Finding Linear Acceleration of a Yo-Yo

[kudoushinichi88]

Homework Statement

A yo-yo is made from two uniform disks, each with mass m and radius R, connected by a light axle of radius b. A light, thin string is wound several times around the axle and then held stationary while the yo-yo is released from rest, dropping as the string unwinds. Find the linear acceleration of the yo-yo.

Homework Equations

<br /> \tau=TR=I\alpha

<br /> F=ma

The Attempt at a Solution

\tau=TR=I\alpha

Tb=2\left(\frac{1}{2}mR^2\right)\alpha

since a_{tan}=r\alpha,substituted into the equation above and simplified,

<br /> Tb=mRa ...1

The yo-yo is accelerating downwards linearly, so

<br /> 2mg-T=2ma ...2

Solving for T in eq.1 and substituting into eq.2,

<br /> 2mg-\frac{mRa}{b}=2ma

Solving for a, I got

<br /> a=\frac{2g}{2+R/b}

which is not the right answer... the correct answer is
<br /> a=\frac{2g}{2+(R/b)^2}

what did I do wrong??

 

[tiny-tim]

Hi kudoushinichi88!

(I haven't actually checked your equations, but …)

wouldn't it be easier to use conservation of energy?

 

[rl.bhat]

String unwinds around the cylinder of radius b. So a(tan) = b*α

 

[willem2]

Tb=2\left(\frac{1}{2}mR^2\right)\alpha

since a_{tan}=r\alpha,substituted into the equation above and simplified,

<br /> Tb=mRa ...1

You substituted the wrong value for r. If the yoyo unwinds with angular speed \omega
the vertical speed of the yoyo is b \omega and not R \omega

 

[kudoushinichi88]

ah... so that's why!
Thank you all! This has also helped me to find the angular acceleration and tension in the string...