1

This seems rather basic to prove mathematically but i am second guess my intuition. Say you have an object in rotation and a force acts on it through the axis of rotation perpendicular to the axis of rotation. would the effect of the perpendicular force on the rotation would be zero?

for example, a Frisbee spins through its center along say the y axis. would a force from above or below at the center of rotation have any effect on the rotation?

or say an american football spins at its point would a force on the opposite point effect the rotational velocity at all?

mitchell graham
  • 11

1 Answers1

2

Rotational equations of motion for an applied force $\vec{F}$ located at $\vec{r}$ relative to the center of mass _C are:

$$ \vec{r} \times \vec{F} = {\rm I}_C \dot{\vec{\omega}} + \vec{\omega} \times {\rm I}_C \vec{\omega} $$

where ${\rm I}_C$ is the 3×3 rotational inertia matrix. So lets look at the frisbee example:

  • $\vec{\omega} = (0,\Omega,0)$ rotation about the y-axis
  • $\vec{r} = (0,y,0)$ along the axis of rotation
  • $\vec{F} = (F,0,0)$ perpendicular to the axis of rotation
  • ${\rm I}_C = \begin{bmatrix} \frac{m}{4} R^2 & 0 & 0 \\ 0 & \frac{m}{2} R^2 & 0 \\ 0 & 0 & \frac{m}{4} R^2 \end{bmatrix}$
  • $\begin{pmatrix} 0 \\ 0 \\ -F y \end{pmatrix} =\begin{bmatrix} \frac{m}{4} R^2 & 0 & 0 \\ 0 & \frac{m}{2} R^2 & 0 \\ 0 & 0 & \frac{m}{4} R^2 \end{bmatrix} \begin{pmatrix} \dot{\Omega}_x \\ \dot{\Omega}_y \\ \dot{\Omega}_z \end{pmatrix} +\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $ equations of motion
  • $\dot{\Omega}_z = -\frac{4 F y}{m R^2} $ and all other angular accelerations are zero

So the components of rotation are going to change over time, starting from the z-axis.

Note that any perturbation is not necessarily stable. See this post: https://physics.stackexchange.com/a/17507/392

Also if the force is through the rotation axis ($y=0$) then there is no rotation. $\dot{\Omega}_z =0$

Community
  • 1
John Alexiou
  • 40,139