Spring 2003 1 Mech
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A mass m is attached to the end of a string and rotating in a circle on a frictionless table, with initial kinetic energy E0. The string passes through a hole in the center of the table, and someone below is keeping the string taut and slowly pulling down, until the radius is halved.
How much work is done?
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Answer
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(1)\begin{align} W = -\int_R^{R/2} F dr \end{align}
(2)
\begin{align} F = \frac{mv^2}{r}=\frac{l^2}{mr^3} \end{align}
(3)
\begin{align} W = \frac{l^2}{2mr^2} \bigg|_{R}^{R/2} = \frac{3l^2}{2mR^2} = \frac{3}{2}mV_o^2 \end{align}
(4)
\begin{equation} W = 3 E_0 \end{equation}
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