Stumped by Centripetal Acceleration?

  • Thread starter planck42
  • Start date
  • #1
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EDIT: I figured out what I did wrong. [tex]T=\frac{2{\pi}r}{v}[/tex], not [tex]\frac{2{\pi}r}{\omega}[/tex]. Now it all falls together.





Homework Statement


A small disk of mass [tex]m_1[/tex] on a frictionless table is
attached to one end of a string. The string passes through a hole in the table and an attached narrow, vertical
plastic tube. An object of mass [tex]m_2[/tex] is hung at the other end of the string. A student holding the tube makes the
disk rotate in a circle of constant radius r, while another student measures the period T(if this is too hard to mentally picture, there is a diagram in the attachment). Derive the equation
[tex]T=2\pi\sqrt{\frac{m_{1}r}{m_{2}g}}[/tex]


Homework Equations


[tex]a_{C}=\omega^{2}r[/tex]


The Attempt at a Solution


First, it is apparent that [tex]T=\frac{2{\pi}r}{\omega}[/tex]. So the question boils to what [tex]\omega[/tex] is. Applying Newton's Second Law in the radial direction gives [tex]m_{1}\omega^{2}r=m_{2}g[/tex]. This simplifies to [tex]\omega=\sqrt{\frac{m_{2}g}{m_{1}r}}[/tex]. Agh! There's an extra r in my answer!
 

Attachments

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Answers and Replies

  • #2
rl.bhat
Homework Helper
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Centripetal force acting on the rotating disc is m1*ω^2*r. It is provided by the weight of m2.
So m2*g = m1*4π^2/T^2*r
Now solve for T.
 
  • #3
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I had figured it out before you posted, rl.bhat. Thanks for offering help anyway.
 

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