Precession of a spinning projectile, other dynamics problems

[aeroegnr]

Ok, I'm using Beer and Johnston's engineering Mechanics/Dynamics book for class, and i don't particularly enjoy it's lack of explanation for the following problem:


The graphic shows a bullet flying to the right with the x coordinate pointing along its long axis and the y-axis perpendicular. The velocity is theta degrees below the x axis

"The 60-lb projectile shown has a radius of gyration of 2.4in. about its axis of symmetry Gx and radius of gyration of 10in. about the transverse axis Gy. Its angular velocity $$\omega$$ can be resolved into two components: One component, directed along Gx, measures the rate of spin of the projectile, while the other component, directed along GD, measures its rate of precession. Knowing that $$\theta$$=5 degrees and the angular momentum of the projectile about its mass center G is Hg= (0.640 lb-ft-s)i - (0.018 lb-ft-s)j, determine:
a) the rate of spin
b) the rate of precession"


Now, the book provides no explanation of precession up to this point, and its absolutely driving me insane not knowing why the answer is what it is.

The answer in the back of the book for b is found by taking the j component of velocity $$\omega_\hat{j}$$ and then using this equation where $$\omega_\hat{p}$$ is the angular velocity in the direction of the moment of precession.

$$\omega_\hat{p}$$$$*sin(\theta)=\omega_\hat{j}$$

The answer given in the back of the book for this therefore is .1596 rad/s.

I do not understand why this is so. The angular velocity in the y direction is
the farthest from the axis of precession, so why does it provide the only source of precession? Why does this precess and not just spin laterally and in horizontally at the same time?

If anyone can provide insight, I would be very thankful.

I also have more questions if you can answer this one.

 

[Clausius2]

I don't know the answer, but perhaps you are very helpful at this discussion: https://www.physicsforums.com/showthread.php?t=17721

 

[ravenprp]

First of all, let's define what precession is. Precession is the motion of a spinning object such as a top or a spinning projectile about an axis that is perpendicular to the axis of rotation. This can be seen in the case of a spinning top, where the axis of rotation is perpendicular to the surface it is spinning on. As the top loses energy, it starts to wobble and precess around this perpendicular axis.

Now, in the case of a spinning projectile, precession occurs due to the torque caused by the projectile's weight acting on its center of mass. This torque causes the projectile to precess around an axis that is perpendicular to the plane of rotation.

To understand why the j component of angular velocity is used in the equation for precession, we need to look at the moment of inertia of the projectile. The moment of inertia is a measure of an object's resistance to changes in its rotational motion. In simple terms, it tells us how hard it is to make an object rotate.

In this problem, the projectile has a radius of gyration of 2.4 inches about its axis of symmetry Gx and 10 inches about the transverse axis Gy. This means that it is easier to make the projectile rotate around the Gx axis compared to the Gy axis. This is why the j component of angular velocity, which is in the direction of the Gy axis, is used in the equation for precession.

Now, let's look at the equation itself. The equation you mentioned is derived from the equation for torque, which is given by:

\tau = I\alpha

Where \tau is the torque, I is the moment of inertia, and \alpha is the angular acceleration. In the case of precession, we can rewrite this equation as:

\tau = I_{p}\omega_{p}

Where I_{p} is the moment of inertia about the axis of precession and \omega_{p} is the angular velocity in the direction of the moment of precession. Since we know that the projectile is precessing around the Gy axis, we can use the j component of angular velocity, which is in the direction of the Gy axis, to calculate the moment of inertia about that axis. This moment of inertia is given by:

I_{p} = mr^{2}

Where m is the mass of the projectile and r is the distance from the Gy axis to the center of mass. This distance is given by the