Find the angular acceleration of the rod

[lion_]

Homework Statement

Given: $\mu_B=0.52$

$\theta=30^{\circ}$

Weight- $25$ lb

$\omega=0$

$l=6$ ft

$r_c=3\sqrt 2$ radius of curvature.

Homework Equations

My Equations of motion are the following:

$\xleftarrow{+}\sum F_x=N_A\sin 60 - F_B=0$

$\downarrow{+} \sum F_y= N_A \cos 60-N_B+mg=m(a_G)_y$

$\circlearrowright + \sum M_G=2.60N_B-1.5F_B+2.60N_A\cos 60=I_G \alpha$

$(a_G)_y=\alpha r$

$I_G=\frac{1}{12}ml^2$

The Attempt at a Solution

I am not sure what $r$ to use for my $(a_g)_Y$ value
I have tried multipule values of $r$ such as 6,3 and $r_c$ none yield the correct answer.

 

[andrewkirk]

Why have you set the horizontal net force equal to zero in your first equation? As the rod falls, won't its centre of gravity move? For instance, the final centre of gravity will be further right than the centre of gravity when the rod's pivoting collar was at the very top of the quarter circle pipe.

Are you using $a_G$ to refer to the acceleration of the rod's centre of mass and $(a_G)_y$ to refer to the vertical component of that acceleration? If so, won't you just have $(a_G)_y=\frac{l\alpha\cos\theta}{2}$, since the centre of mass is at radius $\frac{l}{2}$ from the (moving) point on the ground about which the rod is rotating?

 

[lion_]

Why have you set the horizontal net force equal to zero in your first equation? As the rod falls, won't its centre of gravity move? For instance, the final centre of gravity will be further right than the centre of gravity when the rod's pivoting collar was at the very top of the quarter circle pipe.

Are you using $a_G$ to refer to the acceleration of the rod's centre of mass and $(a_G)_y$ to refer to the vertical component of that acceleration? If so, won't you just have $(a_G)_y=\frac{l\alpha\cos\theta}{2}$, since the centre of mass is at radius $\frac{l}{2}$ from the (moving) point on the ground about which the rod is rotating?

Let me make this a little bit clearer the rod is released from rest at 30 degrees. Now I don't have to worry about horizontal motion.

 

[andrewkirk]

The initial horizontal velocity of the CoM being zero does not entail that the horizontal accel of the CoM is zero.

Further, if we take the centre of the quarter circle as the origin, then the x coordinate of the CoM of the rod is initially $3\cdot(1-\frac{\sqrt{3}}{2})\approx 0.4$, whereas once the rod has fallen flat the x coordinate will be $3\cdot(\sqrt{2}-1)\approx 1.2$. So the CoM must move horizontally.

 

[lion_]

But at the instant theta is 30 degrees it is released from rest. So imagine a hand holding the rod and as soon as it is released the rod has no initial angular velocity. The normal component of the acceleration is obviously 0 at that instant but there is tangential acceleration of the rod.

 

[lion_]

The picture you see is the rod being released from rest at that angle.