Question

An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away Fig. below). The coefficient of static friction between person and wall is "µs", and the radius of the cylinder is R. (a) Show that the maximum period of revolution necessary to keep the person from falling is T= (4p^2Rµs/g)^1/2. (b) Obtain a numerical value for T if R= 4.00m and "µs"=4.00. How many revolutions per minute does the cylinder make?

Answer 1

w = 296 rev / min

Step-by-step explanation:

a) Let's write Newton's second law

Radial axis

N = m a

Vertical axis

fr -W = 0

fr = mg

The friction force equation is

fr = μ N

μN = mg

The acceleration of the body is centripetal

a = v² / r

N = m v² / r

We replace

μ (m v² / r) = mg

v² = g r / miu

The speed module is constant, so we can use

v = d / t

The distance traveled is and length of the circle and the time taken is called the period (T)

d = 2π R

We replace

(2π R / T)² = gR /μ

T² = 4 π² R μ / g

T = √ (4π² R μ / g)

b) let's calculate

μ = 4

T = √ (4 π² 4.00 4 / 9.8)

T = 8 s

Make a complete lap in 8 s, so the angular velocity is

w = θ / t

w = 2π / 8

w = 0.7854 rad / s

Let's reduce to rev / min

w = 0.7854 rad / s (1 rev / 2pi rad) (60s / 1 min)

w = 296 rev / min