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Lagrange - yo-yo with moving support

  • Thread starter cscott
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  • #1
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Homework Statement



A uniform circular cylinder of mass `m' (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 [itex]Z(t)[/itex] at time `t'. Find the upward acceleration of the yo-yo.

Homework Equations



Define `a' as the radius of the cylinder.
Define theta as the rotation angle of the yo-yo.

[tex]V = -mg(a\theta - Z)[/tex]

[tex]T = 3/4 m v^2 = 3/4 m (a \dot{\theta} - \dot{Z})^2[/tex]

[tex]L = T-V[/tex]

Use, [tex]\frac{d}{dt} ( \frac{dL}{d\dot{\theta}} ) - \frac{dL}{d\theta} = 0[/tex]

The Attempt at a Solution



[tex]\frac{d}{dt} ( \frac{dL}{d\dot{\theta}} ) = 3/2m(a\ddot{\theta} - \ddot{Z})[/tex]

[tex]\frac{dL}{d\theta} = mga[/tex]

So solving gives me,
[tex] \ddot{\theta} = \frac{2/3g +\ddot{Z}}{a} [/tex]
(downwards angular acceleration.)

Therefore upwards acceleration of the yo-yo is,

[tex] \ddot{z} = -a \ddot{\theta} = \ddot{Z}-2/3g[/tex].

But I'm missing a factor of 1/3 in front of [tex]\ddot{Z}[/tex] or,

[tex]\ddot{z} = 1/3(\ddot{Z}-2g)[/tex].
 
Last edited:

Answers and Replies

  • #2
diazona
Homework Helper
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It seems that you didn't correctly take into account the rotation of the yo-yo... the kinetic energy of the system consists of two parts, rotation and translation. The latter just arises from the change in height of the yo-yo, with [tex]v = a\dot\theta - \dot{Z}[/tex]. But there is also the rotation of the yo-yo around its axis, specified by [tex]\omega = \dot\theta[/tex] and the moment of inertia [tex]I = \frac{1}{2}ma^2[/tex]. When you put all that together, the kinetic energy comes out to be slightly different than what you wrote.
 
  • #3
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So [tex]L = T-V = [1/2m(a\dot{\theta}-\dot{Z})^2 + 1/4ma^2\dot{\theta}^2] - [-mg(a\theta-Z)][/tex] ?

[tex]\frac{d}{dt}\frac{dL}{d\dot{\theta}} = ma(a\ddot{\theta}-\ddot{Z})+1/2ma^2\dot{\theta}[/tex](*)
[tex]\frac{dL}{d\theta} = mga[/tex](**)

(*)-(**)=0

I'm getting [tex]\ddot{\theta} = \frac{2}{3a} (\ddot{Z} - g)[/tex].

Then the linear acceleration of the yo-yo relative to the support is,

[tex]\ddot{z} = a \ddot{\theta} = 2/3(\ddot{Z}+g)[/tex]

So relative to the ground I'd have acceleration,

[tex]\ddot{z}' = 2/3(\ddot{Z}+g) - \ddot{Z} = 1/3(2g-\ddot{Z})[/tex]

So the upwards acceleration would be,

[tex]- \ddot{z} ' = 1/3(\ddot{Z}-2g)[/tex]

Does this look ok?
 
Last edited:
  • #4
diazona
Homework Helper
2,175
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Just a couple of typos:
[tex]L = T-V = [1/2m(a\dot{\theta}-\dot{Z})^2 + 1/4ma^2\dot{\theta}^2] - [-mg(a\theta-Z)][/tex]
and
[tex]\ddot{z}\,' = 2/3(\ddot{Z}+g) - \ddot{Z} = 1/3(2g-\ddot{Z})[/tex]
Other than that it looks reasonable. Note that the upwards acceleration will be [tex]+\ddot{z}\,'[/tex] if [tex]z'[/tex] is defined such that it increases upwards (which is what it looks like you did), or [tex]-\ddot{z}\,'[/tex] if [tex]z'[/tex] is defined such that it increases downwards.
 
  • #5
782
1
Just a couple of typos:
[tex]L = T-V = [1/2m(a\dot{\theta}-\dot{Z})^2 + 1/4ma^2\dot{\theta}^2] - [-mg(a\theta-Z)][/tex]
and
[tex]\ddot{z}\,' = 2/3(\ddot{Z}+g) - \ddot{Z} = 1/3(2g-\ddot{Z})[/tex]
Other than that it looks reasonable. Note that the upwards acceleration will be [tex]+\ddot{z}\,'[/tex] if [tex]z'[/tex] is defined such that it increases upwards (which is what it looks like you did), or [tex]-\ddot{z}\,'[/tex] if [tex]z'[/tex] is defined such that it increases downwards.
Thanks, fixed them. I just assigned the variable `z' randomly (yeah, sloppy hehe) but thanks for the reminder.

If I wanted to now find total energy at a time `t' should I use the [tex]\ddot{\theta}[/tex] I just found and integrate twice over t'=0..t then sub into T+V as a have defined above? Looks like a lot of algebra heh.
 

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