Physical interpretation of the normal to a curve in $\mathbb{R}^3$ if the tangent is interpreted as velocity

Question

Assume I have a curve $\vec{r}(t) \in \mathbb{R}^3$. Its tangent vector, $\vec{T} = \vec{\dot{r(t)}}$ can be interpreted as the velocity of a particle moving along the curve (as long as my parameterization is chosen correctly). The normal vector, though, doesn't seem to have as nice an interpretation. What is the physical interpretation, assuming there is one, of the normal vector (and the binormal)?

Edit: As has been pointed out in comments, the normal vector is not unique. I shall define "the" normal as: $\vec{N} = \frac{d\hat{T}}{dt}$, where $\hat{T} = \frac{\vec{T}}{||\vec{T}||}$, and the binormal as $\vec{B} = \vec{T}\times\vec{N}$. I am specifically interested in the interpretation of $\vec{N}$, unless $\vec{B}$ has a more obvious interpretation.

Answer 1

The normal vector can be thought of as direction of curvature or, wordily, direction of change of direction. To justify this, let's write the velocity vector as magnitude times direction: $T = v\hat{T}$. Acceleration is the derivative of $T$ with respect to time, so let's apply the product rule: $$ \frac{dT}{dt} = \frac{dv}{dt}\hat{T} + v\frac{d\hat{T}}{dt} = \frac{dv}{dt}\hat{T} + vN $$ Acceleration breaks into two components. The first term corresponds to speeding up in the same direction you were going. The second term corresponds to changing direction. So we can interpret $N$ as pointing in the direction about which the curve is, well, curving.

Answer 2

It is more obvious to define the binormal unit vector as the unit vector orthogonal to the plane in which the velocity $\mathbf{v}$ and the acceleration $\mathbf{a}$ lay, i.e.: $$ \mathbf{b}=\frac{\mathbf{v}\times\mathbf{a}}{|\mathbf{v}\times\mathbf{a}|} $$ The principal normal unit vector is, by consequence, the unit vector of the unique direction orthogonal to $\mathbf{t}$ and $\mathbf{b}$.