# DETERMINDING the axis of rotation

say i have a lever, fixed to an axis in its right end.
now a perpendicular force is exert on the lever, from its left end.
the lever will obviously rotate relatively to the fixed axis.
(the force will rotate with it, and will always be perpendicular to the lever. so the lever has a fixed rotational acceleration)

we know the axis of rotation will be the right end, because we have experience in life.

but if we DIDN'T know that, and only know what forces are exerted on the lever, how could we know where will be the axis of rotation?

another example is,
a body is on a frictionless floor. two forces are exerted on it, where will be the axis of rotation?

picture:
http://img573.imageshack.us/i/pamd.jpg/

timthereaper
You'd have to know where the forces are applied on the beam and you could use dynamics to find where the center of rotation of the beam is.

If you wanted to know where the beam is relative to the ground, that's a different matter. You'd have to either pick a point of interest on the ground and solve it by relative position or connect it to a point on the ground and you'll find that the instant center of the mechanism is at the point where the two points have the same tangential velocity.

yes yes i know

if you know where the forces are applied, what direction they have, and everything you need..

how can you determined the center of rotation?

timthereaper
It really depends on whether you're considering the moment of inertia of the beam. If you're not, then you can use this:

If your forces are located at specific locations on the beam (e.g. if the left side is x=0, then F1 would be located at like x1=L/6 and F2 would be located x2=L/3), then you find the place where the two torques are equal (F1(x1-x)+F2(x2-x) = 0, where x is the position of the center of rotation). Solve for the position and that would be the center of rotation.

Of course, I'm making certain assumptions, like the forces are in opposing directions and no moment of inertia.

It really depends on whether you're considering the moment of inertia of the beam. If you're not, then you can use this:

If your forces are located at specific locations on the beam (e.g. if the left side is x=0, then F1 would be located at like x1=L/6 and F2 would be located x2=L/3), then you find the place where the two torques are equal (F1(x1-x)+F2(x2-x) = 0, where x is the position of the center of rotation). Solve for the position and that would be the center of rotation.

Of course, I'm making certain assumptions, like the forces are in opposing directions and no moment of inertia.

okay thanks a lot..

a few questions though,
i don't see how this calculation is true... the body SHOULD have torque which is not necessary zero....
and i don't understand what's the meaning of ignoring the moment of inertia....
anyway..
what if you are considering the moment of inertia

someone?

markem
Bump, I've got essentially the same question.. Given a body and all the forces acting upon it, apart from resorting to physical intuition, how are we to locate its axis of rotation?

Delta Kilo
Given a body and all the forces acting upon it, apart from resorting to physical intuition, how are we to locate its axis of rotation?

The motion of rigid body can be split into translation of the center of mass (CM) and rotation around the CM and these can be computed independently. You integrate the sum of all forces and get the CM position as a function of time, and you integrate the sum of all torques relative to the CM and get the orientation as a function of time (easy in 2D case, tricky in 3D). See http://en.wikipedia.org/wiki/Rigid_body_dynamics

Points on the axis of rotation in some coordinate system are those instantaneously at rest (by definition). At those points the motion due to rotation around CM is precisely compensated by the translation motion of the CM. Basically you solve $\vec{w} \times (\vec{r}-\vec{r_0})+\vec{v_0}=0$ for $\vec{r}$, where $\vec{r_0}(t), \vec{v_0}(t), \vec{w}(t)$ are coordinates and velocity of the CM and angular velocity around CM.