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Simple Differential equation problem -- Block sliding inside a ring on a table...

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

    A block of mass m slides on a frictionless table, It is constrained to move inside a ring of radius ##l## which is fixed to the table. At ##t = 0## the block is moving along the inside of the ring with velocity ##v_0##. The coefficient of friction between the bock and ring is ##\mu##, Find the veloctiy of the block at later times.

    2. Relevant equations


    3. The attempt at a solution

    Forces due to the ring are ##N^\prime## and ##F_f##

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

    And we know ##r## is constant,

    So,
    ##\vec a = -r\dot \theta^2 \hat r + r\ddot \theta \hat \theta##

    Tangential forces are ##m a_\theta## and ##F_f##

    So, ##F## and ##F_f##, where ##F## is some forces driving the block forward against friction,

    Which gives, ##F - F_f = Mr \ddot \theta = Ml\ddot \theta## ---- (1)

    Radial force is ##N^\prime##

    ##N^\prime = Mr\ddot \theta^2 = Ml\ddot \theta^2##

    Since ##N^\prime \mu = F_f##,

    Substituting for ##N^prime## in (1)

    ##F = Ml \ddot \theta + Ml\dot\theta^2 \mu##

    What should I do now ? I know I need to solve a DE but I don't how to get that DE.
     
    Last edited by a moderator: May 20, 2017
  2. jcsd
  3. May 20, 2017 #2
    What a piece of crap I wrote here, does not make any sense.

    What I actually meant was that tangential forces are ##F## and ##F_f## not ##a_\theta## and ##F## is some forces against friction.

    I should not write late in night :). Sorry for inconvenience.
     
  4. May 20, 2017 #3

    TSny

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    OK
    OK

    As you noted in your second post, ##ma_\theta## is not a force.

    The problem does not mention any force driving the block forward. I believe you are to assume that only the friction force acts in the tangential direction.

    Why the double-dot on ##\theta##?

    Once you take care of the the fact that there is no "driving force" ##F##, you will have your DE. You might try rewriting the equation in terms of the angular velocity ##\omega = \dot{\theta}##.
     
  5. May 21, 2017 #4
    One question still remains is that how it is going to move without a driving force ?
     
  6. May 21, 2017 #5

    TSny

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    The block starts with an initial velocity. It is similar to a block sliding on a table. You give the block an initial velocity and then let friction bring it to rest. In your problem, you will see how long it takes for the block to come to rest.
     
  7. May 21, 2017 #6
    Ok I completed it as you told.

    I got,

    ##F_f = - Ml \ddot \theta \iff N^\prime = -Ml\ddot \theta \iff \mu l\dot \theta^2 M = -Ml\ddot \theta^2 = \iff \mu \omega^2 = -\dot\omega ##

    Which gives ##\displaystyle \int^t_0 \mu dt = -\int^\omega_{\omega_0} {1 \over \omega^2} d\omega \iff \mu t = \left[\frac1\omega - \frac1\omega_0 \right]##

    Now I susbtitute ##\omega_0 = v_0/l## and ##\omega = v/l## to get ##v_0 = \dfrac{v_0}{\dfrac{v_0 \mu t}{l} + 1}##.

    Is this substitution statisfied ?

    The second part of question is to find the position of the block at some other time, to do this I integrate the velocity expression that I just found with time, right ?
     
    Last edited: May 21, 2017
  8. May 21, 2017 #7

    TSny

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    That all looks good.
     
  9. May 21, 2017 #8
    :smile::smile:
     
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