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An amusement park ride consists of a large vertical cylinder that spins about it

ID: 1423295 • Letter: A

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. P6.59). The coefficient of static friction between person and wall is mu_s, and the radius of the cylinder is R. Show that the maximum period of revolution necessary to keep the person from falling is I = (4pi^2R mu_s/g)^1/2. If the rate of revolution of the cylinder is made to be somewhat larger, what happens to the magnitude of each one of the forces acting on the person? What happens in the motion of the person? If the rate of revolution of the cylinder is instead made to be somewhat smaller, what happens to the magnitude of each one of the forces acting on the Person? How does the motion of the person change?

Explanation / Answer

Part a) We use Newton's second law

Vertical Y axis

body does not slide on the wall a = 0

Axis Y vertical

fr – W = 0 fr = W (1)

fr = N

Axis X radial

N = m ac   ac == V2 / R

using the rotational kinematics

V = w R

N = m w2R2/R N = m R w2

w = 2 f = 2 / T

N = m R (2 / T)2

replace it in 1

m R 42/T2= m g

T2= R 42/ g

T = sqrt ( R 42 / g)

Part b) The angular velocity of the cylinder increases,

The first effect is the increase of normal and consequently the friction force

N = m R w2

as the body weight is constant in equation (1) must be a net upward force

fr – w = m a this acceleration in the vertical direction

m R 42 /T2 - m g = m a

remember that increasing w is a decrease of T

In summary

W remains constant

N increase

fr increases

Part c)

if the angular velocity is less than the calculated in part to the weight value is greater than the friction force and the body slides down

Summary

W remains constant

fr decreases

N decreases

Remember that an increase w implies a decrease T

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