Question

Riders in a carnival ride stand with their backs against the wall of a circular room of diameter

Answer 1

Answer:

option C

Explanation:

given,

diameter of circular room = 8 m

rotational velocity of the rider = 45 rev/min

                  = $45 \times \dfrac{2\pi}{60}$

                  =4.712 rad/s

here in this case normal force is equal to centripetal force

N = m r ω²

N = m x 4 x 4.712²

N = 88.83m

frictional force = μ N

    = 88.83m x μ

now, for the body to not to slide

gravity force is equal to frictional force

m g = 88.83 m x μ

g = 88.83 x μ

9.8 = 88.83 x μ

 μ = 0.11

hence, the correct answer  is option C