What happens if I increase the magnitude of the centripetal force? Suppose that a particle is in uniform circular motion and the magnitude of the centripetal component of the net force increases. Does this increase the tangential speed or decrease the radius of the circular path?
One case that comes to mind is moving a ball attached to a string, moving in a circle. In that case, the radius cannot decrease as centripetal force would disappear. But what if the source is the gravitational or electrostatic force?
 A: Both.
Suppose that a particle of mass $m$ is in uniform circular motion with radius $r$ and tangential velocity $v_T$.  We know that there must be a centripetal force maintaining this motion $$F_c = m \frac{v_T^2}{r}.$$
We also know that the system has an associated angular momentum whose magnitude is $$L = rmv_T.$$
Now if we only increase the centripetal force, then we can't produce a torque on our particle.  (Because the force is along the position vector so $\vec \tau = \vec r \times \vec F = \vec 0$.)  So our angular momentum $L$ is a constant.
Solving for $v_T$ from our equation for $L$ we have $$v_T = \frac{L}{rm},$$ and, plugging into our centripetal force equation, $$F_c = \frac{L^2}{mr^3}.$$
Therefore if we increase $F_c$, $r$ must decrease: $$r = \left( \frac{L^2}{m F_c} \right)^{1/3}.$$
Since $r$ is decreasing, our formula for $v_T$ in terms of $L$ tells us that $v_T$ must increase with increasing $F_c$: $$v_T = \frac{L}{rm} = \left( \frac{L F_c}{m^2}\right)^{1/3}.$$
A: For your ball on a string, run the string down through a length of glass tube (heat polished on each end). Swing the ball overhead in a horizontal circle. Then pull down on the string. The ball moves to a smaller circle with an increase in speed. The centripetal force does not disappear.
