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 21.0° with the horizontal. A 37.0 kg piece of luggage is placed on the carousel, 7.10 m from the axis of rotation. The travel bag goes around once in 44.0 s. (a) Calculate the force of static friction between the bag and the carousel. (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 a position 7.88 m from the axis of rotation. The bag is on the verge of slipping as it goes around once every 28.20 s. Calculate the coefficient of static friction between the bag and the carousel.
To solve this problem, we need to analyze the forces acting on the luggage and use the equations of rotational motion.
(a) First, let's calculate the force of static friction between the bag and the carousel.
We start by calculating the acceleration of the luggage using the equation for rotational motion:
a = ω^2 * r
Where:
- a is the acceleration
- ω is the angular velocity, which is given by ω = 2π / T, where T is the period
- r is the distance of the luggage from the axis of rotation
Given:
r = 7.10 m
T = 44.0 s
ω = 2π / T
ω = 2π / 44.0
ω ≈ 0.1429 rad/s
a = ω^2 * r
a = (0.1429)^2 * 7.10
a ≈ 0.1423 m/s^2
Next, we can calculate the gravitational force acting on the luggage:
F_gravity = m * g
Where:
- F_gravity is the gravitational force
- m is the mass of the luggage
- g is the acceleration due to gravity
Given:
m = 37.0 kg
g = 9.8 m/s^2
F_gravity = 37.0 * 9.8
F_gravity ≈ 362.6 N
Now, the force of static friction opposes the downhill motion of the luggage and ensures it stays on the carousel. We can calculate it using the equation:
F_friction = m * a
F_friction = 37.0 * 0.1423
F_friction ≈ 5.268 N
Therefore, the force of static friction between the bag and the carousel is approximately 5.268 N.
(b) Now let's calculate the coefficient of static friction between the bag and the carousel when it is on the verge of slipping.
Using the same equation for the acceleration as before:
a = ω^2 * r
Given:
r = 7.88 m
T = 28.20 s
ω = 2π / T
ω = 2π / 28.20
ω ≈ 0.2229 rad/s
a = ω^2 * r
a = (0.2229)^2 * 7.88
a ≈ 0.3859 m/s^2
Now, redefine the equation for static friction as:
F_friction = μ_s * N
Where:
- F_friction is the force of static friction
- μ_s is the coefficient of static friction
- N is the normal force, which is equal to the gravitational force acting on the luggage
Since the luggage is on the verge of slipping, the force of static friction is equal to the maximum static friction force. Therefore:
F_friction = F_max_static_friction = μ_s * N
Rearranging the equation, we get:
μ_s = F_friction / N
From part (a), F_friction is approximately 5.268 N and N is equal to F_gravity, which we previously calculated as about 362.6 N.
μ_s = 5.268 / 362.6
μ_s ≈ 0.0145
Therefore, the coefficient of static friction between the bag and the carousel is approximately 0.0145.