Calculating Yo-Yo Acceleration, Velocity, and Frequency

In summary, the conversation discusses a yo-yo with a mass of 0.5kg and a solid, homogeneous disc of radius 5cm connected to two strings of length 0.5m via a weightless rod of radius 0.5cm. The yo-yo is in a downward motion and the questions revolve around the acceleration it receives, the forces acting on the suspension, the highest downward velocity and frequency of revolution it reaches, and the average force acting during the reversal process. The conversation also mentions the unconventional design of the yo-yo and the need to consider both translational and rotational accelerations.
  • #1
Icheb
42
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I have a yo-yo of mass 0.5kg, which consists of a solid, homogeneous disc of radius 5cm. It is connected via a rod that is assumed to be weightless of radius 0.5cm to two strings of length 0.5m.
Now the yo-yo is in the motion of falling down.

a) What acceleration does the yo-yo receive?

Wouldn't that just be 9.81m/s^2 since no other forces are acting on the yo-yo?

b) Which force is acting on the suspension?

I'm assuming the yo-yo is still in the downward motion, so why would there be a force acting on the suspension? It's basically falling freely.

c) What's the highest downward velocity and what is the highest frequency of revolution the yo-yo reaches?

To calculate the velocity at the turning point I would use s=1/2 * at^2 and solve it for t so I know how long it took to reach the point. This t I would then insert in v=a*t to get the velocity at that point. Right?

To calculate the frequency I can use the radius of the rod, which is 0.5cm, to calculate its circumference. With the circumference I can calculate what distance the yo-yo travels in one revolution and then I just have to calculate how often it has to turn in one second to reach the velocity at the lowest point. Right?

d) At the end the yo-yo changes its direction. Which average force acts during the reversal of the process, which occurs within half a revolution of the yo-yo, additionally on the suspension?

This is where I'm confused. When the yo-yo hits the end of the strings, it has a force of f=m*a=0.5kg*9.81m/^2 and then this force acts on the suspension because the suspension has to stop the movement. During the reversal process that force gets smaller and smaller until the yo-yo is flying up again.
Is there another formula for the force which I should be using or is my approach flawed? I'm asking because it mentions how long the process takes and I'm not using that information.

e) Which force is acting on the suspension during the upward motion?

Shouldn't this be similar to b), where there is no force acting on the suspension itself because it's just flying upwards?
 
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  • #2
I cannot make sense out of the description of the yo-yo. A yo-yo is not a solid disk. The typical yo-yo is made of two disks connected by a small rod with one string wrapped around the rod. If that is what you have, the questions you are being asked make sense, but I don't see what two strings has to do with the problem, and I can't visulaize the rod described in this problem.

If we can assume this thing really is a yo-yo and that the rod is the usual central rod with one string wrapped around it that is fixed at the other end, then the acceleration will be far less than g. You need to consider the translational and rotational accelerations at the same time.
 
  • #3
It would be something like this:
http://img170.imageshack.us/img170/2010/maxwellradsmallwx2.jpg

For some reason they called it a yo-yo.
 
Last edited by a moderator:
  • #4
Icheb said:
It would be something like this:

For some reason they called it a yo-yo.

Ah... and now I see the reason for two strings. A bit unconventional, but it will still act like any other yo-yo. Do you understand why the acceleration is less than g? Can you figure out what it should be?
 

Related to Calculating Yo-Yo Acceleration, Velocity, and Frequency

1. What is the formula for calculating yo-yo acceleration?

The formula for calculating yo-yo acceleration is a = (v^2 - u^2)/2r, where a is acceleration, v is final velocity, u is initial velocity, and r is the radius of the yo-yo's string.

2. How do you calculate yo-yo velocity?

To calculate yo-yo velocity, you can use the formula v = √(u^2 + 2ar), where v is velocity, u is initial velocity, a is acceleration, and r is the radius of the yo-yo's string.

3. What is the relationship between yo-yo acceleration and velocity?

The relationship between yo-yo acceleration and velocity is that acceleration is the rate of change of velocity. This means that as the yo-yo accelerates, its velocity will increase or decrease depending on the direction of the acceleration.

4. How does the length of the yo-yo string affect its frequency?

The length of the yo-yo string directly affects its frequency. The longer the string, the lower the frequency, and the shorter the string, the higher the frequency. This is because the length of the string determines the distance the yo-yo travels in one cycle, which is directly related to its frequency.

5. Can you calculate yo-yo acceleration, velocity, and frequency without knowing the mass of the yo-yo?

No, the mass of the yo-yo is an essential component in calculating its acceleration, velocity, and frequency. Mass affects the yo-yo's inertia, which is necessary for calculating its acceleration, and it also affects the tension in the string, which is necessary for calculating its velocity and frequency.

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