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planck42
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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.
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}}
a_{C}=\omega^{2}r
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!
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!
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