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 use the concept of rotational equilibrium, centripetal force, and static friction. Let's begin by solving part (a), calculating the force of static friction between the bag and the carousel.
(a) To calculate the force of static friction, we need to consider the forces acting on the luggage. These forces include the force of gravity (mg), the normal force (N), and the force of static friction (f_s). The normal force is perpendicular to the surface of the carousel and cancels out the vertical component of the force of gravity.
Using the concept of centripetal force, we know that the net force acting towards the center of rotation is responsible for the circular motion of the luggage. This centripetal force is given by the equation:
Centripetal Force = m * (v^2 / r)
where m is the mass (37.0 kg), v is the linear velocity (calculated using the distance traveled in one revolution and the time), and r is the radius (7.10 m).
Now, we can calculate the force of static friction using the following equation:
Force of Static Friction = Centripetal Force (in this case)
Let's go step by step to find the force of static friction:
Step 1: Calculate the linear velocity (v):
Linear velocity = Distance / Time
Linear velocity = 2 * π * r / Time (Since one rotation covers a distance of 2 * π * r)
Plugging in the values:
Linear velocity = 2 * π * 7.10 m / 44.0 s
Linear velocity ≈ 1.437 m/s
Step 2: Calculate the centripetal force:
Centripetal Force = (37.0 kg) * (1.437 m/s)^2 / 7.10 m
Centripetal Force ≈ 10.99 N
Therefore, the force of static friction acting between the bag and the carousel is approximately 10.99 N.
Now let's move on to part (b) to calculate the coefficient of static friction:
(b) In this part, the bag is on the verge of slipping, which means the force of static friction is at its maximum value. We need to find the coefficient of static friction (μ_s) using the given information.
Step 1: Calculate the linear velocity (v):
Linear velocity = 2 * π * 7.88 m / 28.20 s
Linear velocity ≈ 1.764 m/s
Step 2: Calculate the centripetal force:
Centripetal Force = (37.0 kg) * (1.764 m/s)^2 / 7.88 m
Centripetal Force ≈ 9.25 N
Step 3: Calculate the force of static friction (f_s):
Force of Static Friction = Centripetal Force
Force of Static Friction = 9.25 N
Now, we can use the equation for the force of static friction:
Force of Static Friction = μ_s * N
Since the normal force (N) cancels out the vertical component of the force of gravity, it is equal to the weight of the luggage (mg):
N = (37.0 kg) * (9.8 m/s^2)
N ≈ 362.6 N
Plugging in the values, we have:
9.25 N = μ_s * 362.6 N
Rearranging the equation to solve for μ_s:
μ_s = 9.25 N / 362.6 N
Therefore, the coefficient of static friction between the bag and the carousel is approximately 0.0255.