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A squirrel in ring

  1. May 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A squirrel of mass m runs with a constant velocity v0 relative to the ring of radius R with moment of inertia I. the friction of the ring is proportional to it's angular velocity and is ##M_f=-k\dot{\phi}##.
    What is the equation of motion relative to the laboratory frame
    From the condition that the squirrel's velocity v0 relative to the ring is constant, find a relation between ##\dot{\phi}## and ##\dot{\theta}##
    Get a single differential equation for θ in order to find the squirrel's motion relative to the ground
    Assume there's no friction and that θ(t)<<1 then solve for θ(t)

    2. Relevant equations
    For the ring: ##M=I\cdot \ddot{\phi}##
    Solution for 2nd order homogeneous differential equation: ##\theta(t)=e^{rt}##

    3. The attempt at a solution
    The restoring force is ##mg\cdot \sin\theta## so:
    $$mg\cdot \sin\theta-k\dot{\phi}=I\cdot \ddot{\phi}$$
    The relation between ##\dot{\phi}## and ##\dot{\theta}##: ##\left(\dot{\phi}+\dot{\theta}\right)R=v_0##
    $$\rightarrow \dot{\phi}=\frac{v_0}{R}-\dot{\theta},\ \ddot{\phi}=-\ddot{\theta}$$
    I combine these two to get a single differential equation:
    $$mg\cdot \sin\theta-k\left(\frac{v_0}{R}-\dot{\theta}\right)=-I\ddot{\phi}$$
    $$\rightarrow I\ddot{\theta}+k\dot{\theta}+mg\cdot\sin\theta=\frac{kv_0}{R}$$
    With the simplifying assumptions: ##\ddot{\theta}+\frac{k}{I}\dot{\theta}=0##
    The solution to this differential equation is ##\theta=e^{rt}##
    $$\rightarrow r^2e^{rt}+\frac{k}{I}r\cdot e^{rt}=e^{rt}\left(r^2+r\frac{k}{I}\right)=0$$
    $$\rightarrow r=-\frac{k}{I},\ \theta(t)=C\cdot e^{-\frac{k}{I}t}$$
    At t=0 θ=0 →C=0 and it can't be
     

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  3. May 16, 2015 #2

    Svein

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    Is the ring vertical or horizontal?
     
  4. May 16, 2015 #3
    Vertical
     
  5. May 16, 2015 #4

    haruspex

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    You seem to have a mixture of a force and two torques there.
    You later have ##\dot {\phi}+\dot {\theta}##. Don't they oppose, as far as v is concerned?
     
  6. May 16, 2015 #5
    $$\left\{ \begin{array}{l} R\cdot mg\cdot \sin\theta-k\left(\frac{v_0}{R}-\dot{\theta}\right)=-I\ddot{\phi} \\
    \left(\dot{\phi}+\dot{\theta}\right)R=v_0 \end{array}\right.\ \rightarrow I\ddot{\theta}+k\dot{\theta}+R\cdot mg\cdot\sin\theta=\frac{kv_0}{R}$$
    But this doesn't change the final problem
     
  7. May 16, 2015 #6
    I don't understand what you mean in oppose, isn't that correct?
     
  8. May 16, 2015 #7

    TSny

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    Looks like you might have gone too far with the simplifying.

    Approximate sinθ for small θ.

    Why did you set the right hand side equal to zero? Is it because you are looking for the solution to the homogeneous equation?
     
  9. May 16, 2015 #8

    TSny

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    The squirrel exerts a tangential force f on the ring and the ring exerts a tangential reaction force on the squirrel. f is not equal to mgsinθ if the squirrel has tangential acceleration relative to the ground (i.e., ##f \neq mg \sin \theta## if ##\ddot{\theta} \neq 0##).
     
    Last edited: May 16, 2015
  10. May 16, 2015 #9

    haruspex

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    Sorry, my mistake.
     
  11. May 17, 2015 #10
    $$\left\{ \begin{array}{l} m(Rg\cdot \sin\theta+\ddot\theta R)=k\dot\phi+I\ddot\phi \\ \left(\dot{\phi}+\dot{\theta}\right)R=v_0 \end{array}\right.\ \rightarrow (mR-I)\ddot{\theta}+k\dot{\theta}+R\cdot mg\cdot\sin\theta=\frac{kv_0}{R}$$
    With the simplification:
    $$(mR-I)\ddot{\theta}+mRg\cdot\theta=0,\ \theta(t)=e^{rt}$$
    $$e^{rt}[(mR-I)r^2+mg]=0\rightarrow r^2=-\frac{mg}{mR-I}$$
    And it's a root of a negative
     
  12. May 18, 2015 #11

    TSny

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    In the first term I believe there is a sign error. Also, note that ##mR## does not have the same dimensions as ##I##.

    What happened to the ##k\dot{\theta}## term?
     
  13. May 18, 2015 #12
    $$\left(\dot{\phi}+\dot{\theta}\right)R=v_0 \ \rightarrow \dot{\phi}=\frac{v_0}{R}-\dot{\theta},\ \ddot{\phi}=-\ddot{\theta}$$
    $$mR(g\cdot\sin\theta+R\ddot\theta)=k\dot\phi+I\ddot\phi$$
    $$mR(g\cdot\sin\theta+R\ddot\theta)=k\left(\frac{v_0}{R}-\dot\theta\right)-I\ddot\phi$$
    $$(mR^2+I)\ddot\theta+k\dot\theta+mRg\cdot\sin\theta=k\frac{v_0}{R}$$
    There is no friction in the simplification so k=0:
    $$(mR^2+I)\ddot\theta+mRg\cdot\theta=0$$
    $$e^{rt}\left[ r^2(mR^2+I)+mRg\right]=0\rightarrow r^2=-\frac{mRg}{mR^2+I}$$
    Not good, negative root
     
  14. May 18, 2015 #13

    TSny

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    OK, I overlooked the part where it says to neglect the friction when solving the equation.

    Your work looks good. Why do you say that it's bad for r2 to be negative? What is the nature of the solution if r is imaginary?
     
  15. May 19, 2015 #14
    I don't know the nature of the solution if r is imaginary but i tried to solve:
    $$r=\pm i\sqrt{\frac{mRg}{mR^2+I}},\ r=\lambda\pm \mu i$$
    $$\theta=e^{\lambda t}(C_1\cdot\cos(\mu t)+C_2\cdot\sin(\mu t))$$
    $$\theta=C_1\cdot\cos\left(\sqrt{\frac{mRg}{mR^2+I}}t\right)+C_2\cdot\sin\left(\sqrt{\frac{mRg}{mR^2+I}}t\right)$$
    Initial condition θ(t=0)=0:
    $$\theta=C_1\cdot\cos\left(\sqrt{\frac{mRg}{mR^2+I}}\cdot 0 \right)+C_2\cdot\sin\left(\sqrt{\frac{mRg}{mR^2+I}}\cdot 0\right)$$
    $$\theta(t=0)=0=C_1\rightarrow C_1=0$$
    $$\theta(t)=C_2\cdot\sin\left(\sqrt{\frac{mRg}{mR^2+I}}\cdot t\right)$$
    $$\dot\theta=C_2\cdot\left(\sqrt{\frac{mRg}{mR^2+I}}\right)\cdot\cos\left(\sqrt{\frac{mRg}{mR^2+I}}\cdot t\right)$$
    Second initial condition: ##\dot\theta(t=0)=0## (at least i think it's so):
    $$0=C_2\cdot\left(\sqrt{\frac{mRg}{mR^2+I}}\right)\rightarrow C_2=0$$
    ?
     
    Last edited: May 19, 2015
  16. May 19, 2015 #15

    TSny

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    OK. In the argument of your trig functions you left out the square root symbol.

    The initial conditions are not specified in the problem. It is natural to take ##\theta(0) = 0##.
    If you also take ##\dot{\theta}(0) = 0## then that would mean the ring is spinning at ##t = 0## so that the relative speed of the squirrel and ring is ##v_0##. In this case the squirrel just remains at the bottom of the ring as the ring spins. So, it is not surprising that you get ##C_1 = C_2 = 0##. That is ##\theta(t) = 0## for all time.

    It is more interesting to take the ring to be at rest at ##t = 0##. Then ##\dot{\theta}(0) \neq 0## and you will get a nonzero value for ##C_2##.
     
  17. May 19, 2015 #16
    From:
    $$\left(\dot{\phi}+\dot{\theta}\right)R=v_0, \ \dot\phi=0\rightarrow\theta=\frac{v_0}{R}$$
    $$\frac{v_0}{R}=C_2\cdot\left(\sqrt{\frac{mRg}{mR^2+I}}\right)\rightarrow C_2=\frac{v_0}{R}\sqrt{mR^2+I}{mRg}$$
    $$\theta=\frac{v_0}{R}\sqrt{mR^2+I}{mRg}\cdot\left(\sqrt{\frac{mRg}{mR^2+I}}\right)\cdot\cos\left(\sqrt{\frac{mRg}{mR^2+I}}\cdot t\right)$$
     
  18. May 19, 2015 #17
    From:
    $$\left(\dot{\phi}+\dot{\theta}\right)R=v_0, \ \dot\phi=0\rightarrow\theta=\frac{v_0}{R}$$
    $$\frac{v_0}{R}=C_2\cdot\left(\sqrt{\frac{mRg}{mR^2+I}}\right)\rightarrow C_2=\frac{v_0}{R}\sqrt{\frac{mR^2+I}{mRg}}$$
    $$\theta=\frac{v_0}{R}\sqrt{\frac{mR^2+I}{mRg}}\cdot\left(\sqrt{\frac{mRg}{mR^2+I}}\right)\cdot\cos\left(\sqrt{\frac{mRg}{mR^2+I}}\cdot t\right)=\frac{v_0}{R}\cdot\cos\left(\sqrt{\frac{mRg}{mR^2+I}}\cdot t\right)$$
    But it's strange that at t=0 the squirrel already has initial velocity. i expected it to have only acceleration
     
  19. May 19, 2015 #18

    TSny

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    The expression for ##\theta(t)## should involve the sine function, not the cosine function.
    Yes, but it's the only way to satisfy the condition that the relative speed of the squirrel and ring is vo if the ring starts at rest. Unfortunately, the initial conditions were not specified in the problem.
     
  20. May 20, 2015 #19
    $$\theta=\frac{v_0}{R}\cdot\sin\left(\sqrt{\frac{mRg}{mR^2+I}}\cdot t\right)$$
    You intend to say that a realistic situation would be with the squirrel accelerating to v0, right?
    So θ involves sin. does it mean that the squirrel will go up and down periodically?
    Up i understand, but why down?
     
  21. May 20, 2015 #20

    TSny

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    The constant factor in front of the sine function is not correct. Note that it does not have the correct dimensions.

    The problem states that the relative speed of the squirrel and ring is a constant ##v_0##. So, I was thinking of an initial condition where the squirrel has a relative speed of ##v_0##. If the ring starts at rest, then the squirrel would need to start with an initial speed of ##v_0##. (Maybe the squirrel jumps onto the ring at t = 0 with an initial speed of ##v_0##.)
    Yes.
    When the friction is set to zero, then the differential equation is independent of ##v_0##. Also the differential equation is linear and homogeneous. So, any two solutions of the differential equation can be added together and the result will still be a solution. This is true even if the two original solutions correspond to different ##v_0##.

    One solution is the solution you found where the squirrel remains at the bottom of the ring while the ring rotates with a constant angular speed ##v_0/R##. In this solution the relative speed of the ring and squirrel is ##v_0##.

    Another solution is to let the squirrel hang on to a fixed point of the ring and let the system swing as a pendulum with an amplitude such that the speed of the squirrel at the bottom is ##v_0##. During this motion, the relative speed of the squirrel and ring is zero.

    If you add these two solutions, you get a solution where the squirrel swings back and forth like a pendulum and the ring spins such that the relative speed of the squirrel and ring is a constant ##v_0##. This corresponds to the solution you got in post #19 (once the constant is corrected).
     
    Last edited: May 20, 2015
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