# Stumped by Centripetal Acceleration?

1. Sep 2, 2009

### planck42

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.

1. The problem statement, all variables and given/known data
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
$$T=2\pi\sqrt{\frac{m_{1}r}{m_{2}g}}$$

2. Relevant equations
$$a_{C}=\omega^{2}r$$

3. 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!

#### Attached Files:

• ###### diagram.doc
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Last edited: Sep 2, 2009
2. Sep 3, 2009

### rl.bhat

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. Sep 3, 2009

### planck42

I had figured it out before you posted, rl.bhat. Thanks for offering help anyway.