0
$\begingroup$

enter image description here

I understand that there is the radial and tangential force components acting on a spinning object, but according to the attached image from my book, there is a third component F_1z that is perpendicular to the other vector components and is along the axis of rotation.

I would understand if this was the vector for torque or angular acceleration, but it still seems to be talking about the components of linear force on the object. Does this force also signify the direction of the object? I've never heard of a third vector component when it comes to describing the linear forces of a rotating object.

$\endgroup$
1
  • $\begingroup$ If the center of mass is accelerating along the axis of rotation, it would be a result of an axial force. $\endgroup$ Commented Dec 25, 2023 at 13:14

2 Answers 2

3
$\begingroup$

The idea is that, any vector in space (e.g. Force) has 3 degrees of freedom to it, and you can choose an arbitrary coordinate to break it down. If the object is rotating around an axis, breaking down the forces the way in the diagram simplifies the problem a lot. It's similar to other physics problem, when choosing a reference frame, there are many different choices, but some specific choices is the easiest to deal with.

If you know linear algebra, the above means that force lives in a 3-dimensional vector space. And breaking it down to individual component means choosing a basis.

$\endgroup$
0
$\begingroup$

I believe you are thinking of motion in two dimensions, and so implicitly assuming that the axial force $F_{1z}$ is zero. However, if $F_{1z}$ is not zero then the rigid body will move up (or down) parallel to the $z$ axis while still rotating about the $z$ axis. In this case, the path of the particle $m_1$ will be a helix rather than a circle.

$\endgroup$
2
  • $\begingroup$ In the diagram, a non-zero Fz is incompatible with the depicted motion in the x-y plane. $\endgroup$
    – john
    Commented Dec 24, 2023 at 13:26
  • $\begingroup$ @john Yes, it is. The diagram is incorrect. $\endgroup$
    – gandalf61
    Commented Dec 24, 2023 at 14:33

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