Answer:
Friction force F₂ after doubling the angular speed is same as the friction force at angular speed ω₁
Explanation:
Consider the fig attached below. Forces acted on person are Centripetal force (-mv²/r) exerted in x direction and reversal normal force N wall exerted on person.
[tex]\sum F_{x} =0\\N+ ma_{x}\\-N=m(-\frac{v^{2}}{r})\\-N=m(-r\omega^{2})N=m(r\omega^{2})[/tex] ---(1)
In y direction there is frictional force Fs exerted in upward direction that keeps the person standing without falling which is balanced by weight of person in downward direction.
[tex]\sum F_{y} = 0\\F_{s}-mg=0\\F_{s}=mg[/tex]----(2)
from eq 2 it can e seen that frictional force is equal to weight of person exerted in upward direction, it does not depends on angular speed ω₁. So when the angular speed is doubled i.e ω₂ = 2ω₁, frictional force Fs remains same.
