What does weight actually do in gyroscopic precession?

[phantomvommand]

I 'get' how the gyroscope works, but I only understand it mathematically. I understand that because Torque = r x F, so the direction of the torque is such that it can change the direction of the Angular momentum, resulting in the gyroscope precessing.

However, I just cannot visualise how this works physically. I can see some significant similarity between this and circular motion, but the case for circular motion is very direct: velocity is being 'pushed' around by the force.

However, for the gyroscope, I cannot get over the fact that the downward force (weight) will act to pull the gyroscope downwards. Can someone highlight the link between the downward force and the shift in planar motion of the gyroscope, in a physical sense, instead of using torque = r x F?

Thanks for all the help

 

[ergospherical]

It's hard to understand gyroscopes without invoking moments! Take the "classic" example of a spinning wheel attached to the end of an axle, for instance. It's acted upon by its weight as well as a contact force acting on the system at the point where the axle is pivoted. When added vectorially to the weight one obtains a resultant centripetal force which causes the centre of mass of the gyro to undergo circular motion.

[In other words, vertical component of contact force balances weight, horizontal component of contact force provides centripetal acceleration].

 

[phantomvommand]

It's hard to understand gyroscopes without invoking moments! Take the "classic" example of a spinning wheel attached to the end of an axle, for instance. It's acted upon by its weight as well as a contact force acting on the system at the point where the axle is pivoted. When added vectorially to the weight one obtains a resultant centripetal force which causes the centre of mass of the gyro to undergo circular motion.

[In other words, vertical component of contact force balances weight, horizontal component of contact force provides centripetal acceleration].

Yes I realize torque is crucial to this. What I can't seem to get over is the fact that there would be a downward moment due to weight, which by definition of torque, is translated to a change in the direction of the angular momentum vector. But unless I use this mathematical definition of torque, my intuition always tells me the gyroscope would fall due to the downward moment!

Is there some more physical way of seeing how the downward moment due to weight 'translates' into precession without dragging the gyroscope down?

 

[ergospherical]

The crucial part is that the disc is spinning about its axis. If it weren't, then when released (with zero initial angular momentum) the angular momentum of the system would grow in the (constant) direction of the moment of the weight force until the system reached the lowest point of its swing.

When the disc is initially spinning, the initial angular momentum is along the axis of the gyroscope $\boldsymbol{L} = I \boldsymbol{\omega}$. When the moment of the weight force (which is perpendicular to the angular momentum) acts for a time $\delta t$ it now causes $\boldsymbol{L}$ to rotate about the vertical axis, instead of change its magnitude. And the contact forces at the axle ensure that the centre of mass doesn't accelerate vertically.

(Mathematically, if $\boldsymbol{n}$ is a unit vector along the axis of the gyroscope, and $\boldsymbol{k}$ is a vertical unit vector, then the equation of motion is $I\omega \dot{\boldsymbol{n}} = -mgR \boldsymbol{n} \times \boldsymbol{k}$. This is of the form $\dot{\boldsymbol{n}} = \boldsymbol{\Omega} \times \boldsymbol{n}$, i.e. rotation of $\boldsymbol{n}$ about the vertical axis at a rate $| \boldsymbol{\Omega} |$).

 

[italicus]

Yes I realize torque is crucial to this. What I can't seem to get over is the fact that there would be a downward moment due to weight, which by definition of torque, is translated to a change in the direction of the angular momentum vector. But unless I use this mathematical definition of torque, my intuition always tells me the gyroscope would fall due to the downward moment!

Is there some more physical way of seeing how the downward moment due to weight 'translates' into precession without dragging the gyroscope down?

Pay attention. The moment of the weight (force) wrt. the point of suspension of the wheel is not downward, it is horizontal . Without considering angular momentum vector $\vec L$ and the moment of the force as a vector, it is rather difficult to answer your question. The intrinsic reason is that angular velocity $\vec\Omega$ is essentially a vector quantity, not only in this issue but always. You cannot think of an angular velocity without thinking of rotation. But one has to be careful : in general , a vector$\vec\omega$ of a free rigid body doesn’t define the position of the rotation axis. Only when a rigid body has a fixed point , or else a fixed axis of rotation , you can speak of "rotation of the rigid body wrt. the fixed point , or wrt, the fixed axis. It is a very delicate part of the rigid body mechanics.

But now, coming back to the spinning wheel , suspended by a vertical rope at the end of its rotation axis, with rope not passing through the center of gravity , it happens that the vector $\vec L$ tries to “reach" the angular spin vector , that is, tries to become parallel to it. The following lesson by Walter Lewin is a masterpiece on the subject. He begins to speak of the gyroscope effect at about min 15:00

 

[ergospherical]

it happens that the vector $\vec L$ tries to “reach" the angular spin vector

Not sure what you mean, because in the gyroscopic approximation the relationship $\boldsymbol{L} = I \boldsymbol{\omega}$ holds, i.e. $\boldsymbol{L}$ is always parallel to $\boldsymbol{\omega}$, the spin angular velocity.

 

[italicus]

Not sure what you mean, because in the gyroscopic approximation the relationship $\boldsymbol{L} = I \boldsymbol{\omega}$ holds, i.e. $\boldsymbol{L}$ is always parallel to $\boldsymbol{\omega}$, the spin angular velocity.

Sorry, a banal mistake of mine. The angular momentum $\boldsymbol{L}$ tries to “reach “ the moment of the weight , rotating wrt. the point of the vertical axis where the wheel is attached, in the horizontal plane. Of course, without ever reaching it.
This is well explained by W. Lewin.
Thank you for having noticed my mistake.

 

[A.T.]