# 2nd Law of Newton in translation and rotation combined. Is this really a necessary condition?

Serway's 10ed says that Newton's second law of rotational ($$\sum \tau_{z} = I_{z}\alpha_z$$) is true when there is combined translation and rotation as long as the moving axis (1) passes through the center of mass and (2) is an axis of symmetry. Sears-Zemanski (edition 12) also adds that (3) the axis must not change direction.

I don't know how to prove it, but I do not find any of these conditions strictly necessary. In this post it has been mentioned that at a given moment, the movement of a rigid body can be described by a translation of any chosen point, plus a rotation about that point, so the axis of rotation is an arbitrary matter. Condition 3 doesn't seem necessary to me either ... only that a time-varying direction axis would make the counts much more complex, because the equations would change at every instant.

My hypothesis is that these conditions are more a description of the type of exercises the books are limited to rather than a necessity for applying the equations. I am right?

• Where is $\tau_z$ summed up at? May 28, 2020 at 18:06
• Also reference here for combined motion about arbitraty point. physics.stackexchange.com/questions/80433/… May 28, 2020 at 18:43
• z is an arbitrary axis May 28, 2020 at 19:34

Euler's law of rotational motion is expressed in vector form at the center of mass as

$$\sum \vec{\tau}_{\rm C} = \mathbf{I}_{\rm C} \dot{\vec{\omega}} + \vec{\omega} \times \mathbf{I}_{\rm C} \vec{\omega} \tag{1}$$

It is derived from treating the rigid body as a collection of particles, each with velocity $$\vec{v}_i = \vec{\omega} \times \vec{r}_i$$ which is a rotation about the center of mass. Then the force acting on each particle is $$\vec{F}_i = \frac{\rm d}{{\rm d}t} (m_i v_i)$$ and the torque about the center of mass $$\vec{\tau}_i = \vec{r}_i \times \vec{F}$$.

The book reference states the conditions where $$\vec{\omega} \times \mathbf{I}_{\rm C} \vec{\omega} = 0$$ in general, or the component along the $$\hat{z}$$ axis is zero $$\hat{z} \cdot (\vec{\omega} \times \mathbf{I}_{\rm C} \vec{\omega}) = 0$$

• The above law is valid regardless of where the axis of rotation is, but all quantities must be measured about the center of mass. The motion of the center of mass is described by $$\sum \vec{F} = m \dot{\vec{v}_{\rm C}}$$, and about the center of mass by (1). See this post for how these equations change when not measured about the center of mass.

• For the second term (called the gyroscopic term) to be zero consider the direction of rotation $$\hat{z}$$ such that $$\vec{\omega} = \omega\, \hat{z}$$. If the axis of rotation _is one of the 3 principal axes of rotations then $$\mathbf{I}_{\rm C} \hat{z} = I_z \hat{z}$$ where $$I_z$$ is the scalar mass moment of inertia and $$\hat{z} \times I_z \hat{z} =0$$. This also happens if $$\hat{z}$$ is an axis of symmetry.

• If the axis does not change direction the gyroscopic term does not vanish in general, causing a dynamic imbalance in the system. This is the reason balance weights need to be added to the wheel after a new tire is installed. The gyroscopic terms would cause a terrible wobble at speed as out of plane alternating torques were applied. The gyroscopic terms along the axis of rotation are zero and thus only along the axis of rotation $$\tau_z = I_z \dot{\omega}_z$$ is valid.