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Angular acceleration of a rotating point mass

  1. Dec 2, 2014 #1

    guv

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    Hi,first time posting here, so please be gentle... I am studying a point mass doing uniform circular motion on a horizontal frictionless table with tension in a string fixed at one end as the center. Everything is clear if center of circle is treated as center of rotation 'O' for the $$\tau = I \alpha$$ dynamics equation. It shows that relative to the center, angular acceleration is 0.

    However, if I select an arbitrary point 'S' on the edge of the circle and re-establish the equation, I have a contradiction. From geometry consideration, $$\theta_S = 1/2 \theta_O$$, $$\omega_S = 1/2 \omega_O$$, finally since it's UCM, $$\omega_O$$ is a constant, therefore, $$\alpha_S = 0$$. But when I use the dynamics equation $$\tau_S = I_S \alpha_S$$, the torque term is clearly non-zero $$\tau_S = \vec r \times \vec T$$, therefore $$\alpha_S$$ cannot be zero...

    Could someone enlighten me where my thoughts are flawed? Thanks,
     
    Last edited: Dec 2, 2014
  2. jcsd
  3. Dec 2, 2014 #2

    A.T.

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    This case is really not that hard to visualize, and the above is obviously wrong.
     
  4. Dec 2, 2014 #3

    guv

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    I am not seeing it for UCM... The angular velocity is constant... I know alpha is not zero but the geometrical view suggests otherwise... What is missing?
     
  5. Dec 3, 2014 #4

    guv

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    The kinematics/geometric derivation of the angular acceleration of a point mass wrt S on the edge of the circle for uniform circular motion

    $$\theta_S = 1/2 \theta_O$$
    $$\dot \theta_S = 1/2 \dot \theta_O$$
    $$\ddot \theta_S = 1/2 \ddot \theta_O = 0$$

    The dynamics derivation:
    $$ \tau_S = I_S \cdot \alpha_S$$
    $$ TR sin(\theta_S) = m (R^2 + R^2 + 2 R R cos(\theta_O)) \alpha_S $$
    $$ \ddot \theta_S = \alpha_S \neq 0 $$
     
  6. Dec 3, 2014 #5

    A.T.

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    Explain that please.
     
  7. Dec 3, 2014 #6

    guv

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    Because the angle on the edge corresponding to an arc is half of the central angle corresponding to the same arc. It's a geometry property of circle. It can be proven by using facts of external angle and isosceles of triangle.
     
  8. Dec 3, 2014 #7

    A.T.

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    That is only true if the point on the edge is opposite of the arc. You need the edge angle as function of the central angle for the entire circle, not just at one specific point.
     
    Last edited: Dec 3, 2014
  9. Dec 3, 2014 #8

    guv

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    It's true in general. You can prove that or find the proff online.
     
  10. Dec 3, 2014 #9

    Nathanael

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    I just want to point out, I didn't believe [itex]\theta_s=\frac{1}{2}\theta_0[/itex] and I still get the same conclusion:
    The angle as measured from the circumference (measured relative to the tangent) would be [itex]\theta_s=arctan(\frac{1-cos(\theta_0)}{sin(\theta_0)})[/itex] which differentiates to give you [itex]\omega _s=\frac{1}{2}\omega_0[/itex]

    Perhaps it is true that there is no torque? There would still be a force that changes the radial distance (unlike in the original frame of reference where the radius is constant) but why must the torque necessarily be nonzero?
     
  11. Dec 3, 2014 #10

    guv

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    Thanks for all the replies, I found the problem. The geometric/kinematics approach is correct $$\alpha_S \equiv 0$$ for UCM.

    The problem is the dynamics equation:
    $$\tau_S = I_S \alpha_S$$

    This is only true if I_S is a constant, remember it is like inertia. In general you could only say:

    $$\tau_S = \frac{d (I_S \omega_S)}{dt}$$

    It's like newton's second law has a more general form:
    $$ F = \frac{d (m v)}{dt}$$

    There is no contradiction. We cannot forget the condition to apply an equation...


    Nathanael, if you use trig double angle identity in your derived equation, you will discover $$\theta_S = 1/2 \theta_o$$
     
    Last edited: Dec 3, 2014
  12. Dec 3, 2014 #11

    Nathanael

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    Thanks for explaining that. I just wanted to point out one small typo,
    You wrote:
    Instead of:
    $$\tau_S = \frac{d (I_S \omega_S)}{dt}$$
     
  13. Dec 4, 2014 #12

    A.T.

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    I realized the same thing, when I applied this equation to a particle moving linearly, with no torque at all.
     
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