# Is a reaction moment/force created in off-center constrained rotation?

1. Apr 15, 2014

### diffusegrey

Assume that the arbitrary-shape object shown has a torque applied at the black X with a circle, and is free to rotate about the X axis. Assume that the object is also constrained by a pin at this torque input point, parallel to the X-axis. The CG of the object is the blue plus sign, and there is an otherwise uniform mass/density distribution. Ignore gravity.

When torque is applied, is there a reaction moment/force created, acting on the constraint pin, simply from the inertia of the body being rotated (effective from the CG point, about the x-axis, against rotation)?

Or put another way, is there a constant tendancy of the object to want to naturally rotate about its CG point (instead of its constrained point), that creates a constant moment against the direction of rotation, about the CG, that acts as a force on the constraint pin?

And would such a moment exist only during acceleration (during application of torque and increase of angular velocity), or would it exist at steady state too? Would such a "reaction moment" absorb or reduce any of the kinetic energy of the rotational motion?

2. Apr 15, 2014

### Matterwave

You might be interested in the parallel axis theorem: http://en.wikipedia.org/wiki/Parallel_axis_theorem

The moment of inertial is minimum around an axis that passes through the CoM, so the moment of inertia is higher around your different axis. Therefore, it's "harder" to get the object to rotate around your off-CoM axis than it is to get it to rotate from an axis passing through the CoM. I hope that answers your question?

3. Apr 15, 2014

### diffusegrey

Thanks, but it doesn't quite answer what I'm looking for.

The parallel axis theorem will tell me the higher total moment inertia I might have from rotation centered away from the cg, which of course will require more torque to equivalently accelerate to a given angular velocity.

But I'm looking to find what the force or moment is that is acting on the off-center constraint point itself, on account of the inertia of the eccentric mass, and if such force is constantly present, and also for what effect it might have on the object's motion or total kinetic energy.

4. Apr 15, 2014

### Matterwave

There's no difference in that analysis than if the constraint axis went through the Center of Gravity. In the absence of gravity, the center of mass is not going to provide any torques by itself...why would it? There's always a constraint force, present whenever there's a constraint, and to find that one needs to use Lagrange multipliers, but the analysis is the same for this new axis as an axis through the CoM.

The parallel axis theorem tells you exactly how the situation changes. The situation changes because the moment of inertia changes. I'm not sure what else you expect to change...

5. Apr 15, 2014

### FactChecker

The torque at the pivot point causes motion at the CG. The CG is going in a circle, so it is always accelerating even if the rate of rotation is constant. Assuming that the source of the original torque does not apply any linear force, all the linear force required to cause that acceleration at the CG is being supplied by the pivot point. It's not too hard to calculate the amount of force required.

Here is a real-world example that you may have experienced some time -- Suppose you are drilling a hole at one end of a long board. If the drill suddenly jams, the board will start to swing around and the drill will want to swing with it. That is the force on the pivot point of the drill.

6. Apr 15, 2014

### diffusegrey

"Suppose you are drilling a hole at one end of a long board. If the drill suddenly jams, the board will start to swing around and the drill will want to swing with it. That is the force on the pivot point of the drill. "

That's an interesting example of a similar situation... where here the inertia of the long side of the board pushes back against the drill (pivot point or constraint point) as the drill applies torque to the board, as the combined board-drill system seeks to rotate about the cg of the board.

How would we then find the force applied onto the drill? Is it just the force corresponding to the difference in system inertia via the parallel axis theorem, comparing between the pivot point and the cg point? Like, take the torque needed in the off-center case, minus the at-the-cg case, and find the equivalent moment-arm-force using the distance between those two points?

And if the drill did permanently jam, and held its position, and the board started rotating around this point, is this situation analogous to the board's center of mass point itself applying a constant torque to the system, with the drill resisting this torque if it doesn't move?

7. Apr 15, 2014

### FactChecker

Figure out the rotational rate of the object and calculate the circular motion of the CG. The acceleration of the CG can be calculated and then F=mA gives the linear force that must be causing that acceleration. That force is being applied at the pivot point.