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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.

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]

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

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!

## 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!

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