# Stumped by Centripetal Acceleration?

EDIT: I figured out what I did wrong. $$T=\frac{2{\pi}r}{v}$$, not $$\frac{2{\pi}r}{\omega}$$. Now it all falls together.

## Homework Statement

A small disk of mass $$m_1$$ 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 $$m_2$$ 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
$$T=2\pi\sqrt{\frac{m_{1}r}{m_{2}g}}$$

## Homework Equations

$$a_{C}=\omega^{2}r$$

## The Attempt at a Solution

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

#### Attachments

• diagram.doc
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