What happens to the velocity if you decrease the radius in uniform circular motion? PROBLEM
Suppose we have a frictionless table and we have a mass attached to a string rotating in a circular motion. If we decrease the radius by pulling string downwards (from a hole in the table), what will be the effect on the speed of the mass?

I came up with two solutions for this-

*

*$$T = \frac{mv^2}{r}$$
Acc to this solution $v^2$ is directly proportional to $rT$. So if the radius decreases, so should the speed.

*Other approach is since we pull the string down slowly then the Torque at the center point will be zero and L(angular momentum) will be conserved. So we can conveniently write-
$$L_1 = L_2 (L_1\;for\;initial)$$
$$mv_1r_1 = mv_2r_2$$
$$v_2 = \frac{v_1r_1}{r_2}$$
As r2 is less so v2 should increase.
Please tell me which one is correct and WHY?

 A: Conservation of angular momentum dictates that it will speed up as the area swept out per unit time is a constant.
as for your first solution i have no idea where u got that the velocity square is directly proportional to rT as in this case r is a variable not a constant
A: The equation in option 1, $T = \frac{mv^2}{r}$, can't be used to state that the velocity on a circle with a smaller radius is smaller than that on a circle with a bigger $r$. In the way you envision. Why not? You write:
"So if the radius decreases, so should the speed",
which you infer from $rT = mv^2$ (which follows from $T = \frac{mv^2}{r}$). If you lower $r$ though (by making the string shorter) then you exert a force on the mass which makes its kinetic energy increase. While the last formula holds for two separate circular orbits, it can't be used to say that because $r$ gets lower, so must $v$. By lowering $r$, $T$ gets bigger and so does $v$. The relation $rT = mv^2$ will of course hold for all circular orbits.
Option two is not right either. The tension force, $\vec{T}$, acting on the mass gives rise to a non-zero torque. So momentum is not conserved. Just as momentum is not conserved for a mass that is accelerated in a gravitational force field. The kinetic energy of the mass increases and so does its momentum (of course, the total momentum of the mass that gives rise to the gravitational force and the accelerated mass has to be conserved).
And now for real. The torque $\vec{\tau}=\vec{T}\times\vec{r}$, the exterior product of $\vec{T}$ and $\vec{r}$. It points out of the table and perpendicular to it. If zero torque is applied to a circular moving mass then the angular momentum of the mass stays constant. The force only does work to increase the velocity of the mass, but the angular momentum is conserved. Just as applying zero force on a linearly moving mass conserves linear momentum.
I should get very lucky now!!!
