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Vexing equations of motion problem

  1. Dec 11, 2004 #1
    The question:

    A 2-kg point mass is welded on the interior of a 2-kg thin ring at point P. The ring has a radius R = 160 mm and rolls on the surface without slipping.

    (a) Draw a free body diagram for the ring and point mass. Develop the equations of motion for the system.

    (b) Combine the equations of motion derived in (a) into a single (nonlinear) ODE governing θ (t). Assuming a small angle θ (t), develop a linearized equation of motion for the system and calculate the natural frequency and also the free response when θ (0) = 25° and dθ/ dt (0) = 0.

    With the origin placed at the center of the ring (axis of rotation), theta is measured from the -y axis in the ccw direction (from south to east if you will)

    My approach:
    (a) FBD was drawn.

    Forces:
    [tex]\vec N=N\hat J [/tex] Normal force acting up.

    [tex]\vec W=-mg\hat J [/tex] weight force acting down from the center of mass.

    [tex]\vec F_f=f\hat I [/tex] force of friction acting at the contact point.

    Point C is defined as the center of mass

    Point O is the origin

    Point P is the point on the ring where the point mass is located

    Point A is the point where the normal force and frictional forces act.

    The point mass is on the ring thus the center of mass of the ring/mass body is at [itex]R/2[/itex]

    [tex]\vec r_{C/O}=(\sin\theta\hat I-\cos\theta\hat J)R/2[/tex]

    M=total mass=m1+m2

    Acceleration of point C was determined thusly
    [tex]\vec a_C=\vec a_O+ \ddot \theta \times \vec r_{C/O}-\dot \theta^2\vec r_{C/O} [/tex]


    [tex]\vec a_O=\ddot \theta r \hat I [/tex]


    [tex]\ddot \theta \times \vec r_{C/O}=\begin{vmatrix} \hat I & \hat J & \hat K\\ 0 & 0 & -\ddot \theta\\ \sin \theta R/2&-\cos \theta R/2& 0 \end{vmatrix}=-\ddot \theta\cos\theta R/2\hat I+\ddot\theta\sin\theta R/2\hat J [/tex]

    [tex]-\dot \theta^2\vec r_{C/O}=-\dot \theta^2\sin\theta R/2\hat I+\dot \theta^2\cos\theta R/2\hat J [/tex]

    Summing the forces and moments about C:

    I: [tex]f=M(\ddot \theta R -\ddot \theta\cos\theta R/2-\dot \theta^2\sin\theta R/2)[/tex]

    J: [tex]N-Mg=M(\ddot\theta\sin\theta R/2+\dot \theta^2\cos\theta R/2)[/tex]

    K: [tex]f(1-\frac{\cos\theta}{2})R-N\sin\theta R=\ddot\theta I_T [/tex]

    Two questions thus far: Is my approach correct and how do I determine the total moment of inertia([itex]I_T[/itex])?

    Thanks

    [edit] Silly me: the moment of inertia about c should be [itex]I_T=m_1R^2+m_1\frac{R^2}{4}+m_2R^2[/itex]

    thus [tex]I_T=m\frac{9R^2}{4}[/tex] Is this correct?
     
    Last edited: Dec 11, 2004
  2. jcsd
  3. Dec 11, 2004 #2
    Ok, simplifying the above I get:

    [tex]\ddot\theta+\dot\theta^2\frac{4\sin\theta+4\cos\theta}{1+4\sin\theta+4\cos\theta}+\frac{8g}{R(1+4\sin\theta+4\cos\theta)}=0[/tex]

    Thus the linear function for small angles of theta becomes:

    [tex]\ddot\theta+\frac{8g}{R(5+4\theta)}=0[/tex]

    Is this correct?

    Thanks
     
    Last edited: Dec 11, 2004
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