Question

A "Barrel of Fun" consists of a large vertical cylinder that spins about its axis fast enough so that any person inside will be held against the wall. At an angular speed ω1, an upward frictional force F1 holds a person against the wall without slippingIf you double the angular speed (i.e., ω2 = 2 ω1) what is the new friction force F2?

Answer 1

Friction force F₂ after doubling the angular speed is same as the friction force at angular speed ω₁

Step-by-step 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.

`\sum F_(x) =0\\N+ ma_(x)\\-N=m(-(v^(2))/(r))\\-N=m(-r\omega^(2))N=m(r\omega^(2))` ---(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.

`\sum F_(y) = 0\\F_(s)-mg=0\\F_(s)=mg`----(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.