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Torque on a rod

  1. Jul 24, 2012 #1
    Consider a free rod lying horizontically in the air. Gravity produces no torque around the center of mass. Now let the rod be attached to a hinge like of that one the picture. Now the hinge provides an upwards force on the left end of the rod, whilst there is a torque effectively acting at the center of mass, which will make it rotate downwards. Correct so far right?
    Now what I dont understand is this: Obviously the rod doesnt rotate about its center of mass. Yet as the hinge provides an upwards force on the left end equal to mg, m being the mass of the rod, there must be a torque about the center of mass, since we saw that the rod just lying horizontally in the air had no torque about the cm. So therefore: Why doesnt the rod rotate about the center of mass?
     

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  3. Jul 24, 2012 #2

    CWatters

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    Objects only rotate about their center of mass when unconstrained. The hinge constrains the rod to rotate about a different point.
     
  4. Jul 25, 2012 #3
    It is obvious there must be gravitational force, W, acting on the rod. The hinge can provide a supporting force, R, on the rod. In a state of static equilibrium (i.e. no translation plus rotation), the torque by the supportive force (T1) and the torque by gravitational force (T2), relative to a same point (any point), should obey "T1 = -T2"

    Therefore, if the torques produced by W and R are same in magnitude and opposite in direction (i.e. T1 = -T2), the rod will not rotate.
     
  5. Aug 6, 2012 #4
    hmm you will have to elaborate on that. Im not so good at torques.
    We saw for a free rod that the torque around the center of mass is zero since the torques produced at each free end is of the same size but opposite direction.

    Now attach it to the hinge: The same torques due to gravity must exist around the center of mass but only now there is also a supportive upwards force. Why is that NOT equal to a net torque that rotates the rod clockwise?

    I do realize that the question above is identical to my original question, i.e. that I havent realized anything yet, but can you point out exactly where I am drawing a wrong conclusion?
     
    Last edited: Aug 6, 2012
  6. Aug 6, 2012 #5

    CWatters

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    It might seem an obvious statement but the force due to gravity is a force not a torque. To convert a force to a torque you need a pivot (or at least a force not acting on the center of mass)

    The free rod doesn't rotate BUT not for the reasons you give. In the case of the free rod there is no pivot to convert the force of gravity into a torque. Without a pivot the rod is in freefall.

    Lets add a pivot under the center of mass. NOW the two parts on either side of the pivot do produce torques that balance.

    Then lets move the pivot towards one end. The torques become unbalanced and the rod will try to rotate.
     
    Last edited: Aug 6, 2012
  7. Aug 6, 2012 #6

    CWatters

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    Perhaps another way look at it... To produce a torque there must be pair (couple) of forces seperated by a distance. One force on it's own isn't enough.

    In the case of a free rod there is a force acting at the centre of mass and that's it.

    In the case of a rod with a pivot under the center of mass there are two forces (one due to gravity and one due to the reaction force of the pivot) but again there is no distance seperation.

    With the pivot at one end there are two forces seperated by a distance.
     
  8. Aug 6, 2012 #7
    At every instant of time the rod is rotating about its center of mass. But the center of mass itself is moving along the arc of a circle, so the combined motion looks like the rod rotates about its end.

    This is a misconception. If the rod is not uniform, say one half is heavier than the other, it will rotate. It will oscillate like a pendulum, trying to orient the heavy half downward - all this while falling down.
     
  9. Aug 6, 2012 #8

    CWatters

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    Are you sure? The feather and hammer dropped on the moon landed at the same time. They would still land at the same time if connected together by a rod.
     
  10. Aug 6, 2012 #9
    In an approximation, yes. In real gravity, there is always a torque (unless the object has identical principal momenta of inertia). Some satellites are vertically stabilized this way.
     
  11. Aug 6, 2012 #10
    hmm I cant say I agree with you one the heavy pendulum part. If the other end is heavy then it will still just translate since the shifting of the center of mass exactly balances the extra mass that produces the torque.
    But in general I am unsure of when you can use torques to determine motion of an object. I believe I heard that you can use any point and calculate the torques around that, but If I do that for an object with for instance a massless bar of a specific length with a mass m in each end and try to calculate the torque around a general point closer to one of the ends I get a net torque around that point, which is obviously nonsense. It seems you can only use the center of mass and get something sensible. I believe I heard that an object rotates around the cm according to only the external torques. How do you show that?
     
  12. Aug 6, 2012 #11

    TSny

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    For any fixed origin in an inertial frame, the net torque about the origin acting on any system of particles will equal the rate of change of total angular momentum about the origin. So, suppose you pick a fixed point 5 ft above the floor as your origin. Hold your massless rod with the two masses such that the rod is horizontal and positioned so that the origin is located on the rod closer to one of the masses. Let the system fall freely from that position. Can you see that as the system falls, there is a net torque about the fixed origin (5 ft above the floor)? Can you see how the total angular momentum about the origin is changing at a rate equal to the net torque (even though the rod remains horizontal)?
    The center of mass is special in that the net external torque about the center of mass equals the rate of change of the angular momentum about the center of mass even if the center of mass is accelerating with respect to an inertial frame. Can you see how that holds true for your falling rod?
     
  13. Aug 7, 2012 #12
    Precisely correct. The torques and angular momenta are defined relative to an arbitrary origin of a reference frame.

    You get this paradox because you look at just one body, the rod, without looking at the other body, the Earth. If you take a gravitating system of two bodies, you will have conservation of angular momentum (and of linear momentum for that matter), and you will have identical results no matter what frame of reference you use.

    This is shown in any course of mechanics, by looking at a system of material points and applying the laws of Newton to it. The third law implies that internal forces and torques cancel out, leaving only the external sources. You can try for yourself, it just takes some vector multiplication.
     
  14. Aug 7, 2012 #13
    The hinge doesn't provide an upward force on the left end equal to mg. That is only true if the rod is at equilibrium. There is an upward force on the left end, but it is less than mg. Why don't you draw a free body diagram, and show the various forces acting on the rod as it is rotating? The center of mass of the rod is accelerating downward, but not with a vertical component of g. Why don't you do an angular momentum balance on the rod around the hinge, taking into account the rotational inertia? Incidentally, there is also a horizontal component of force at the hinge.
     
  15. Aug 12, 2012 #14
    So all in all I get the correct thing for the center of mass because it only cares about external torques. Therefore I can restrict myself to only the rod and regard the earth as an external system.
    I can't do the same for an arbitrary point on the rod, because that also moves according to internal torques in the rod.
    But how does taking the earth and the rod as a system as a whole incorporate those internal torques? And can you give me an example of how to start my calculation of the torques relative to an arbitrary point on the rod? I don't really see how I put that into my calculation.
     
  16. Aug 12, 2012 #15
    We can replace the Earth with a point mass at distance R from either end of the rod. So it is an isosceles triangle with the sides R, R and l, where the latter is the length of the rod. I suggest you draw this so you can follow the argument easily. The rod is two identical masses m at its ends. Let's compute the torques about one of the masses of the rod. The torque acting on the other mass is [itex]T_1 = -\frac {\gamma mM} {R^2} l sin \theta[/itex], where [itex]\gamma[/itex] is the gravitational constant and [itex]\theta[/itex] is the angle between the axis of the rod and the line to the center of the Earth (at the other mass). The torque acting on the Earth is [itex]T_2 = \frac {\gamma mM} {R^2} R sin (\pi - 2\theta) = \frac {\gamma mM} {R^2} R sin 2\theta = \frac {\gamma mM} {R^2} R 2 cos \theta sin \theta[/itex]. Now, [itex]\frac {l}{2} = R cos \theta[/itex], so [itex]T_2 = \frac {\gamma mM} {R^2} l sin \theta = -T_1[/itex], so the net torque is zero.
     
  17. Aug 12, 2012 #16

    PhanthomJay

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    When you use the hinge as the rotation point, you should note that the net torque is mgl/2 and that the moment of inertia of a rod rotating about one end is ml^2/3. Using Newton's 2nd law for rotational motion , you can solve for the angular acceleration of the center of mass, and it's tangential acceleration as well. Now, if you insist instead in using net torque about the center of mass, you must first find the reaction force at the hinge, which has already been pointed out is NOT mg. Once you find that reaction force, you can then calculate the net torque about the center of mass, and solve for the angular acceleration of the center of mass, this time using I = ml^2/12, and get the same result. The hard way.
     
  18. Aug 14, 2012 #17
    Thank you so much! Very helpful and easy to follow. Now I must ask though:

    What principle in the formulation of the mathematics behind rigid bodies is it that enables you to use an arbitrary reference point in an inertial system? I have actually wondered about the same thing in translational mechanics, why can you pick an arbitrary coordinate frame and still always get the same?

    And, as Im not 100% sure yet, why did the calculations with the center of mass not have to involve the torque on the earth? In your example there wouldn't be any torque on the earth because the center of mass is in the middle of the rod, but what if it wasnt?

    And overall it is a little weird for me that you have to also take the torque ON the earth into the calculation. Why does the force on the earth matter? It is not a part of the rigid body and therefore why should a force on that add to any rotation of it?
     
  19. Aug 15, 2012 #18
    This follows from the Third Law given by Newton. Any two masses interacting have acting on them forces equal in magnitude and opposite in direction. That means that the forces can depend only on the position of the masses with respect to one another, not on any arbitrary point of reference. When this is formulated in terms of vectors, only vector differences of the kind [itex]\vec{r_1} - \vec{r_2}[/itex] appear in equations, where [itex]\vec{r_1}[/itex] and [itex]\vec{r_2}[/itex] are vectors to the interacting masses from an arbitrary point.

    The point is that the torque of gravity about the center of gravity (if one exists) is always zero - pretty much by definition. Now, the center of gravity is not necessarily the same as the center of mass, but when gravity is "uniform" in certain ways, they are!

    I did that to illustrate that the net torque is always zero no matter what point of reference is considered, if the entire system is analyzed. If we analyze some part of the system, i.e., there is an external force acting on it (from other parts), then the torque in general depends on the point of reference. Torques about two different points of reference are equal only if the sum of all external forces is zero.
     
  20. Aug 15, 2012 #19
    But isn't it a bit weird that the torque on the earth is used to determine the total torque about our point and thus whether it rotates or not? For me it is a bit like saying that one body wont move because a counterforce is exerted on a completely different one. But maybe that is a too simplified view. How do you see it?

    Also, I think I have at least learnt that you should always look at the whole system before saying that the torque about a point is zero. But then again, going back to the original problem with a rod and the hinge. Then we did not have too look at the torque on earth in that situation even though the rotation point was constrained to be exactly one of the ends of a bar of mass. Why did the torque on the earth not come into question?
     
  21. Aug 15, 2012 #20
    Please keep in mind that a body may rotate if it has zero torque acting on it about any point. Torque changes rotation. Secondly, the term rotation is itself a bit ambiguous. The torque is related with angular momentum, and even a dimensionless point moving along a straight line can have changing angular momentum, when hardly anyone would speak of rotation. Incidentally, this is just the case of a rod falling in uniform gravity.

    A rotating body is the body where at least one point has zero velocity in an inertial reference frame, and at least one point with a non-zero velocity in the same frame.

    Thus, by looking at the Earth, we did not conclude that the rod was not (or was) rotating.

    By considering the Earth here, we will end up again with zero net torque. But here the Earth would get a tiny rotation countering the rod's rotation, thus keeping the net angular momentum constant.
     
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