Rotation of rigid bodies: yo-yo

In summary, the conversation revolved around finding the final velocity of a yo-yo after falling a certain height using Newton's second law. The yo-yo consists of two cylinders with different radii and masses, and the string is assumed to be vertical. The formula for acceleration was derived and it was noted that it is constant. The question of how to obtain the final velocity was raised, and it was suggested to consider the relationship between acceleration and velocity when an object is falling under gravity.
  • #1
syang9
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consider a yo-yo consisting of two cylinders of radius R1, (combined) mass M1, glued to another smaller cylinder of radius R2, mass M2. find the final velocity after falling a height h using Newton's second law. assume the string is vertical.

Chapter1229.gif


the inner radius R2 is R0, the outer radius R1 is R in this picture

so, i started off by applying f = ma to M2..

Fnet, y = (M1 + M2) - T = (M1 + M2)(a)cm

tnet, ext = T(R1) = I(alpha), alpha = (a)cm/R1

i calculated the moment of inertial to be the sum of the moments of inertia of the cylinders..

I = 1/2*(M1*R1^2 + M2*R2^2)

solving for (a)cm, i get a very nasty formula..

a = (M1 + M2)g / ( ( I/(2R1^2)) + (M1 + M2))

my original plan was to integrate this wrt time to get a velocity.
but this formula doesn't involve anything that changes with time.. so how do i obtain the final velocity? am i even on the right track?
 
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  • #2
It looks ok.
As you noticed, the expression for a doesn't involve anything that changes wrt time. In other words, a is constant!

If you had an object that was falling under gravity, what would its velocity be after having fallen a distance h?

So then, what if g = a?
 
  • #3


Your approach is correct, but you need to consider the conservation of energy in addition to Newton's second law. Initially, the yo-yo has gravitational potential energy given by mgh, where m is the combined mass of the yo-yo cylinders. As the yo-yo falls, this potential energy is converted into kinetic energy, given by (1/2)(m)(v^2), where v is the final velocity. Therefore, we can set these two energies equal to each other:

mgh = (1/2)(m)(v^2)

Solving for v, we get:

v = sqrt(2gh)

This means that the final velocity of the yo-yo will depend on the height from which it is dropped, but it will not depend on the masses or radii of the cylinders. This makes sense intuitively, as the yo-yo will accelerate at the same rate regardless of its mass or shape.
 

1. What is the basic concept of rotation in yo-yo?

The basic concept of rotation in yo-yo is that the yo-yo is able to rotate around its central axis while maintaining its circular shape. This rotation is caused by the force of gravity acting on the yo-yo as it is dropped and pulled back up.

2. How does the string affect the rotation of a yo-yo?

The string plays a crucial role in the rotation of a yo-yo. As the yo-yo is dropped, the string unwinds and provides the necessary torque for the yo-yo to rotate. As the yo-yo is pulled back up, the string winds back up and maintains tension, allowing the yo-yo to maintain its rotation.

3. How does the mass distribution affect the yo-yo's rotation?

The mass distribution of a yo-yo can greatly affect its rotation. A yo-yo with a more evenly distributed mass will rotate more smoothly and consistently, while a yo-yo with a heavier side will wobble and be more difficult to control.

4. What is the difference between a fixed axle and a ball bearing yo-yo in terms of rotation?

A fixed axle yo-yo has a fixed central axle that the yo-yo rotates around, while a ball bearing yo-yo has a small ball bearing that allows the yo-yo to spin more freely and for a longer period of time. This means that a ball bearing yo-yo will have a longer and smoother rotation compared to a fixed axle yo-yo.

5. How does the yo-yo's rotation affect its stability?

The rotation of a yo-yo is essential for its stability. As the yo-yo spins, it creates a gyroscopic effect that helps to stabilize it and keep it spinning in a straight line. This is why a yo-yo will wobble and become less stable as it slows down and loses its rotation.

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