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When a top is spinning in the counterclockwise direction as seen from above.Suppose its axis of rotation making an angle of 15º with the vertical.I am trying to find the angular momentum of this top and its direction neglecting any friction on it.

The forces acting on the top are its weight and the normal reaction at its tip.Apart from that due to the rotational motion the top has an angular momentum along its spinning axis in the upward direction.As the top is tilted the the weight does work on the top until the top gets slanted to the vertical as much as it could get.This work done by the weight of the top increases its angular momentum.

That is what I think about the angular momentum. But I have no idea to figure out the precession of the top.

Would it be clockwise or anti-clockwise when seen from above?

AfiJaabb
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2 Answers2

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enter image description here (image source: Wikipedia)

There is a net torque (The green vector) about the point of contact due to the gravitational force, which will make the top (and hence its angular momentum vector) precess in the counterclockwise direction as seen from above.

The precession occures due to the following reason: The torque is perpendicular to the axial angular momentum $\vec{L}_s$ (corresponding to $\omega_s$), so it cannot change the magnitude of the $\omega_s$ (hence, that of $\vec{L}_s$). Rather, it will keep changing its direction (just as the centripetal force keeps changing the direction of velocity in uniform circular motion), and the the direction of $\vec{L}_s$ will keep rotating about the vertical axis, i.e. to top will undergo precession.

The angular momentum due to axial rotation does not change in magnitude (as the torque is perpendicular to it), but as precision begins, there will be an additional precessional angular momentum.

Archisman Panigrahi
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There is an ambiguity here that needs to be addressed.

The simplest case of angular velocity of a top is the straightforward spinning along the axis of symmetry. That angular velocity has a corresponding angular momentum.

When in addition to that spinning motion there is precessing motion there is a total angular momentum of the top.

For the purpose of calculation is it common to decompose that total angular momentum into:

  • angular momentum of spinning motion
  • angular momentum of precessing motion

(In fact, when the spinning motion is very fast it is common to treat the angular momentum of the precessing motion as negligable.)

Since angular momentum is represented as a vector the decomposition into components is performed as a vector decomposition

The implications of the above:
When statements are made about 'the angular momentum of the top' it should be made clear whether that is about the total angular momentum or the spinning angular momentum.

The case presented in the part of the text in bold (in the question) is one of the precessing motion already in motion, with unchanging tilt.

However, in your subsequent text, offering your own thoughts, you seem to be mulling the case of a spinning top at the very instant it is released at an initial tilt angle, resulting in a change of the tilt angle.

I'll get to the intended case further down, let me first talk a bit about the case of a spinning top right at the instant that it is released. As you point out: at the instant of release (at an initial tilt angle) the center of mass of the top will go down a little. As you point out: that is work done.

(Please note that while work done always changes the kinetic energy, it doesn't necessarily change angular momentum. For example: two celestial bodies, orbiting each other, in eccentric orbit. During the phase in the orbital motion where the the distance between the two bodies decreases their kinetic energy increases, but the angular momentum of the system does not increase. As we know: the angular momentum of a rotating system is conserved.)

At the very instant that you release a spinning top (at an initial tilt angle) I think the subsequent change of the magnitude of the spinning angular momentum will be negligable.

(The kinetic energy from the work done goes into precession and nutation, but that's another story.)

Anyway, the text in bold is about the case of steady precessing motion. That is, steady precession motion such that the magnitude of the tilt angle remains the same.

During steady precessing motion the magnitude of the total angular momentum remains the same, and the magnitude of the spinning angular momentum remains the same. (Both are changing direction of course.)

How to understand the direction of the precessing motion (clockwise/counterclockwise) is discussed by me in a 2012 answer here on physics.stackexchange, to a question titled: What determines the direction of precession of a gyroscope?

Cleonis
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