Question

Derive an expression for the minimum speed required to perform stunts in the well of death.

Answer 1

In a well of death, the forces exerted on the (rider + motorcycle) are (Fig.)

1. the normal force `vec(N)` exerted by the wall, directed radially inward, is the centripetal force.
2. the upward frictional force `vec(f_s)` exerted by the wall, since the motorcycle has a tendency to slide down.
3. the downward gravitational force `mvec(g)`.

∴ N = `(mv^2)/r` and f_s = μ_sN = μ_s(mv²/r)

where m is the mass and v is the speed of the (rider + motorcycle). For the rider not to fall, `vec(f_s)` must balance `mvec(g)`.

∴ μ_s(mv²/r) = mg

∴ `v^2 = (rg)/μ_s`

∴ The minimum speed necessary for not falling is

`υ_min = sqrt((rg)/μ_s)`