# Deriving $T = F\ r = I\alpha$ for a rigid body

For a single point mass : $\tau=F_{t}r=ma_tr=(m r^2)\alpha = I\alpha$

For multiple point masses bound together : $\sum \tau_i = (m_ir_i^2)\alpha = I\alpha$

But how do we go from that to $I\alpha = F_tr_{app}$ (for the multiple masses case), where $r_{app}$ is the point where force is applied? That is saying that the entire sum of torques of individual masses $\sum \tau_i$ is physically equivalent to (i.e. produces the same angular acceleration of the bound rigid object as) a single torque $F_tr_{app}$ applied to the point $r_{app}$. (Or putting it another way, that applying a single force $F_t r_{app}$ will result in many $F_{t\_i}$ forces on each mass element such that $\sum F_{t\_i}r_{i}$ will be equal to $I\alpha$.)

Every place I've looked for a derivation just jumps from $\sum \tau_i =I\alpha$, to $\tau=I\alpha$, as if they are the same thing.

• Any chance you are going to award this question soon? – ja72 Dec 6 '13 at 15:26

It sounds to me like you're worried about the following:

Suppose I swing a baseball bat. In the process, I put a lot of torque on it, and this makes it rotate. But what if I just look at the last 1cm of the bat? That part is far away from my hands. I didn't touch it. Nonetheless, there are forces on that part. So how can I look at only the forces from my hands and figure out what happens with the entire bat? How do I know I don't have to worry about other forces, like those on the last cm of the bat?

The answer is that if you want to find the net torque on the bat, there are two ways to do it. One is to find the net force on each little part of the bat and integrate. Another is to take only the force exerted by your hands and calculate the torque from that. You'll get the same answer either way.

This is simply due to Newton's third law. The end of the bat rotates due to internal stresses - forces the bat exerts on itself. But the bat can exert no net force on itself. If there is a net force of 10N accelerating the last cm of the bat forward, that last cm exerts a force of 10N backward on the rest of the bat, and these two forces, which exist at the same place, produce opposite torques that cancel.

By the way, any good book on mechanics will indeed discuss this and prove it. See, for example, page VII - 13 of Morin's Introductory Classical Mechanics, which is the book I have on hand.

To arrive at the rotational equations of motion first you must derive the angular momentum equation $\vec{H} = \bar{I} \vec{\omega}$ and then differentiate with time to get $\vec{T} = \bar{I} \vec{\alpha} + \vec{\omega}\times \bar{I} \vec{\omega}$ (see link at bottom for more info).

Starting from Newton's law of linear momentum for a paticle $\vec{L}_i = m_i \vec{v}_i$ and recognizing that the motion is due to a pure rotation (the translation part cancels out later, so I ignore it here) you have $\vec{v}_i = \vec{\omega}\times\vec{r}_i$. Also note that the linear momentum has an axis through the particle center of mass and its equivalent (or equipollent) angular momentum about the origin is $\vec{H}_i = \vec{r}_i \times \vec{L}_i$. So all together we have

$$\vec{H}_i = \vec{r}_i \times m_i (\vec{\omega}\times\vec{r}_i) = \mbox{-}m_i \vec{r}_i \times \vec{r}_i \times \vec{\omega} = \bar{I}_i \vec{\omega}$$

And the parallel axis theorem

$$\bar{I} = \sum \mbox{-} m_i [\vec{r}_i \times] [\vec{r}_i \times]$$

where $[\vec{r}_i \times] = \begin{pmatrix} 0 & -r_z & r_y \\ r_z & 0 & -r_x \\ -r_y & r_x & 0 \end{pmatrix}$ is the cross product matrix operator.

So it is a matter of definition of $\bar{I}$ that gives as the rotational laws of motion

$$\vec{T} = \bar{I} \vec{\alpha} + \vec{\omega} \times \bar{I} \vec{\omega}$$