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.