Question

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

8.0 m. The room is spinning horizontally about an axis through its center at a rate of 45 rev/min

when the ?oor drops so that it no longer provides any support for the riders. What is the minimum

coe?cient of static friction between the wall and the rider required so that the rider does not slide

down the wall?

(a) 0.0012

(b) 0.056

(c) 0.11

(d) 0.53

(e) 8.9

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