Playing with a yoyo (Angular momentum problem)

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In summary, the problem involves a uniform circular cylinder, or yo-yo, with a string wrapped around it and attached to a moving support. The goal is to find the acceleration of the yo-yo. By using equations for moment of inertia, torque, and velocity, and making a substitution, the correct solution of v' = -2/3g + 1/3Z''(t) is found. The mistake was in the calculation of the angular velocity, which was corrected to be omega = (v'-z')/a.
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blalien
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[SOLVED] Playing with a yoyo (Angular momentum problem)

Homework Statement


This problem is from Gregory. This is only the first part of the problem, since I understand how the solution to the first part implies the second part.

A uniform circular cylinder (a yo-yo) has a light inextensible string wrapped around it so that it does not slip. The free end of the string is fastened to a support and the yo-yo moves in a vertical straight line with the straight part of the string also vertical. At the same time the support is made to move vertically having upward displacement Z(t) at time t. Find the acceleration of the yo-yo.

a: the radius of the yo-yo
m: the mass of the yo-yo
I: moment of inertia of the yo-yo
v: velocity of the center of mass of the yo-yo
[tex]\omega[/tex]: angular velocity of the yo-yo
g: acceleration of gravity
Z(t): the displacement of the support
T: the tension force the string exerts on the yo-yo
[tex]\tau[/tex]: torque of the yo-yo

Homework Equations


Since the yo-yo is a cylinder, I = ma^2/2
[tex]\tau[/tex] = T*a = I [tex]\omega[/tex]'
m v' = -mg + T + mZ''(t)
v = -a[tex]\omega[/tex]

The Attempt at a Solution


Work a little substitution magic:
m v' = -mg + I [tex]\omega[/tex]'/a + mZ''(t)
m v' = -mg - ma^2/2 v'/a^2 + mZ''(t)
m v' = -mg - m v'/2 + mZ''(t)
3/2m v' = -mg + mZ''(t)
v' = -2/3g + 2/3Z''(t)

The solution given in the book is v' = -2/3g + 1/3Z''(t). So, somewhere along the line I missed something with Z''(t). Can anybody guess where I went wrong?
 
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  • #2
Nevermind, I got it. Turns out it was omega = (v'-z')/a. Sometimes just typing it out seems to help, eh?
 

Related to Playing with a yoyo (Angular momentum problem)

1. What is angular momentum?

Angular momentum is a measure of the rotational motion of an object. It is calculated by multiplying the moment of inertia (resistance to change in rotational motion) by the angular velocity (speed of rotation).

2. How does playing with a yoyo involve angular momentum?

When playing with a yoyo, the string is wrapped around the yoyo's axle, causing it to spin as it is thrown and returned. This spinning motion creates angular momentum.

3. Why is it difficult to keep a yoyo spinning?

It can be difficult to keep a yoyo spinning because it is affected by external forces such as friction, air resistance, and gravity. These forces can slow down the spinning motion and cause the yoyo to stop.

4. How does the length of the string affect the angular momentum of a yoyo?

The length of the string can affect the angular momentum of a yoyo. A longer string will have a larger moment of inertia, making it more difficult to change the rotational motion and thus maintaining the yoyo's angular momentum for a longer period of time.

5. What is the relationship between angular momentum and kinetic energy in a yoyo?

Angular momentum and kinetic energy are both measures of an object's motion. In a yoyo, as the angular momentum decreases due to external forces, the kinetic energy also decreases. This is because the kinetic energy is directly proportional to the square of the angular velocity, which decreases as the yoyo slows down.

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