faust9
Dec11-04, 01:49 PM
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?
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?