Centripetal force yo-yo problem

In summary, the conversation is about finding the speed at which a yo-yo clutch opens against a spring to its maximum. The question is whether F cos 90 - θ or Fc = F / cos 90 - θ should be used to determine the centripetal force. The solution involves resolving the force from the ball bearing using moments and a lever model, and then using the relationship F_s = m \omega^2 r to find the speed. The question is clarified with the use of diagrams and the conversation ends with the suggestion to use F cos (abc) or F / cos (abc) to resolve the force through angle (abc).
  • #1
gilboy64
6
0
Hi,

I am trying to find the speed it takes to open a yo-yo clutch, against a spring, to its maximum, the problem I am stuck on is whether you have to use F cos 90 - θ to get the centripetal force or if you have to do this in reverse eg. Fc = F / cos 90 - θ. I ask this because if you start with v using Fc = ω r you can find the centripetal force. You then can resolve it through cos 90 - θ to get the vertical component acting on the ball bearing. Then using moments you can model the clutch as a lever, and find the force exerted by the spring on the lever when it is opened to maximum. So, if you know the force exerted by the spring, working backwards, you should be able to find the speed. So, when you come to resolving the the force from the ball bearing, i think you should use the second one, Centripetal force = F / cos 90 - θ but i am not sure. Please help

thank you,
Gilmore
 

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  • #2
Hi, can anybody help?
 
  • #3
I would be delighted to help, but I really don't know what you are asking. I have no idea what a "yo-yo clutch" is, and I can't relate your question to your diagram.

If you can describe it a bit more, that might help.
 
  • #4
Its quite hard to explain without drawings, but basicly it is a lever that opens when the yo-yo is at a certain speed, there is a spring that stops it from opening until the speed is reached, and the force acting on the end of the lever (the ball bearing) is greater than the force exerted by the spring on the lever. I am tring to predict that speed, i have found the spring constant of the spring, and what force is exerted on the lever by the spring when the lever is open to its maximum. I have first used the moments on the lever to find the vertical force acting on the ball bearing. where x and y are distances.
x * force exerted by spring = y * force acting on ball bearing.
I know that the centripetal force acts directly from the center of the ball bearing into the center of the yo-yo, I also know that the centripetal force = mv2/r. So if I resolve the force on the ball bearing i can find the centripetal force and therefore the speed. But do I have to use centripetal force = F2 cos angle or (because the Fc is a 'real' force, and if I started with this force i definatly would cos angle) use Fc = F2 / Cos angle??
i hope this makes a little more sence
 

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  • #5
I'm still not quite getting it. (Maybe I need another cup of coffee?) But let me take a guess on what the problem is and make some comments. The forces on the ball bearing are two: the spring pulling it towards the center and the normal force of the yo-yo pushing it away from the center. The net force is: [itex]F_s - N[/itex]. Since the yo-yo is spinning, the ball bearing is centripetally accelerated. Thus, Newton's 2nd law tells us that: [itex]F_s - N = m \omega^2 r[/itex]. The ball bearing will separate from the yo-yo when the normal force = 0, which is at the speed where the spring is providing the full centripetal force: [itex]F_s = m \omega^2 r[/itex].

Let me know if I am getting closer to your question. :smile:
 
  • #6
1 picture = 1000 words?

Ah... Now I think I see what you are talking about:
http://stuffo.howstuffworks.com/yo-yo3.htm

So to analyze the ball bearing from the reference frame of the yo-yo (an accelerating frame): You need to find the speed at which the centrifugal force on the bearing is enough to match the force of the spring on the lever. Then, yes, you need the component of the centrifugal force perpendicular to the lever arm. Equate the torque due to the spring to the torque due to the centrifugal force on the ball bearing.

Am I getting closer? :smile:
 
  • #7
[itex]F_s = m \omega^2 r[/itex] is the same as [itex]F_s = m v^2 / r[/itex], so if i know the Fs, I can find v. I know the force the spring is exerting on the lever, so using moments on the lever I can find the force acting vertically upwards on the ball bearing, right? So if I resolve this force so that it is now acting directly away from center of the yoyo it will be the same as the Fs, because the yoyo is spinning constanly the lever will be in equilibrium so the force = the same.
I have drawn one more picture, black arrows = the forces.
If I find angle (abc) I can resolve the vertical force (F) through this, my problem is to resolve this force through (abc) do i have to use F cos (abc) or F / cos (abc)??
 

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1. How does centripetal force affect a yo-yo?

Centripetal force is the force that acts on an object moving in a circular path, always pointing towards the center of the circle. In the case of a yo-yo, the string exerts a centripetal force on the yo-yo as it is being spun around, keeping it in a circular path.

2. What factors affect the centripetal force in a yo-yo?

The centripetal force in a yo-yo is affected by the mass of the yo-yo, the speed at which it is being spun, and the length of the string. A heavier yo-yo or a faster spinning speed will result in a greater centripetal force, while a longer string will decrease the centripetal force.

3. How does the radius of the yo-yo's circular path affect the centripetal force?

The radius of the yo-yo's circular path has a direct effect on the centripetal force. The smaller the radius, the greater the centripetal force needed to keep the yo-yo in its circular path. This is why yo-yos with a smaller radius require more force to spin compared to yo-yos with a larger radius.

4. Can the centripetal force in a yo-yo ever be greater than the weight of the yo-yo?

Yes, it is possible for the centripetal force in a yo-yo to be greater than its weight. This can happen when the yo-yo is spun at a high enough speed or when the radius of its circular path is small enough. In this case, the yo-yo will experience a net force in the direction of the circular path, causing it to accelerate and move in a circular motion.

5. How is the concept of centripetal force used in other real-world applications?

The concept of centripetal force is used in many real-world applications, such as amusement park rides, car racing, and satellite orbits. In these scenarios, the centripetal force is used to keep objects moving in a circular path and prevent them from flying off in a straight line. Understanding centripetal force is crucial in designing and operating these systems safely and effectively.

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