I'm trying to solve the following problem:
A bead $P$ of mass $m$ slides on a smooth wire. The wire lies in a horizontal plane, and its shape is given parametrically by $\mathbf{r}=(x(s),y(s),0)$, for $0\leq s \leq 1$, where $\mathbf{r}$ is measured relative to a point $O$ in the plane and $s$ is arclength. The wire is forced to rotate at a constant angular speed $\omega$ about the vertical axis through $O$, which remains fixed. Show that $$ \mathbf{T} \cdot \left(\frac{d^2 \mathbf{r}}{dt^2}\right)' = \omega^2\mathbf{T} \cdot \mathbf{r} $$ where $\mathbf{T}$ is the tangent to the wire at $P$.
by $\left(\frac{d^2 \mathbf{r}}{dt^2}\right)'$ they mean the second derivative of $\mathbf{r}$ but with respect to some other rotating frame, I'm assuming one which rotates with the wire.
Not really sure how to proceed with this question to be honest.
The relation between accelerations measured with respect to an inertial frame centered in $O$ and with respect to the rotating one (non inertial, constant angular velocity) is
$\vec a_{ni}=\vec a_{i}-2\vec\omega\times\vec v_{ni}-\vec\omega\times(\vec\omega\times\vec r)$ with $\vec a_{ni}=\left(\dfrac{d^2 \vec{r}}{dt^2}\right)'$
Now compute $\vec T·a_{ni}$ with $\vec T=\dfrac{d\vec r}{ds}$, a vector tangent to the wire.
$\vec T·a_{ni}=\vec T·\vec a_{i}-2\vec T·\vec\omega\times\vec v_{ni}-\vec T·\vec\omega\times(\omega\times\vec r)$
Some terms vanish:
The forces acting on the bead are all orthogonal to the wire: the wire is "smooth", meaning that no friction force acting along the wire, thus from the wire the only force is the normal reaction and the gravity is too orthogonal as the wire is in a horizontal plane. This means that in an inertial reference frame, where the Newton's second law holds (acceleration proportional to the sum all applied forces), the acceleration is always orthogonal to the wire. Being $\vec T$ a vector tangent to the wire, its dot product is zero: $\vec T·\vec a_{i}=0$
Now, $2\vec T·\vec\omega\times\vec v_{ni}=0$ considering that,
$\vec T=\dfrac{d\vec r}{ds}=\dfrac{d\vec r}{dt}\dfrac{dt}{ds}=\dfrac{dt}{ds}\vec v_{ni}$
we have that,
$2\vec T·\vec\omega\times\vec v_{ni}=2\dfrac{dt}{ds}\vec v_{ni}·\vec\omega\times\vec v_{ni}=0$ As $\vec v_{ni}$ is orthogonal to $\vec\omega\times\vec v_{ni}$
Thus, we've got,
$\vec T·a_{ni}=-\vec T·\vec\omega\times(\omega\times\vec r)=$
$=-\vec T·(\vec\omega(\vec\omega·\vec r)-\vec r(\vec\omega·\vec\omega))$
The term $\vec\omega(\vec\omega·\vec r)$ vanishes because the bead's position vector $\vec r$ is always in a plane orthogonal to $\vec\omega$ ($\vec\omega$ is along the $z$ axis and $\vec r$ has $z=0$)
$\vec T·a_{ni}=\vec T·\vec r(\vec\omega·\vec\omega)$
$\vec T·a_{ni}=\vec T·\vec r\omega^2$