# Driving shaft - Zero net torque yet linear acceleration exists

Imagine a scenario where we are talking about a driving shaft placed on a truck.

I am doing kinematic analysis on the subject and faced the following paradox:

The input shaft rotating speed is constant ( let's name it $\omega$ ), which implies a constant magnitude linear velocity for a point rotating @ a fixed radius.

Also, this implies zero angular acceleration. This means that the linear acceleration is zero as well.

Now, from my understanding of vector math:

in order for the magnitude of the derivative of a vector to be zero, the derivative of each component has to be zero, i.e

$\ |a| == 0 => \ a_x == \ a_y == \ a_z == 0$

The problem is that with my analysis, the velocity components of v are resembling sin ofrms of functions of time, thus their time derivative is by no means zero.

I use principal rotations and coordinate transformations matrices to move from one coordinate frame to another. The one frame i am looking at, has a fixed axis ( the same one as the World reference ) w.r.t which it is rotating.

is incorrect. You mention that the magnitude of the velocity is constant, and indeed it is, but the direction of the velocity is changing. That means the acceleration is not zero. The acceleration is of course the usual centripetal acceleration i.e. a vector directed radially inwards and with a magnitude of $r\omega$.
• Hello, thanks for your answer. Let's take baby steps here to clarify: 1) Constant $\ omega => \ a_a == 0$ 2) $\ a_a== \ a_l / \ r => \ a_l == 0$ I agree with you that the direction of the velocity changes, and that's why i am wondering how this thing works. Sep 27 '16 at 10:53
• @ParaskevasDimitris What you call $a_l$ is only the component of the linear acceleration vector that is tangent to the circular path. This is indeed zero. However, there is another component to the linear acceleration vector: the radial component that points towards the center of the circular path. The magnitude of this component (and the magnitude of the total linear acceleration) is $a_l = |v|^2/r$. Sep 27 '16 at 11:50
The partial derivative of velocity with time is called spatial acceleration and it indeed is zero in this case. The total derivative of velocity (as the particle moves) is called material acceleration and it includes the components of the change in direction of velocity.
$$\vec{a} = \frac{{\rm d} \vec{v}}{{\rm d}t} = \frac{\partial \vec{v}}{\partial t} + \vec{\omega}\times \vec{v}$$