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Force on a bead due to a rotating wire

  1. Jul 22, 2017 #1
    1. The problem statement, all variables and given/known data
    A bead slides without friction on a horizontal rigid wire rotating at
    constant angular speed ω. The problem is to find the force exerted on
    the bead by the wire. Neglect gravity.
    upload_2017-7-22_20-34-2.png

    2. Relevant equations


    3. The attempt at a solution

    The physical force acting on the bead is the force exerted by the rigid wire.Let's call this force ## \vec F_{ph} ##.
    Wrt. inertial frame,
    ## \vec F_{ph} = m { a_r } {\hat r } + m { a_{\theta}} {\hat {\theta} }##

    constant angular speed ω gives ## \ddot {\theta} = 0 ##
    ##a_r = \ddot r - r {\dot {\theta}}^2 ##
    ## a_{\theta} = 2{ \dot r }{\dot {\theta}} + r {\ddot {\theta}} = 2{ \dot r }{\dot {\theta}} ##
    So,
    ## \vec F_{ph} =## m( (## \ddot r - r { {\dot {\theta}}^2 }##) ## {\hat r} + 2 { \dot r } {\dot {\theta}} {\hat {\theta}} ## )

    Is this correct?

    Please, tell me where is the mistake in Latex code? I edited and corrected the code.
     
    Last edited: Jul 22, 2017
  2. jcsd
  3. Jul 22, 2017 #2

    TSny

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    It's correct so far, but there's quite a bit more you can do that will lead to an explicit expression for the force as a function of time.
     
  4. Jul 22, 2017 #3
    I corrected the Latex code.
     
    Last edited: Jul 22, 2017
  5. Jul 22, 2017 #4
    I have no more information about ## \dot r ## or ## \ddot r##. So, what more could be done?
     
    Last edited: Jul 22, 2017
  6. Jul 22, 2017 #5

    TSny

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    The wire is frictionless.
     
  7. Jul 22, 2017 #6
    This means that ##\vec F_{ph} ## has no component in the ## \hat r ## direction.

    [ Earlier, I thought that there must be some force to provide ## \ddot r##. So, I was stuck. I often do this mistake. Thanks for pointing out that there is no friction. ]

    So, ## a_r = \ddot r - r {\dot {\theta}}^2 = 0##
    ## \int_{t_i}^{t_f} \left ({\ddot r} - r {\dot {\theta}}^2 \right ) \, dt = 0 ##
    ##{\dot r}_f - {\dot r }_i = r \left (t_f - t_i \right ) {\dot {\theta}}^2 ##

    Taking ## {\dot r }_i = t_i =0##

    ##{\dot r}_f = r t_f {\dot {\theta}}^2 ##

    ##{\dot r} = r t {\dot {\theta}}^2 ##

    So,
    ## \vec F_{ph} = 2 m { \dot r } {\dot {\theta}} {\hat {\theta}} = 2 m r t {\dot {\theta}}^3 {\hat {\theta}}##

    Is this correct?

    I think the following step is wrong as r itself is a function ot t.
    ##{\dot r}_f - {\dot r }_i = r \left (t_f - t_i \right ) {\dot {\theta}}^2 ##
     
    Last edited: Jul 22, 2017
  8. Jul 22, 2017 #7

    TSny

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    Yes. Can you solve this differential equation?
    You treated ##r## as constant when integrating ##r {\dot {\theta}}^2##. But ##r## is some function of time that you are trying to find.
     
  9. Jul 22, 2017 #8
    ## a_r = \ddot r - r {\dot {\theta}}^2 = 0 ##
    ## \frac{d{\dot r}} {dt} = \frac {d{\dot r}}{dr} \frac {dr} {dt} = r {\dot {\theta}}^2 ##
    ##\dot r {d{\dot r}} = r {\dot {\theta}}^2 {dr} ##
    Integrating both sides with limit ## \dot r ## → 0 to ## \dot r ## and r →0 to r
    ## \frac {{\dot r}^2} 2 = \frac {r^2 {{\dot {\theta}}^2} }{2} ##
    ## \dot r = r \dot {\theta}##

    Is this correct so far?

    ## \vec F_{ph} = 2 m { \dot r } {\dot {\theta}} {\hat {\theta}} = 2 m r {\dot {\theta}}^2 {\hat {\theta}}##
     
    Last edited: Jul 22, 2017
  10. Jul 22, 2017 #9
    The question doesn't say to express the force as a function of time?
    So, how do you get to know beforehand that the force should be expressed as a function of time?
    Is it because the only one independent variable in this problem is time?
     
  11. Jul 22, 2017 #10

    haruspex

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    We cannot tell if you posted the whole problem as given to you.
    Clearly the force will depend on r and r will depend on t. Is there any hint as to whether you are to find the force as a function of r or of t (or....)?
     
  12. Jul 23, 2017 #11
    Actually , this is an example given in the book. But, I wonder how can one know that the force will depend on r before calculating it?

    To solve the same question w.r.t. wire frame i.e. non- inertial frame,

    we have,
    ## \vec F_{n-in }= \vec F_{ph} + \vec F_{pseudo} ##

    where ## \vec F_{n-in }## is net force acting on the system measured wrt. the non - inertial frame
    ## \vec F_{ph }## is the net physical force acting on the system measured wrt. the inertial frame
    ## \vec F_{pseudo }## is pseudo force acting on the system

    Here,
    ## \vec F_{ph }## is the contact force exerted by the wire on the bead
    ## \vec F_{n-in } = m \vec a_{n-in }##
    ## \vec a_{n-in } = \ddot r {\hat r}##
    In this frame, the bead is moving along one direction. Now, How to specify this direction? Should I say that ## \hat r ## is constant here?


    ## \vec F_{pseudo } = -m \vec \omega \times \vec \omega \times \vec r - 2 m \vec \omega \times \vec v_{n-in}##

    where ## \vec \omega ## is the angular velocity with which the non -inertial frame (here wire) is rotating
    ##\vec v_{n-in} ## is the velocity of the system wrt. non-inertial frame
    Here, ##\vec v_{n-in} = \dot r \hat r ##

    So, now using ## \vec F_{n-in }= \vec F_{ph} + \vec F_{pseudo} ## we have,

    ## m \ddot r~ {\hat r} = \vec F_{ph} - m \vec \omega \times \vec \omega \times \vec r - 2 m \vec \omega \times\dot r \hat r ##
    ## m \ddot r~ {\hat r} = \vec F_{ph} + m { \omega }^2 r~ \hat r - 2 m \omega \dot r~ \hat \theta ##

    From the above equation, it's clear that ## \vec F_{ph}## can have components only in ##\hat r## or ##\hat \theta ## direction.
    ## \vec F_{ph} ## is a contact force. Its component in ## \hat r ## direction is known as friction. And it's given that friction is 0. Thus, the direction of ## \vec F_{ph} ## is known from the equation of motion.
    So, we have ## \vec F_{ph} = 2 m \omega \dot r \hat \theta ##

    Considering eqn. of motion in ## \hat r ## direction, we have the following differential equation,
    ## \ddot r = { \omega }^2 r ##

    now , follow post # 8.
     
    Last edited: Jul 23, 2017
  13. Jul 23, 2017 #12

    haruspex

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    Looks ok.
    Now solve that to get r as a function of theta (and hence of t).
     
  14. Jul 23, 2017 #13
    ##\dot r = r\dot \theta ##
    ##\frac {dr} {dt} =r \frac {d \theta} {dt} ##
    Integrating both sides wrt t,
    ##\int \frac 1 r \frac {dr} {dt} \, dt =\int \frac {d \theta} {dt} \, dt ##
    ## \ln r = \theta + C ##
    ## r =A~ e^{\theta} = A~ e^{\omega t} ##

    ##\vec F_{ph} = 2 m { \dot r } {\dot {\theta}} {\hat {\theta}} = 2 m \omega ^2 ~ A e^{\omega t} ~{\hat {\theta}} ##
     
  15. Jul 23, 2017 #14
    What about this?
     
    Last edited: Jul 23, 2017
  16. Jul 23, 2017 #15

    TSny

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    OK.
    This is not generally true. When you integrated, your lower limits imply that ##\dot r = 0## at ##r = 0##. But this was not given in the problem as either an initial condition or a boundary condition. (If the particle starts at ##r = 0## with ##\dot r = 0## then it would remain at rest there.)

    Whether you want the force as a function of ##t## or as a function of ##r##, the answer will depend on initial conditions (or boundary conditions).
     
  17. Jul 24, 2017 #16
    Yes, what I did is wrong.
    ## a_r = \ddot r - r {\dot {\theta}}^2 = 0##
    The initial condition which I took makes ar 0.
    Hence, If the particle starts at r = 0 with ##\dot r = 0 ## then it would remain at rest there wrt. the wire frame.

    ##a_r = \ddot r - r {\dot {\theta}}^2 = 0 ##
    ##\dot r {d{\dot r}} = r {\dot {\theta}}^2 {dr}##
    Integrating both sides with limit ## \dot r ##→ vi to ##\dot r ##and r →ri to r,
    ## \frac {\{{\dot r}^2 - v_i^2\}} 2 = \frac {\{r^2 - r_i^2 \} {{\dot {\theta}}^2} }{2} ##
    ## \dot r = \sqrt{ \frac {v_i^2} 2 + \frac {\{r^2 - r_i^2 \} {{\dot {\theta}}^2} }{2} } ##

    Is this correct so far?

    But, then, I don't get r(t) as an exponential function of t.
     
    Last edited: Jul 24, 2017
  18. Jul 24, 2017 #17

    TSny

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    Yes. Note the 2's will cancel.

    If you want ##r## as a function of time, go back to the differential equation

    ##\ddot r - r {\dot {\theta}}^2 = 0##

    or

    ##\ddot r - \omega^2 r = 0##

    This is a second order differential equation for ##r(t)##. If you have studied differential equations,you should find this equation to be easy to solve
     
  19. Jul 24, 2017 #18
    So,##\dot r = \sqrt{ {v_i^2} + {\{r^2 - r_i^2 \} {{\dot {\theta}}^2} }}##
    ##\frac {dr} {dt} =\dot \theta \sqrt{c^2 + r^2} ##, where ## c^2 = \frac {{v_i}^2} {{\dot \theta}^2} -{ r_i}^2 ##
    Integrating both sides wrt t gives,
    ##\int_{r_i} ^{r_f}\frac 1 {\sqrt{c^2 + r^2} } \, dr = \int_{t_i}^{t_f} \dot \theta\, dt ##
    ## \frac {r_f} {\sqrt {{r_f}^2 + c^2} } - \frac {r_i} {\sqrt {{r_i}^2 + c^2} } =\dot \theta \{ t_f - t_i\}##

    Is this correct so far?
    Taking c2 =0 and ri=0, I should get the old result. But I don't get so. So, is there anything wrong here?
     
    Last edited: Jul 24, 2017
  20. Jul 24, 2017 #19

    TSny

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    If you want ##r(t)##, it will be much easier to solve ##\ddot r - \omega^2 r= 0## directly rather than using ##\dot r {d{\dot r}} = r {\dot {\theta}}^2 {dr}##.
     
  21. Jul 24, 2017 #20
    Yes, that I know. But, I want to know : is there anything wrong with the way I am solving it here?
     
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