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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
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    As another intuitive example, how much force is required to open a door if you push/pull on the hinge? The answer is an infinite amount of force. No amount of force—regardless of its magnitude—can engender the rotation of an object if it has no moment arm. – Bryson S. Nov 13 '14 at 19:08
  • Actually the answer depends if the object is symmetric or not. But the OP needs to define what the mean by "effect on rotation". Is it magnitude only, or magnitude and direction? – John Alexiou Nov 13 '14 at 19:52

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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. $\dot{\Omega}_z =0$

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John Alexiou
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  • Worth making explicit that if the force is through the axis of rotation, in the case of the (symmetrical) frisbee $y=0$ and there will be no change in the axis of rotation. Also, did you lose a factor $I_C$ in your "equations of motion" line (bullet 5)? – Floris Sep 29 '15 at 12:11
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    Thanks. See corrected now. If it is through then $y=0$ then, yes there isn't any rotation. – John Alexiou Sep 29 '15 at 14:12
  • Edit that last point into your answer for completeness. Either way you got the upvote... – Floris Sep 29 '15 at 14:14