# Whirling mass

## Homework Statement

Mass m whirls on a frictionless table, held to a circular motion by a string which passes through a hole in the table. The string is slowly pulled through the hole so that the radius of the circle changes from $l_1$ to $l_2$ Show that the work done in pulling the string equals the increase in kinetic energy of the mass

Work and energy

## The Attempt at a Solution

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I guess that the statement "the string is slowly pulled" means that the distance of the mass from the hole cannot change much in a short time, so I assume that $\dot r = v_r = cst$ and $\ddot r = 0$.

By Newton second law in the radial direction:
$\vec T = m\ \vec a_r = m ( \ddot r - r {\dot \theta} ^2)\ \hat r= - mr {\dot \theta} ^2\hat r$

There no tangential acceleration because there is no force in that direction so :
$0 = a_\theta = r\ddot \theta + 2 \dot r \dot \theta = \frac{1}{r}\frac{d}{dt}(r^2\dot\theta)$

So the quantity $r^2\dot\theta$ is constant, and ${\dot \theta} ^2 = \frac{c^2}{r^4}$ where $c$ is constant, and rope tension is now $\vec T = -mc^2\frac{\hat r}{r^3}$

Since rope tension is a central force, its work from $l_1$ to $l_2$ is :
$W_{21} = \int_{l_1}^{l_2} T(r) dr = -mc^2 \int_{l_1}^{l_2} \frac{1}{r^3} dr = \frac{1}{2} m c^2 (\frac{1}{l_2^2} - \frac{1}{l_1^2})$

Because $c = r^2\dot\theta = r v_\theta$, and because $\dot r = v_r$ is constant, then \begin{align} W_{21} =& \frac{1}{2} m ( v_\theta^2(l_2) - v_\theta^2(l_1)) \\ = & \frac{1}{2} m ( v_r^2(l_2) + v_\theta^2(l_2) - (v_r^2(l_1) +v_\theta^2(l_1))) \\ = & \frac{1}{2} m ( v^2(l_2) - v^2(l_1)) \\ =& \triangle K \end{align}

Is that correct ?