# Net force when deriving relation between torque and angular acceleration

When deriving the equation $\tau = I\alpha$, it is assumed that $$F_{tangential}=ma_{tangential}$$$$a_{tangential}=\alpha r$$$$\tau=rF_{tangential}$$$$\tau=mr^2\alpha=I\alpha$$

However, doesn’t Newton’s second law describe that $\Sigma F=ma$? And isn’t $\Sigma F$ in this case equal to $F_{centripetal}+F_{tangential}$? Why don’t you have to take into account of centripetal force?

Or you can just arbitrarily throw around $F=ma$ for any force without minding that it’s not the net force?

You are confusing a 2D problem with a 1D problem. In 2D, force and motion also have direction (other than 1 or -1). The two forces you are trying to sum are orthogonal to each other and should not appear in each other's force equilibrium.

To put it in coordinates, let's choose a $(r,\theta)$ coordinate system. In that system $$\vec{F}=F_r\hat{r}+F_{\theta}\hat{\theta}.$$

To calculate the torque about the origin that this force exerts on a particle located at position $\vec{r}$ we calculate a cross product $$\vec{\tau}=\vec{r}\times\vec{F}.$$

In case you don't know about cross products, one of the results is that any component of $\vec{F}$ that is parallel to $\vec{r}$ vanishes, and because $\vec{r}$ points in the $\hat{r}$ direction, the $F_r$ component doesn't contribute to the torque.$$\vec{\tau}=rF_{\theta}\hat{k}$$

For a more intuitive idea, consider this. If you have a particle at rest some distance from the origin and you exert a force on it directly toward the origin, there is no angular acceleration around the origin. By definition, angular accelerations must be tied to a torque, and vice versa, so zero angular acceleration implies zero torque. Radial forces don't cause torques about the central point.

$$1^{st}$$ method

The torque is only generated by the perpendicular component of force acting along the radius vector (as can be seen from this formula $$\mathbf \tau = \mathbf r \times \mathbf F$$)

$$2^{nd}$$ method

You may know that $$v \neq \omega r$$ but $$\mathbf v = \mathbf \omega \times \mathbf r$$. Therefore $$\mathbf a = \mathbf \alpha \times \mathbf r$$. Now you might notice that since the component of $$\mathbf a$$ which is parallel to $$\mathbf r$$ can never be generated via a cross product therefore the component of acceleration parallel to the radius isn't accounted for.