Finding the safe rotational speed of a carousel?

AI Thread Summary
To determine the safe rotational speed of a carousel, it's essential to calculate the maximum speed at which a person can remain on the ride without being thrown off due to centrifugal force. The maximum diameter of the carousel is 2 meters, with a standard maximum peripheral speed of 5 m/s. However, at a 1-meter radius, the centripetal acceleration calculated from this speed suggests an unrealistic friction coefficient of 2.5, indicating that the 5 m/s standard may not be safe. The critical case for safety occurs when a person is at the edge of the carousel, where the friction force must counteract the centripetal acceleration. Reevaluation of the maximum speed is necessary to ensure rider safety, especially for unseated individuals.
MarkH748
Messages
8
Reaction score
0
Hi,

I'm currently working on a project for a braking system of a carousel. Basically in order to start designing the brake I need to know what the maximum rotational speed of the carousel should be. The max diameter of the surface will be 2 meters and the maximum speed at the periphery should not be greater than 5m/s from standards. However as the diameter gets smaller and smaller the rotational speed will still increase taking these standards.

What I want to do is work out the speed that will throw someone off the carousel using maybe the centrifugal force or some such method. I assume that this is most likely on the outside of the circle? (i.e. max radius of 1). I was wondering if anyone had any suggestions as to how I would calculate this?

Any help would be greatly appreciated.

Mark.
 
Physics news on Phys.org
Assuming the person is not holding onto a pole (or a horse), then the critical case is when the person is standing at the edge of the circle, where the centripetal acceleration is the greatest, and where the only force capable of providing that acceleration is the friction force between the floor of the carousel and the person's shoes (or feet, if barefoot, etc.). I imagine that the safe speed of a carousel assumes a rather low friction value, especially in consideration of the unseated ticket taker worker. In your example, using your 5m/s tangential speed at the periphery, the centripetal acceleration for a 1 m radius would be v^2/r = 25m/s^2, implying a friction coefficient of 2.5, which is way higher than the friction coefficient between the feet and the surface, meaning that the person would fly off at that speed. So I would question that 5m/s value you got from "Standards".
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...

Similar threads

Back
Top