Calculating Time to Complete One Revolution of Airport Carousel

[gmunoz18]

Homework Statement

The drawing shows a baggage carousel at an airport. Your suitcase has not slid all the way down the slope and is going around at a constant speed on a circle (r = 12.8 m) as the carousel turns. The coefficient of static friction between the suitcase and the carousel is 0.760, and the angle in the drawing is 32.6°. How much time is required for your suitcase to go around once?

Homework Equations

v=sqrt(R*g*tan(theta))

2*pi*r

v=sqrt(r*friction*9.8)

The Attempt at a Solution

2*pi*12.8=80.424 meters

v=sqrt(125.44*tan(32.6)

v=sqrt(80.222)= 8.95 m/s


v=sqrt(8.95*.760*9.8)


this is incorrect, i know it has something to do with the frictional force but I am not sure where it goes into play

 

[Doc Al]

Rather than plugging into a formula that neglects friction, analyze the problem using first principles. Identify the forces acting on the suitcase, draw a free body diagram, and apply Newton's laws.

 

[Moetasim]

.

I would approach this problem by first identifying the known values and variables, as well as the relevant equations. From the given information, we know the radius of the carousel (r = 12.8 m), the coefficient of static friction (μ = 0.760), and the angle at which the suitcase is sliding (θ = 32.6°). The relevant equations for this problem include the formula for centripetal acceleration (a = v^2/r) and the formula for frictional force (F = μN).

Next, I would analyze the problem and determine what information is needed to solve it. In this case, we are looking for the time it takes for the suitcase to complete one revolution, which can be found using the formula T = 2πr/v, where T is the time, r is the radius of the circle, and v is the velocity.

To find the velocity (v), we can use the formula for centripetal acceleration and set it equal to the formula for frictional force, since these two forces must be equal in magnitude for the suitcase to maintain a constant speed. This gives us the equation a = v^2/r = μN, where N is the normal force acting on the suitcase.

Using trigonometry, we can determine the normal force as N = mgcosθ, where m is the mass of the suitcase and g is the acceleration due to gravity. Substituting this into our equation for frictional force, we get F = μmgcosθ.

Now, we can substitute this value for frictional force into our equation for centripetal acceleration, giving us a = μmgcosθ/m = μgcosθ. Solving for v, we get v = sqrt(μgrcosθ).

Finally, we can substitute this value for velocity into the equation for time, giving us T = 2πr/sqrt(μgrcosθ). Plugging in the known values, we get T = 2π(12.8)/sqrt(0.760*9.8*12.8*cos(32.6)) = 12.2 seconds.

Therefore, it would take approximately 12.2 seconds for your suitcase to complete one revolution on the airport carousel.