Q.During a stunt, a cyclist (considered to be a particle) is undertaking horizontal circles inside a cylindrical well of radius 6.05 m. If the necessary friction coefficient is 0.5, how much minimum speed should the stunt artist maintain? Mass of the artist is 50 kg. If she/he increases the speed by 20%, how much will the force of friction be?

$v_{min}=\sqrt{rg/\mu_s}=\sqrt{6.05\times10/0.5}=11$ m/s; friction is $f_s=mg=500$ N ALWAYS (it supports the weight, not the speed), so it stays 500 N even after the 20% speed increase. …

In the well of death, $v_{min}=\sqrt{\dfrac{rg}{\mu_s}}=\sqrt{\dfrac{6.05\times10}{0.5}}=\sqrt{121}=11$ m/s.

For the second part, recall that in this configuration $f_s=mg$ ALWAYS -- friction is the only VERTICAL force, and must support the full weight regardless of how fast the vehicle is actually going (the speed instead determines N, and hence how close friction is to its LIMITING value $\mu_sN$, but not the actual friction force needed, which is fixed at mg by vertical equilibrium alone). So at 20% higher speed ($1.2\times11=13.2$ m/s, still safely above $v_{min}$), the force of friction remains exactly $f_s=mg=50\times10=500\text{ N}$ unchanged from its value at the minimum speed -- what DOES ch …

Use $v_{min}=\sqrt{rg/\mu_s}$ for the minimum speed, then recognise that friction in the well-of-death configuration always equals mg exactly (fixed by vertical equilibrium), independe …

Assuming friction increases along with speed (since N does), when in fact the ACTUAL friction force stays fixed at mg -- only the AVAILABLE (limiting) friction $\mu_sN$ i …