragraph shows that you have even more freedom than that. You can even keep moving your definition of the origin as long as you move it in just one direction and at a constant speed; you will still get the same equation of motion.
     
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