The yo-yo

  • #1

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


[itex]
\tau=TR=I\alpha[/itex]

[itex]
F=ma[/itex]


The Attempt at a Solution


[itex]\tau=TR=I\alpha[/itex]

[itex]Tb=2\left(\frac{1}{2}mR^2\right)\alpha[/itex]

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

[itex]
Tb=mRa[/itex] ...1

The yo-yo is accelerating downwards linearly, so

[itex]
2mg-T=2ma[/itex] ...2

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

[itex]
2mg-\frac{mRa}{b}=2ma[/itex]

Solving for a, I got

[itex]
a=\frac{2g}{2+R/b}[/itex]

which is not the right answer... the correct answer is
[itex]
a=\frac{2g}{2+(R/b)^2}[/itex]

what did I do wrong??
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
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Hi kudoushinichi88! :smile:

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

wouldn't it be easier to use conservation of energy? :wink:
 
  • #3
rl.bhat
Homework Helper
4,433
7
String unwinds around the cylinder of radius b. So a(tan) = b
 
  • #4
1,982
267
[itex]Tb=2\left(\frac{1}{2}mR^2\right)\alpha[/itex]

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

[itex]
Tb=mRa[/itex] ...1
You substituted the wrong value for r. If the yoyo unwinds with angular speed [tex]\omega[/tex]
the vertical speed of the yoyo is [tex]b \omega[/tex] and not [tex]R \omega[/tex]
 
  • #5
ah... so that's why!
Thank you all! This has also helped me to find the angular acceleration and tension in the string...
 
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