0
$\begingroup$

I realize there is another question on the centripetal force involved in a gyroscope's motion, but I found the answer to not be very complete.

With respect to a gyroscope attached to a vertical spring - I understand that precession of a spinning gyroscope is caused due to a torque (due to weight of the wheel and tension in the string which act as a couple) which changes the direction of angular momentum of the wheel about its central axis. However, in a question I solved, the string was said to be at an angle to the vertical, such that the horizontal component of tension provided the centripetal force necessary for the circular motion. I am a bit confused about this - centripetal force in the case of a ball attached to a string rotating in a horizontal circle acts so as to maintain the circular motion and ensure the ball does not fly off in a straight line path with its linear velocity at that instant.

However, in the gyroscope case, doesn't the torque constantly keep changing the axis of rotation such that the gyroscope moves in a horizontal circle? It seems like the torque is what keeps the wheel moving in a horizontal circle by shifting the spinning axis in a horizontal plane, so why is there an extra centripetal force needed to be provided by a component of tension?

$\endgroup$
3
  • $\begingroup$ can you please write the equations that you solved $\endgroup$
    – Eli
    Jan 14, 2021 at 14:38
  • $\begingroup$ Which other question? $\endgroup$
    – Qmechanic
    Jan 14, 2021 at 18:31
  • $\begingroup$ @Qmechanic this one - physics.stackexchange.com/q/238126 $\endgroup$
    – ani
    Jan 18, 2021 at 19:39

1 Answer 1

1
$\begingroup$

I'm not sure I understand your question, but let me present some remarks, hopefully they will be helpful.

Let me first discuss the simplest case, a gimbal mounted gyroscope wheel.

gimbal mounted gyroscope

(This image is from an 2012 answer by me explaining the mechanism of gyroscopic precession

The gimbal mounting supports the center of mass of the gyroscope wheel.

If a weight is added, such that it tends to pitch the gyroscope wheel, then the resulting precessing motion is around the center of mass of the gyroscope wheel.

With the center of mass of the gyroscope wheel remaining at the same place there is of course no need for a centripetal force.


However, often the setup is without a gimbal mounting. Most commonly a bicyle wheel is used, with a vertical string attached to one end of the axle.

Suppose you want that point of suspension to remain in the same place. In that case the precessing motion has the center of mass of the wheel going round in a circle. That circular motion requires a centripetal force to sustain it.

So:
A centripetal force, if it needs to be provided, is not involved in the mechanism of gyroscopic precesssion.

$\endgroup$
3
  • $\begingroup$ The case of a bicycle wheel mounted onto a string with the point of suspension staying in the same place - If the precessing motion here causes the center of mass to go around in a circle, I don't understand why a centripetal force is needed to sustain this motion. Why would a centripetal force be required to sustain circular motion that is being continuously caused/driven by something? $\endgroup$
    – ani
    Jan 18, 2021 at 19:53
  • $\begingroup$ @ani This is independent of whether the circumnavigating motion is occuring in the course of precessing motion, or circumnavigating motion of a non-spinning object. Either way: if no centripetal force is provided then a moving object will move in a straigh line. Motion along a circle requires a centripetal force. Precessing motion is caused, but not continuously caused. Precessing motion is caused and subsequently sustained. See the 2012 answer that I had already linked to: the mechanism of gyroscopic precession $\endgroup$
    – Cleonis
    Jan 18, 2021 at 20:11
  • $\begingroup$ In case of a bicycle wheel with a vertical string attached to one end of the axle: is it the centripetal force associated with the precession that makes a nutating wheel move back up above the equilibrium angle? $\endgroup$
    – anoniem
    Jan 5 at 10:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.