# Vexing equations of motion problem

1. Dec 11, 2004

### 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

 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?

Last edited: Dec 11, 2004
2. Dec 11, 2004

### 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

Last edited: Dec 11, 2004