1. Nov 30, 2012

### cmmcnamara

1. The problem statement, all variables and given/known data

A disk or radius b rotates about a rod in its center with a constant angular velocity, Ω. At the disk's edge A, a pin is attached allowing for a body to be attached an freely swing during the rotation. Determine the differential equation for the angle β between the attached body and the vertical for a) A slender rod b) A half disk.

This problem is driving me mad because my answer only differ's from that of my professor's solution by a minus sign (versus plus). However I assume my solution must be wrong because the professor has used a CAS to derive the final differential equation (for the rod only).

2. Relevant equations

Newton's laws
Galilaen relativity equations
Mass moment of inertia
Product of inertia
etc

3. The attempt at a solution

For the pin at the disk's edge, it experience's simple circular motion and therefore has:

$$\vec{v_A}=-Ωb\hat{i}$$
$$\vec{a_A}=-Ω^2b\hat{j}$$

Additionally the attached body (2) experiences compound rotation defined in terms of the rates of change of the swing angle:

$$\vec{ω_2}=\dot{β}\hat{i}+Ω\hat{k}$$
$$\vec{α_2}=\ddot{β}\hat{i}+Ω\dot{β}\hat{j}$$

These are simply the kinematics of the body and should be generally true.

If I look at the free body and inertial force diagrams I can see on the FBD that the only applied forces are those contributed from the pin reactions and the weight of the body. Likewise on the IFD, only the inertial force and inertial moment are present meaning that the only way to avoid the pin reactions is to take equate the moments about the pin A on the FBD to the inertial moment on the IFD. Furthermore assuming the body is symmetric about y-z the relation can be reduced to:

$$-Mgr_{A/G(Y)}=iC_{Ax}$$

Where M is the body's mass, g is gravitational acceleration, r_A/Gy is the y component of the position vector of the pin relative to the body's mass center, and iCx represents the inertial moment's x component.

Using the definition of the inertial moment vector we have:

$$\vec{iC}=[I_A]\vec{α_2}^τ+\vec{ω_2}\times[I_A]\vec{α_2}^τ+M(\vec{r_{G/A}}\times\vec{a_A})$$

Where [I_A] is the inertia tensor. Computing all of this out results in the x-component of the inertial moment to be:

$$iC_{Ax}=I_x\ddot{β}+Ω^2I_{yz}+MΩb^2r_{G/A(z)}$$

Which when equated to the previous relation and simplified gives the following general solution which only requires modification based on the attached body's geometry:

$$I_x\ddot{β}+MΩb^2r_{G/A(z)}+Mgr_{G/A(y)}+Ω^2I_{yz}$$

From here I proceeded to plug in the geometrical information about a slender rod for part A. I found that for a rod, it's mass center would be in the middle meaning the y, and z components should be half the length times either cosine or sine of the swing angle. Ix about the pin would be the Ix about the rod's edge. Finally for the Iyz, I assumed a bar with z along its length and y perpendicular at the centroid would mean Iyz for those axes would be zero (symmetry). Since the coordinates being used are those standard, then the Iyz could be derived by transformation equations through an angle of -β. Putting all this together I got:

$$r_{G/A(y)}=\frac{1}{2}Lsinβ$$
$$r_{G/A(z)}=\frac{1}{2}Lcosβ$$
$$I_x=\frac{1}{3}ML^2$$
$$I_{yz}=-\frac{1}{6}ML^2sin(2β)$$

Plugging all this into the general solution and simplifying I find that for a slender rod:

$$\ddot{β}=Ω^2cosβ\left(sinβ-\frac{3b}{2L}\right)-\frac{3g}{2L}sinβ$$

This is exactly what the teacher has found for his solution, EXCEPT the term in parenthesis....the - should be a plus. I can't seem to find an error but I don't understand why I am wrong....I've checked all through my math and it all works out, I was hoping maybe someone could point out a problem in my reasoning throughout this that would bring me to the + sign. If I can get this to match, the second part should just be plug an chug.

NOTE: I traced the term in the parenthesis with the negative sign back to the z- distance between the centroid and pin but it can't have a negative sign, it represents the distance.....

EDIT: Nevermind, I let the distances between the pin and centroid actually be the component value of the position vector meaning that the sense has to be considered when configuring the geometry!! Delete this if you'd like!

Last edited: Nov 30, 2012
2. Nov 30, 2012

### ehild

Can you saw a picture?

ehild