# Forces on a Rotating Rigid Body

This question has been puzzling me for some days now and I thought I might ask you guys for some help.

Say that we have a rigid body (like a rod) that is in a vacuum in space. Furthermore, say that we apply some perpendicular force $$F_{app}$$ at some point $$d$$ to the right of the center of mass. We know that the rigid body must accelerate with the same angular acceleration $$\alpha$$ and that the body will rotate around the center of mass. Therefore the sum of all the forces in the perpendicular direction are $$\sum_{i=0}^NF_{i,\theta} = \sum_{i=0}^Nm_ia_i = \alpha\sum_{i=0}^Nm_ir_i$$

where $$r_i$$ is the measured distance from the center of mass to each particle and $$m_i$$ is the mass of that infinitesimal particle.

We also know that sum of forces to a system of particles can be decomposed into internal and external forces such that:

$$\sum_{i=0}^NF_{i,\theta} = \sum_{i=0}^NF_{ext} + \sum_{i=0}^NF_{int}$$

Since the sum of internal forces cancel each other out (Newton's 3rd Law), this is just:

$$\sum_{i=0}^NF_{i,\theta} = \sum_{i=0}^NF_{ext} + 0 = F_{app}$$

This means that:

$$F_{app} = \alpha\sum_{i=0}^Nm_ir_i$$

which implies that the angular acceleration is independent of the point of application $$d$$, which we know to be false by the torque-angular acceleration equation $$\tau = I\alpha$$

Can you guys help me figure out which step of my reasoning is flawed?

• You forgot the fact that the force also creates a translational acceleration of the whole body. So your first equation is missing the translational acceleration terms, and everything after that point is wrong. – alephzero Feb 22 at 0:18