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Why is torque on a frictionless axle zero?

  1. Oct 9, 2013 #1
    I'm currently reading John Taylor's Classical Mechanics. Near the end of Chapter 3, he states "Because the table is mounted on a frictionless axle, there is no torque in the z direction. Therefore, the z component of the external torque on the system is zero and the [system's angular momentum in the z direction] is conserved.

    I'm not sure why being on a frictionless axle means there's no torque. The problem is this:

    A unifrom circular turntable (mass M, radius R, center O) is at rest in the x y plane and is mounted on a frictionless axle, which lies along the vertical z axis. I throw a lump of putty (mass m) with speed v toward the edge of the turntable, so it approaches along a line that passes within a distance b of O. When the putty hits the turntable, it sticks to the edge, and the two rotate together with angular velocity w. Find w.

    Is it because the center of the mass of the turntable is at O? (But when the putty hits it, it will no longer be at O.)
  2. jcsd
  3. Oct 10, 2013 #2


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    It is because the chosen reference point for the angular momentum calculation is at the axle. That reference point does not change position just because a lump of putty hits a turntable.

    The turntable may be out of balance after the lump of putty attaches itself. And the axle may no longer coincide with the center of mass of the turntable+lump assembly. But that's OK. Unbalanced shapes can rotate too.
  4. Oct 10, 2013 #3
    Friction is a force directed against motion. For rotation in the XY plane, friction acts in the XY plane. Hence, it produces torque in the Z direction. If friction is absent in the axle, then it produces zero torque in the Z direction.
  5. Oct 10, 2013 #4


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    yeah, it says the table is mounted on a frictionless axle along z axis. This means there is no z-component of torque on the table due to this contact. There will be torque on the table due to the putty.
  6. Oct 10, 2013 #5
    So there will be torque when the putty hits the turntable?

    If that is the case, then how do we still know angular momentum is conserved? Or is it that—there is torque only when the putty hits the table? The moment the putty actually hits, the torque is gone and we can assume that, from that state on, angular momentum is conserved.

    ... right?
  7. Oct 10, 2013 #6
    The angular momentum of the system is conserved. The system includes the putty, before and after the impact.
  8. Oct 10, 2013 #7

    D H

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    Not if the axle is frictionless.

    Forget about rotational motion for the moment. Consider what happens when an object is tossed onto a horizontal frictionless surface. The normal force will stop the object's vertical motion when the object hits the surface. The horizontal component of the object's will remain unchanged as the surface is frictionless. Think of contact friction as a force that impedes motion parallel to the surface. That a surface impedes motion normal to the surface does not mean the surface is not frictionless.

    Now let's look at your turntable with a blob of putty tossed onto it. That the turntable axle is frictionless means that the force at any point on the surface of the axle is purely radial. It does not mean that there is no force. This now unbalanced turntable can still rotate frictionlessly about the turntable axle. All that's needed to keep the turntable rotating is a radial force, and our frictionless axle can exert a radial force.
  9. Oct 10, 2013 #8


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    If you consider the table and the putty as two separate objects, then yes there will be a torque acting on the table due to the putty. And it is true that this torque will act only during the collision. And as voko said, the total angular momentum (in the z direction) is conserved. So there is equal and opposite torque on the putty when the collision occurs. By this, I mean that the force on the putty is equal and opposite to the force on the table, and since they are contact forces, they both occur in the same place, therefore the torque on the putty is equal and opposite to the torque on the table.

    Now, instead, you can think of the system as initially being made of the two separate objects of table and putty, then in the final configuration, being made of one combined object which is the table and putty stuck together. This is essentially the same as what I said in the last paragraph, but maybe more intuitive. The important thing is that the total angular momentum in the z direction is the same before as it was afterwards.

    The reason the total angular momentum in the x and y directions is not conserved is because the axle can exert a torque on the table in these directions. And by definition, the axle cannot exert a torque in the z direction. This is what D H means by 'there will not be a torque' I think. (i.e. there will not be a torque in the z direction due to the axle).
  10. Oct 10, 2013 #9
    Ah! Ok. I was mixing up "axle" and "axis" (I know, a stupid mistake; an axis can't have... well... physical properties like those that create friction). I think I understand now.

    So, say the axle weren't frictionless. If I were to grab one end and twist, the axle would exert a torque on the turntable and cause it to spin—but since the axle is frictionless, no matter how fast that axle spins, it has absolutely no effect on the table or the table/putty combination.

    If one considers the whole system as the turntable and the putty, then there can be no possible external torque; while the putty causes some torque on the turntable, it was part of the internal system in the first place, so angular momentum is still conserved.

    I know I'm basically rewording what you guys said, but more so because it helps me internalize it.

    Thanks for all of your help. :)
  11. Oct 11, 2013 #10


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    All the book is saying that in the absence of external torques, angular momentum is conserved. If the axle is frictionless, there is no external torque from friction, and thus angular momentum is conserved. If there was friction in the axle, there would be an external torque from friction, and thus angular momentum would not not be conserved, and thus the problem could not be solved using conservation of momentum equations.
  12. Oct 11, 2013 #11


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    Just to throw a little complication in...

    If the physicle axle and the analyst's chosen reference axis do not coincide then the fact that the axle is frictionless is not enough. A side-to-side force provided by the axle could have a non-zero offset from the axis and thereby produce a non-zero torque.

    The analyst chooses to make the reference axis coincide with the axle so that any such torque is zero, thereby simplifying the problem.
  13. Oct 11, 2013 #12
    What do you mean a "side to side" force?

    If the axle were actually moving side to side (in which case, wouldn't the axis not coincide with the axle anyway?) or if the axle itself were tilted w/ respect to gravity so there was some normal force or something?
  14. Oct 11, 2013 #13


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    A force in the x-y plane. The axle is aligned with the z axis.

    The axle need not move from side to side in order to exert a side to side force. By Newton's third law, if you push the turntable into the axle, the axle will exert an equal and opposite force on the turntable. In the problem at hand, there will be a horizontal force exerted by the axle on the turntable, even though the axle remains motionless.
  15. Oct 11, 2013 #14


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    yeah, the main point is that we can choose our axis so that it points in the z-direction, but does not go through the axle. In this case, the axle will exert forces that can be seen as a torque when using this choice for our axis.

    In other words, we have a frictionless axle, so it only exerts a force radially away (or towards) itself (or in the z direction). So if we choose our axis to go through the axle, the axle produces no torque. But when our axis does not go through the axle, the axle can produce a torque, since 'radially away from the axle' is no longer 'radially away from our axis'.

    So, torques will be different depending on where you choose your axis to be. As long as our axis points in the z-direction, we still have conservation of angular momentum. So even though the axis will exert a torque on the table, the table will exert a torque back on the axle. But it is important to remember that the angular momentum will depend on where you choose the axis to be.
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