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Whirling mass

  1. Jan 8, 2015 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    Work and energy

    3. The attempt at a solution

    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 ?
     
  2. jcsd
  3. Jan 8, 2015 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    YES! :)
     
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