Torque is nothing but moment of force, i.e
$\vec{τ} = \vec{r} × \vec{F}$
That is not centripetal force, centripetal force is the force which keeps the object in circular motion and which is always directed radially inwards. And torque is something which is present when there is $Angular$ $Acceleration$ in rotational motion.
Moment of something is defined as measure of a force's tendency to cause a body to rotate about a specific point or axis "or" a moment is an expression involving the product of a distance and physical quantity and that's why we multiply the term "$r$" with it. let's say if we want to study about moment of momentum then we define it as:-
$\vec{L} = \vec{r} × \vec{P}$
Which is also known as angular momentum.
The reason for "why does an object move faster as it comes near the center" is that according to the relation
$V = ωr$
$\frac{V}{r} = ω$
As we can see that from here the angular velocity and the radius is inversely proportional to each other and if we decrease the radius then the angular velocity of the particle increases and that is the reason as the angular velocity increase with decrease in radius.
Edit :- as torque is the moment of force it can be represented as shown above i.e:-
$\vec{τ} = \vec{r} × \vec{F}$
Let's understand this with an example, let the torque be some constant $\vec{τ_0}$ till the end.
And let's consider a rod with some length $l_1$ which is attached to a frictionless hinge (since it may provide an external torque other than the torque which we are providing to the system if it is present).
And exert a force equal to some constant $F_1$ on then end of the rod and mutually perpendicular to the axis of rotation and the rod as well (to keep it simple for further elaboration and your understanding, although you can put the force in any direction you want) hence, if it is perpendicular to the rod By solving the above equation for the torque we get:
$\vec{τ_0} = l_1 F_1 sin \theta \hat{n}$
Note : $\hat{n}$ is the direction perpendicular to the force and the radius vector.
In this case as I already mentioned that the radius vector and force vector are perpendicular to eachother then the theta must be $90°$ so:-
$\vec{τ_0} = l_1 F_1 sin 90° \hat{n}$
Then the torque on the system will come out to be:-
$\vec{τ_0} = l_1 F_1 \hat{n}$
And now let a consider a length equal to $l_2$ Where $l_2 < l_1$
Following the same procedure for $l_2$ as for the length $l_1$ and now for this case exerting the force equal to $F_2$ at the length $l_2$ you will get:
$\vec{τ_0} = l_2 F_2 \hat{n}$
$\hat{n}$ will be in the same direction as before.
Now from the general for the torque i. 544