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Rod Rotational Motion Question

  1. Feb 13, 2004 #1
    I think I have asked this question before, but I either forgot the answer, or never got one.

    A uniform rod is floating in otherwise empty space and a force is exerted on one end of the rod. What will be the acceleration of the rod's rotation and translation? Where will the center of rotation be?
     
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  3. Feb 14, 2004 #2

    HallsofIvy

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    What direction is the force? The rotation, if there is any, will be about the center of mass of the rod. Break the force into two components- one perpendicular to the length of the rod (the line from the point at which the force is applied and the center of mass) and one parallel to it. Use the torque given by the perpendicular component to find the angular acceleration and (with the length of time the force is applied) to find the final angular velocity. The component of force parallel to the axis is applied through the center of mass and give the linear acceleration.
     
  4. Feb 14, 2004 #3
    Why is the center of rotation about the center of mass?
     
  5. Feb 15, 2004 #4

    Doc Al

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    The center of rotation need not be at the center of mass of the rod. The motion is complicated, but it can be treated as a combination of 1) a translational acceleration of the center of mass (F = ma) plus 2) a rotational acceleration of the object about its center of mass (T = I α).
     
  6. Feb 15, 2004 #5
    If I understand you correctly, I should break the force vector into two components, one perpendicular to the rod, and one parallel to the rod. Then solve for the acceleration of the CM of the rod, and the angular acceleration of the rod about the CM.

    I'm still having issues with this explanation, because it implies that if a force perpendicular to the rod was to be applied at one end of the rod, the force would not affect the center of mass at all (since there is no component of the force parallel to the rod). This surely can not be true.
     
  7. Feb 15, 2004 #6

    Doc Al

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    It isn't true. The acceleration of the center of mass is given by Fnet/m.

    I believe Halls made an error in his explanation, unless I am misinterpreting what he said.
     
  8. Feb 15, 2004 #7
    If I am interpreting correctly, you are saying that a force applied (at a point other than the CM) to a body will accelerate the CM as well as accelerate the body's rotation (about the CM). This makes a much more sense to me!

    Thanks very much!
     
    Last edited: Feb 15, 2004
  9. Feb 15, 2004 #8

    Integral

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    Ideally it is indeed the case, if you were able to apply a force normal to the rod, precisely at the end of the rod, then all motion would be rotational.

    In the real world this is nearly impossible to accomplish. Since any real force must be applied over a time interval, if the force is in a constant direction, the rod will have rotated during the application of the force, resulting in a component of the force along the rod(the force has not changed direction, the rod has). This component of force along the rod will result in translational motion. If this could be done in zero gravity, the result would be a rapidly spinning rod slowly moving across the room.

    Have you ever seen the circus act where spinning plates or disks, (This used to be a common TV variety show fare also) are balanced on sticks? The performer keeps them going by striking them sharply on the edge? Is this not an example of exactly what we are talking about? A quick sharp blow tangent to the edge,in the direction of motion, will result in rotational motion only.
     
  10. Feb 15, 2004 #9
    Integral,

    My issue with HallsofIvy's explanation is the following.

    Suppose that a force is exert to one end of the rod. Again, it is exactly perpendicular, so there is no compononent of the force affecting the translation of the CM. Suppose another perpendicular force of different magnitutde is exerted on another location on the rod. This force is exerted so that the torque it produces cancels out the torque of the initial force. Since both of these forces are perpendicular and do not affect the CM, this system would thoeritically be in static equilibrium.

    However, it seems to me that this violates the laws of statics, which requires the sum of the net external forces to be equal to 0 for the system to be in equilibrium.

    I'm not sure if this is even valid, but it seems to me that Doc Al's explanation covers this apparent contradiction.

    Thanks a lot.

    (edited for many grammar errors)
     
    Last edited: Feb 15, 2004
  11. Feb 15, 2004 #10

    Integral

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    Here you go changing the problem. If there is no component though the CM then there would be no translational motion. If there is translational motion then there was a non tangential component though the CM.

    Why is this so hard to understand?

    You are toying with a concept called a couple.

    One very applied area for this type of force is in a standard T bar tap wrench. The idea is to carefully apply tangential forces to each end of the handle. This is very critical for small taps (0-80 for example) any non tangential force results in a broken tap.
     
  12. Feb 15, 2004 #11

    Doc Al

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    Yes, that's exactly what I'm saying. It doesn't matter where the force is applied to a rigid body: Fnet = m acm.
     
  13. Feb 15, 2004 #12

    Doc Al

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    An excellent example. (Good thinking!) For translational equilibrium (no acceleration of the cm) the net force must be zero. It's as simple as that.
     
  14. Feb 15, 2004 #13
    Integral,

    I'm not sure how I am changing the problem. You are saying:

    If there were no component through the CM then there would be no translation motion. If there is translation motion then there was a non-tangential component though the CM.

    This is a hard concept to understand for me because the statement seems false. Consider 2 equal, tangential forces are applied to either end of the rod, both pointing in the same direction. According to what you say, there would be no translation of the CM, since there are no components of these forces going through the CM. This is very clearly not true.

    Doc Al, your explanation seems to be correct to me, but I am not sure why Integral and HallsofIvy, both of whom I hold in high regard, say otherwise.
     
  15. Feb 15, 2004 #14
    Ah, both me and Doc Al replied at the same time. Thanks for the clarification Doc!
     
  16. Feb 15, 2004 #15

    Doc Al

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    Sorry, Integral, but Moose352 is correct; your statement is not.
    An excellent question!
     
  17. Feb 15, 2004 #16

    Integral

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    Please show me how I am wrong...

    In math speak.
     
  18. Feb 15, 2004 #17

    Integral

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    This is incorrect, 2 forces so applied would result in no rotational motion, The net vector would be as if were applied at the CM, thus it would be pure translation.

    How did you change the problem? You added a 2nd force, this is a different problem.
     
  19. Feb 15, 2004 #18
    Integral,

    Exactly. This is just an example (like the other one I posted) where your statements do not hold true. It is clear that forces have no components acting on the CM, yet the CM translates. Note that if the two forces were not equal, the CM would still translate, as well as rotate. This clearly proves you wrong.

    I added the second force to the problem to better see the effects of the first force.
     
  20. Feb 15, 2004 #19

    Doc Al

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  21. Feb 15, 2004 #20

    Integral

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    As I said above, 2 forces acting are different then a single force. The 2 acting parallel are a single force acting through the cm. Since forces are parallel there would be no rotational motion.

    If what you are saying is a simple application of Newtons then it will be simple to write out to show where the translational force comes from when a SINGLE tangential force is applied.


    Your link proves nothing as it does not go into resolution of applied forces. I am not satisfied.
     
    Last edited: Feb 15, 2004
  22. Feb 16, 2004 #21

    Doc Al

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    The two forces combined have the same effect as a single force (of twice the magnitude) acting through the cm, since the torque is zero. But note that the acceleration of the cm is double what it would be if only one force were acting.
    It is simple and I've written it out several times now. The "translational" force is the net force. That's Newton's second law.
    It was not meant as a "proof". It's just a statement of Newton's law applied to a rigid body. I had presumed this to be common knowledge, not something that I would have to "prove". Silly me.
     
  23. Feb 16, 2004 #22

    Integral

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    You restate the obvious and have avoided the question at issue.

    How does a single ideal tangential force result in non rotational motion? The situation with 2 parallel forces proves nothing other then a distraction to the original question.

    A non tangential force can be resolved into a tangential component which applies a torque resulting in rotation about the cm and a component through the cm which results in translational motion. If a pure torque is applied rotation results with no translational motion. I have provided 2 separate physical examples of this type of force, yet you continue to claim otherwise with no physical arguments to back your words.
     
  24. Feb 16, 2004 #23

    Doc Al

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    I admit that what I am stating should be obvious, but you still don't get it. I don't think I have avoided the question; rather I've addressed the question directly by appealing to established physical laws.
    Apparently you don't realize that any net force, regardless of direction, will produce an acceleration of the center of mass. You seem to think that the force must be along a line intersecting the cm in order to produce an acceleration of the cm. Now where did you get that quaint notion?
    So ignore it.
    Half right. It is the tangential component that will contribute to the torque about the cm in this case. But the "component through the cm" causing a translational acceleration is nonsense.
    If by "pure torque" you mean a couple of forces that add to zero, then you are correct. A single applied force---as is the case in question here---cannot provide "pure torque".
    What examples have you provided?? If you can provide a single example of a net force applied to a body without causing an acceleration of the cm, I would love to see it.

    Just because you imagine that a force applied tangent to the rod will only cause rotational, but not translational, acceleration is hardly a physical argument.

    Every statement I have made in this thread is backed by Newton's Laws. Show me where I'm wrong.
     
  25. Feb 16, 2004 #24

    Integral

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    Been thinking about it, and after reading your posts, I think I am seeing my error. First of all if we consider an unanchored rod in a gravity free environment which is delivered a sharp blow to one end, what is the resulting motion?

    It seems that the resulting rotation will not be about the cm but the opposite end of the rod, this will result in displacement of the cm therefore translational motion.

    Care to derive the equation of motion for this system? I will work on it off line see what I can come up with.
     
  26. Feb 17, 2004 #25

    Doc Al

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    Here's my analysis. Let's take the simplest case. A sharp blow is delivered to one end of a thin rigid rod (length 2r; mass m) exactly perpendicular to the rod. Assume the blow imparts an impulse FΔt. What is the resulting motion of the rod?

    The motion of any rigid body can be analyzed as a combination of the motion of its center of mass plus its rotation about that center of mass.

    The impulse does two things. It imparts a translational momentum, thus giving the cm of the rod a speed v = FΔt/m. It also imparts an angular momentum L = rFΔt. The resulting angular speed about the center of mass is ω = rFΔt/I. Since I = mr2/3, ω = 3FΔt/mr. The linear speed of the ends (with respect to the center) is ωr = 3v.

    Of course, after the impact, the cm will continue in a straight line with constant speed, and the rod will continue to rotate with constant ω.

    Immediately after the blow, the motion of the rod can be described as:
    - the center moves with speed v (given above)
    - the end that was hit moves with speed v + ωr = 4v
    - the other end moves with speed v - ωr = -2v

    (Of course, hitting the rod at some other angle would result in a different angular momentum and direction of motion.)
    I think the above analysis is correct. Let me know what you find.
     
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