Given: $\mu_B=0.52$
$\theta=30^{\circ}$
Weight- $25$ lb
$\omega=0$
$l=6$ ft
$1/\kappa=3\sqrt 2$ radius of curvature.
Find $\alpha$
My Equations of motion are the following:
$\xleftarrow{+}\sum F_x=N_A\sin 60 - F_B=0$
$\downarrow{+} \sum F_y= N_A \cos 60-N_B+mg=m(a_G)_y$
$\circlearrowright \sum M=2.60N_B-1.5F_B+2.60N_A\cos 60=I_G \alpha$
$(a_G)_y=\alpha r$
$I_G=\frac{1}{12}ml^2$
I would like to know what $r$ to use for $(a_G)_y$ I am not completely sure what point is rotating about. Did I miss anything in my equations?
Problems like this are most easily solved using conservation of energy. Let $U(\theta)$ be the potential energy and $K(\theta)$ bet the kinetic energy. I've written both expressions as functions of $\theta$ because ultimately we want $\ddot{\theta}$. Also, it is assumed that all energy is either potential or kinetic in this problem.
Due to conservation of energy, we know that $\frac{d}{dt}(K(\theta)-U(\theta)) = 0$. Hopefully, in evaluating that expression, we'll get an expression for $\ddot{\theta}$.
If the mass is distributed evenly along the bar, we can write:
$U(\theta) = \frac{Mg}{l}\int_0^l l\sin\theta ds = Mgl\sin\theta$
where the integral is over the mass through the relation $dm = \frac{M}{l}ds$.
Similarly, we can write an expression for the kinetic energy:
$K(\theta) = \frac{1}{2}\frac{M}{l}\int_0^l v(s,\theta)^2 ds$
In order to evaluate the expression for kinetic energy, it is useful to write expressions for the positions of the endpoints of the rod. If we take the first endpoint to be the one on the ground, then we have $x_1 = r-l\cos\theta$ and $y_1 = 0$ using an origin described in the next paragraph.
The second endpoint can be described by introducing an angle $\phi$ measured at an origin located at the center of curvature of the curved rail from the positive $x$ axis to the point $A$. If we let $r=1/\kappa$ be the radius of curvature, then we can write $x_2 = r\cos\phi$ and $y_2 = r\sin\phi$.
Given the definitions of the endpoints, we can write the position of any point along the rod as
$(x(s),y(s)) = (\frac{s}{l}x_1 + \frac{l-s}{l}x_2, \frac{l-s}{l}y_2) = (\frac{sr}{l} - s\cos\theta + \frac{(l-s)r}{l}\cos\phi, \frac{(l-s)r}{l}\sin\phi)$.
The angle $\phi=\phi(\theta)$ can be found using the law of sines. If we draw a straight line from the point $A$ to the base of the curved rail and label that point $C$, then we can define the angle $\gamma=\angle BCA$ and a length $b$ as the distance from $A$ to $C$. From the law of sines, we have $\frac{\sin\phi}{b} = \frac{\sin\gamma}{r}$ and $\frac{\sin\theta}{b} = \frac{\sin\gamma}{l}$. Solving for $\sin\phi$ yields $\sin\phi = \frac{l}{r}\sin\theta$.
So, to be explicit, the kinetic energy is given by
$K(\theta)=\frac{1}{2}\frac{M}{l}\int_0^l (\frac{d}{dt}x(s,\theta,\phi(\theta)))^2 + (\frac{d}{dt}(y(s,\theta,\phi(\theta)))^2 ds$
Since we have explicit expressions for $x$, $y$, and $\phi$ in terms of $s$ and $\theta$, it should be possible to carry out the integral and perform the differentiation making sure to evaluate all of the partials with respect to $\theta$. Then, all you will need to do is perform the differentiation, $\frac{d}{dt}(K(\theta)-U(\theta))$ . Also, since the expression for $K$ will contain $\dot{\theta}$, the expression for $\frac{d}{dt}(K(\theta)-U(\theta))$ should contain $\ddot{\theta}$.
Hope that gets you started in the right direction.
P.S. I just realized I neglected the work due to friction, so you'll have to take that into account as well.