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Spinning block with decreasing radius

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  1. May 20, 2017 #1
    1. The problem statement, all variables and given/known data

    A mass ##m## whirls around on a string which passes through a ring. Given that gravity is not present and initially the mass is at a distance ##r_0## from the center and is revolving at angular velocity ##\omega_0##. The string is pulled with constant velocity ##V## starting ##t = 0## so that radial distance to the mass decreases. Find ##\omega (t)##

    upload_2017-5-21_5-28-4.png

    2. Relevant equations


    3. The attempt at a solution

    There is no tangential forces on the block and the only radial is tension which is not required.

    Anyhow, ##r = Vt## since the velocity is constant, ##\dot r = V## and ##\ddot r= 0##.

    Pluggin those in,

    ##\vec a = \hat r(\ddot r - r \dot \theta^2 ) + \hat \theta (r \ddot \theta + 2\dot r \dot \theta)##

    ##\vec a = -\hat r (Vt \dot\theta^2) + \hat \theta (Vt \ddot \theta + 2V\dot \theta)##

    Since there are no tangential forces,

    ##Vt \ddot \theta + 2 V\dot \theta = 0 \iff t \ddot \theta = - 2\dot \theta##

    Since ##\dot \theta = \omega##

    ##\displaystyle t \dot \omega = -2 \omega \iff {\dot \omega \over \omega} = {-2 \over t} \iff \int^{\omega}_{\omega_0} {d\omega \over \omega} = \int^t_0 {-2 \over t}dt##

    Now the second integral has the closed form ##\ln |t|## which is not defined at ##t = 0##,

    Where did I go wrong ?
     
  2. jcsd
  3. May 21, 2017 #2

    TSny

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    Does this give the correct value of ##r## at ##t = 0##?
     
  4. May 21, 2017 #3
    No. But why should it since after ##t = 0 ## the rope is pulled, no ?
     
  5. May 21, 2017 #4

    ehild

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    The radius decreases, so it is not Vt.
     
  6. May 21, 2017 #5
    Angular momentum about the hole is conserved
     
    Last edited: May 21, 2017
  7. May 21, 2017 #6

    haruspex

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    t=0 is the instant at which the pulling starts. At that time, r should be its initial value still.
     
  8. May 21, 2017 #7
    Ok I did it.

    At any instance ##r = r_0 - Vt##

    Which gives ##\vec a = -(r_0 - Vt)\hat r + \hat \theta((r_0 - Vt ) \dot \theta + -2V\ddot\theta)##

    So, we get ##(r_0 - Vt ) \dot \theta = 2V\ddot\theta##

    Substituting for omega,

    ##\dfrac{\dot \omega}{\omega} = \dfrac{2V}{r_0 - Vt}##

    Integrating from ##\omega_0 \to \omega## and LHS and from ##0 \to t## on right,

    I get ##\omega = \dfrac{\omega_0 r_0^2}{(r_0 - Vt)^2}##

    Is this correct ?

    Second part of the question was to find the radial force on the block,

    Which I think is ##F_r =m(r_0 - Vt)\omega^2## and I plug from above in this formula,right ?
     
  9. May 21, 2017 #8
    Ok I did it.

    At any instance ##r = r_0 - Vt##

    Which gives ##\vec a = -(r_0 - Vt)\hat r + \hat \theta((r_0 - Vt ) \dot \theta + -2V\ddot\theta)##

    So, we get ##(r_0 - Vt ) \dot \theta = 2V\ddot\theta##

    Substituting for omega,

    ##\dfrac{\dot \omega}{\omega} = \dfrac{2V}{r_0 - Vt}##

    Integrating from ##\omega_0 \to \omega## and LHS and from ##0 \to t## on right,

    I get ##\omega = \dfrac{\omega_0 r_0^2}{(r_0 - Vt)^2}##

    Is this correct ?

    Second part of the question was to find the radial force on the block,

    Which I think is ##F_r =m(r_0 - Vt)\omega^2## and I plug from above in this formula,right ?
     
  10. May 21, 2017 #9

    haruspex

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    Yes, that all looks right.
     
  11. May 21, 2017 #10
    Will this solve the question without solving a DE ?
     
  12. May 21, 2017 #11
    I guess so
     
  13. May 21, 2017 #12
    Ok I will try.
     
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