[Help] Amusement Park Physics Problem

[helium-4]

Homework Statement

You are designing a new amusement park ride called Barrel o’ Fun. The idea is that people will stand inside a barrel with their backs against the wall. It takes 1.5 seconds for the barrel to complete one revolution. The diameter of the barrel is 15 meters. What coefficient of friction will be necessary for the people to stick to the wall so that when the barrel is spinning the floor can drop away?

Homework Equations

Force of Static Friction = Coefficient of Static Friction × Normal Force = mv^2/r
Fs = μs×N = mv^2/r
μs = mv^2/rN
μs = v^2/rg
μs = 4∏^2r^2/rgt^2

The Attempt at a Solution

μs = 4∏^2r^2/rgt^2
μs = 13.4143


This does not seem right. I would appreciate your help.

 

[frogjg2003]

Draw a force body diagram.

 

[NasuSama]

As one user just said, you need to draw the free body diagram. You need to include the normal force, the weight vector and the corresponding forces that take place during the ride.

You might want to show us the free body diagram, so we can check to see if you are at the right track!

 

[helium-4]

I already solved the problem. I am sorry for the delay in response - I did not have access to internet for much of yesterday.

I misplaced the Force of Static Friction on the FBD. In this particular problem, the Fs points up, and the Fg points down (y direction).

Fs - Fg = ma_y = 0. So μsN - mg = 0.
μs = mg/N.

The N force points towards the center of the barrel (x direction).

N = ma_x = mv^2/R = 4∏^2mR/T^2

So, μs = (mg)/(4∏^2mR/T^2)
μs = ((g)(T^2)) / ((4)(∏^2)(R)) = 0.74547

 

[Nickie]

I can offer some guidance in solving this physics problem. First, we need to understand the concept of static friction and how it relates to the situation. Static friction is the force that prevents an object from sliding when there is no external force acting on it. In this case, the people standing in the barrel will experience a force of static friction against the wall of the barrel, which will keep them from sliding when the floor drops away.

To calculate the coefficient of static friction, we can use the equation Fs = μs×N, where Fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force. In this case, the normal force is the force exerted by the wall on the people, which is equal to their weight (mg) since they are standing against the wall with their backs.

To find the normal force, we need to consider the centripetal force acting on the people, which is given by mv^2/r, where m is the mass of the person, v is the speed of rotation, and r is the radius of the barrel (which is half the diameter). Since the people are standing against the wall, their weight must be equal to the centripetal force, so we can set mg = mv^2/r and solve for v.

v = √(rg)

Now, we can substitute this value for v into our equation for μs:

μs = Fs/N = mv^2/rN = (m√(rg))^2/(mr) = rg/r = g

Therefore, the coefficient of static friction needed for the people to stick to the wall is equal to the acceleration due to gravity (g). This means that as long as the barrel is spinning at a constant speed, the people will stick to the wall without any additional friction needed.

I hope this explanation helps you solve the problem. It is always important to understand the concepts and equations involved before plugging in numbers to find a solution. Good luck with your amusement park design!