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Math Help - [SOLVED] Find natural frequency?

  1. #1
    Super Member fardeen_gen's Avatar
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    [SOLVED] Find natural frequency?

    Find the natural frequency of the semi circular ring of mass m and radius r which rolls without slipping as shown in the figure.

    Answer:
    Spoiler:
    \frac{1}{2\pi}\sqrt{\frac{g}{(\pi - 2)}}
    Attached Thumbnails Attached Thumbnails [SOLVED] Find natural frequency?-ring.jpg  
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  2. #2
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    I haven't worked this out but I know your answer cannot be correct as it is not dimensionally consistent. You need an r in the denominator of your square root at the least.
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  3. #3
    Super Member fardeen_gen's Avatar
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    You are right. Anyone with ideas how to proceed?
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  4. #4
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    Here's how to do it:

    Spoiler:


    First compute the centre of mass of the object.

    As the object is symmetrical we know the CM lies on this line of symmetry at distance r_0 from O and so using polar coordinates we have that

    m r_0 = 2 \int_0^{\frac{\pi}{2}} \rho \, r^2 \,\sin \theta \, \mathrm{d} \theta

    where \rho is the mass per unit length of the semicircular ring. Consequently we have

    \color[rgb]{0,0,1} r_0 = \frac{2r}{\pi}

    Consider conservation of energy

    If \theta_0 is the angle by which the ring is initially displaced (about O) then subsequently we have

    \text{P.E. lost} = mg r_0 \left( \cos \theta - \cos \theta_0 \right)

    and the total K.E. is given by the addition of the rotational energy (about the C.M) with the translational energy of the C.M. Now, it is easy to show that the angular velocity about the CM is equal to that around O and that the translational velocity v = r \omega is experienced by all parts of the semicircular ring.

    So here's the tricky part: the translational energy of the CM is not simply due to a linear movement at speed v but movement on a curve given by the sum of its linear translation (due to rolling) and its rotational motion about O.

    The rotational motion of the C.M (if O were fixed) would be given by:

    v_{CM} = r_0 \omega

    and so the net component of the velocity of CM parallel to the ground, v' , is

    v' = v- v_{CM} \cos \theta = (r - r_0 \cos \theta) \omega

    and so the net translational K.E. of the CM is given by

    \text{Trans. K.E.of CM} = \frac{1}{2} m \left( \left( v_{CM} \sin \theta \right)^2 +<br />
\left(r- r_0 \cos \theta \right)^2 \omega^2 \right)
    = \frac{1}{2} m \omega^2 \left( r^2 +r_0^2 - 2 r r_0 \cos \theta \right) .

    Note that I have defined \omega to be in the opposite sense to \dot{\theta} .

    Now using the parallel axis theorem we have that

    I_{CM} = m (r^2 - r_0^2 )

    so equating PE with KE we have

    2 mg r_0 \left( \cos \theta - \cos \theta_0 \right) = m (r^2 - r_0^2 ) \omega^2 +<br />
m \omega^2 \left( r^2 +r_0^2 - 2 r r_0 \cos \theta \right) .

    After tidying up and substituting for r_0 we have

    \color[rgb]{0,0,1} \omega^2 = \frac{2g}{r} \, \frac{\cos \theta - \cos \theta_0}{\pi - 2 \cos \theta}.

    Deriving equation for \ddot{\theta}

    Differentiating wrt t and recalling \dot{\theta} = - \omega gives us:

    2 \omega \dot{\omega} = \frac{2g}{r} \frac{(\pi - 2 \cos \theta)-(\cos \theta - \cos \theta_0)\cdot (-2)}{(\pi - 2 \cos \theta)^2} \cdot (-\sin \theta) \, \dot{\theta}

    = -\frac{2g}{r} \frac{\pi - 2 \cos \theta_0}{(\pi - 2 \cos \theta)^2}  \sin \theta \, \dot{\theta}

    so

    \color[rgb]{0,0,1} \ddot{\theta} = -\frac{g}{r} \frac{\pi - 2 \cos \theta_0}{(\pi - 2 \cos \theta)^2}  \sin \theta

    which for small \theta and \theta_0 gives us:

    \ddot{\theta} = - \frac{g}{(\pi-2)r} \theta

    which is of the form

    \ddot{\theta} = -\omega_0^2 \theta

    so then

    f = \frac{1}{2 \pi} \omega_0 = {\color[rgb]{1,0,0} \frac{1}{2 \pi} \sqrt{ \frac{g}{(\pi -2)r}  }} .
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