Calculating g-force of spinning a toddler I sometimes spin my nephew on the ground (we have a very slippery floor). So as my nephew is $90 \; cm$ tall and I rotate him roughly $60$ times a minute, I'm applying a g-force of $1.8$ according to the formula
$$RCF \; \text{or} \; g-force = 0.00001118 \cdot \text{rotor radius} \cdot (rpm)^2 = 0.00001118\cdot45\cdot60\cdot60.$$ Does $1.8 g$ sound right?
 A: If you want to find the force that you're applying, it's best experimentally like this.
Ask someone to stand at the side and estimate the angle $\theta$ to the vertical, that your nephew's body makes as you spin him.  Estimate this angle at the point where you're applying the force, near your hands.
If you are applying a force $F$ at the angle $\theta$
the vertical component of this force matches his weight
$$F\cos\theta=mg$$
$$F= \frac{mg}{\cos\theta}$$
If you wanted to find the centripetal acceleration that your nephew is subject to, it's $$gtan\theta $$ the acceleration and angle will vary along his body, being greater further from the centre of rotation.  Depending where you measure the angle, you should expect about 60 degrees if your calculation is correct, careful of the slippy floor!
A: The relative centrifugal force is defined as $RCF:=\frac{\omega^2 r}{g}$ where $\omega$, $r$ and $g=9.81 \frac{m^2}{s}$ denote the angular speed of rotation, the radius of rotation and the acceleration due to gravity at the earth's surface respectively. Assuming that the point about which the rotation occurs is the midpoint of the length $90 \; cm$, the (correctly calculated in the OP) $RCF=1.118 \cdot 10^{-5} \cdot 45 \cdot 60^2=1.81116$ would be observed at the furthest locations (edges of head and feet) from the midpoint.
