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## 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

## Homework Equations

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 ?