(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 29.0° with the horizontal. A piece of luggage having mass 30.0 kg is placed on the carousel at a position 7.46 m measured horizontally from the axis of rotation. The travel bag goes around once in 37.0 s. Calculate the force of static friction exerted by the carousel on the bag.


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The correct answer is not zero. N

(b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, 7.94 m from the axis of rotation. Now going around once in every 32.9 s, the bag is on the verge of slipping down the sloped surface. Calculate the coefficient of static friction between the bag and the carousel.

To calculate the force of static friction exerted by the carousel on the bag, we need to use the formula for centripetal force.

(a) Given information:
- Mass of the luggage (m) = 30.0 kg
- Distance from the axis of rotation (r) = 7.46 m
- Angle of the metallic surface (θ) = 29.0°
- Time for one revolution (T) = 37.0 s

First, we need to find the speed of the luggage:

The formula for the linear velocity (v) is:
v = (2πr) / T

Plugging in the values:
v = (2π * 7.46) / 37.0

Calculate: v = 1.006 m/s (approx)

Next, we need to find the acceleration of the luggage:

The formula for acceleration (a) is:
a = v^2 / r

Plugging in the values:
a = (1.006^2) / 7.46

Calculate: a = 0.136 m/s^2 (approx)

To find the centripetal force (F_f) exerted by the luggage, we use Newton's second law:

F_f = m * a

Plugging in the values:
F_f = 30.0 * 0.136

Calculate: F_f = 4.08 N (approx)

Therefore, the force of static friction exerted by the carousel on the bag is 4.08 N.

(b) To calculate the coefficient of static friction, we need to find the critical velocity at which the bag is on the verge of slipping.

Given information:
- Distance from the axis of rotation (r) = 7.94 m
- Time for one revolution (T) = 32.9 s

First, calculate the speed of the luggage using the formula for linear velocity (v):
v = (2πr) / T

Plugging in the values:
v = (2π * 7.94) / 32.9

Calculate: v = 1.920 m/s (approx)

Next, find the acceleration of the luggage using the formula for acceleration (a):
a = v^2 / r

Plugging in the values:
a = (1.920^2) / 7.94

Calculate: a = 0.466 m/s^2 (approx)

To be on the verge of slipping, the force of static friction (F_f) needs to be equal to the maximum static friction, which is equal to the normal force (F_N) multiplied by the coefficient of static friction (μ_s).

F_f = F_N * μ_s

Since the force of static friction (F_f) is equal to the product of the mass (m) and acceleration (a), and the normal force (F_N) is equal to the product of the mass (m) and gravity (g), we can write the equation as:

m * a = m * g * μ_s

The mass (m) cancels out, leaving:

a = g * μ_s

Rearrange the equation to solve for the coefficient of static friction (μ_s):

μ_s = a / g

Plugging in the values:
μ_s = 0.466 / 9.8

Calculate: μ_s = 0.047 (approx)

Therefore, the coefficient of static friction between the bag and the carousel is approximately 0.047.