Lagrange - yo-yo with moving support

[cscott]

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 Z(t) 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.

V = -mg(a\theta - Z)

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

L = T-V

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

The Attempt at a Solution

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

\frac{dL}{d\theta} = mga

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

Therefore upwards acceleration of the yo-yo is,

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

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

\ddot{z} = 1/3(\ddot{Z}-2g).

 

[diazona]

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 v = a\dot\theta - \dot{Z}. But there is also the rotation of the yo-yo around its axis, specified by \omega = \dot\theta and the moment of inertia I = \frac{1}{2}ma^2. When you put all that together, the kinetic energy comes out to be slightly different than what you wrote.

 

[cscott]

So L = T-V = [1/2m(a\dot{\theta}-\dot{Z})^2 + 1/4ma^2\dot{\theta}^2] - [-mg(a\theta-Z)] ?

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

(*)-(**)=0

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

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

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

So relative to the ground I'd have acceleration,

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

So the upwards acceleration would be,

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

Does this look ok?

 

[diazona]

Just a couple of typos:
L = T-V = [1/2m(a\dot{\theta}-\dot{Z})^2 + 1/4ma^2\dot{\theta}^2] - [-mg(a\theta-Z)]
and
\ddot{z}\,' = 2/3(\ddot{Z}+g) - \ddot{Z} = 1/3(2g-\ddot{Z})
Other than that it looks reasonable. Note that the upwards acceleration will be +\ddot{z}\,' if z' is defined such that it increases upwards (which is what it looks like you did), or -\ddot{z}\,' if z' is defined such that it increases downwards.

 

[cscott]

Just a couple of typos:
L = T-V = [1/2m(a\dot{\theta}-\dot{Z})^2 + 1/4ma^2\dot{\theta}^2] - [-mg(a\theta-Z)]
and
\ddot{z}\,' = 2/3(\ddot{Z}+g) - \ddot{Z} = 1/3(2g-\ddot{Z})
Other than that it looks reasonable. Note that the upwards acceleration will be +\ddot{z}\,' if z' is defined such that it increases upwards (which is what it looks like you did), or -\ddot{z}\,' if z' 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 \ddot{\theta} 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.