Vexing equations of motion problem

[faust9]

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:
\vec N=N\hat J Normal force acting up.

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

\vec F_f=f\hat I 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 R/2

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

M=total mass=m1+m2

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


\vec a_O=\ddot \theta r \hat I


\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

-\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

Summing the forces and moments about C:

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

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

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

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

Thanks

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

thus I_T=m\frac{9R^2}{4} Is this correct?

 

[faust9]

Ok, simplifying the above I get:

\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

Thus the linear function for small angles of theta becomes:

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

Is this correct?

Thanks

 

[wais]

As for the approach, it seems mostly correct. However, there are a few things that could be clarified or corrected.

1. The normal force and frictional force should both act at point A, not just the normal force. This will not affect the equations of motion, but it is important for accuracy.

2. The expression for the acceleration of point C is not correct. It should be \vec{a}_C = \vec{a}_O + \ddot{\theta} \times \vec{r}_{C/O} - \dot{\theta}^2 \vec{r}_{C/O}. The term \vec{a}_O should have a magnitude of \ddot{\theta}R, not just \ddot{\theta}.

3. The expression for the moment about point C in the i direction should be f - M(\ddot{\theta}R - \ddot{\theta}\cos\theta R/2 - \dot{\theta}^2\sin\theta R/2), not what is currently written. The moment about point C in the j direction is correct.

4. The expression for the moment about point C in the k direction is not correct. It should be f(1-\cos\theta/2)R - N\sin\theta R = \ddot{\theta}I_T. This can be derived by taking the moment about point C in the k direction and setting it equal to the moment of inertia about point C times the angular acceleration.

5. The moment of inertia about point C should be m_1R^2 + m_2(R/2)^2. This can be derived by using the parallel axis theorem and calculating the moment of inertia about the center of mass of the ring and then adding the moment of inertia of the point mass about its own center of mass.

6. The moment of inertia about point C can also be written as I_T = mR^2 + m(R/2)^2 = m(9R^2/4). This is correct.

Overall, your approach is correct, but there are a few small errors that should be corrected. Once these are fixed, you should be able to solve for the equations of motion and proceed with finding the natural frequency and free response. Good luck!