Dynamics: Angular Acceleration Problem

[sandmanvgc]

Homework Statement

At this point both point A and block B are moving in parallel, as well as the question stating angular velocity is 0, my assumption was due to both ends moving in same direction parallel that the rod is not rotating at this instance and the answer is a. angular acceleration = 0. This does not line up with the answer key and I'm at a loss at this point. Answer Key is b. Am I misinterpreting the question?

Relevant Equations

Angular Acceleration = r*sqr(angular velocity)

Both point A and B are moving in parallel in same direction, therefore rod is not rotating at this instance and angular acceleration is 0. Question states angular velocity is equal to zero.

Plugging into Angular Acceleration"AB" = r*sqr(angular velocity"AB") = 0.26m* sqr(0) = 0

[Moderator's note: Approved.]

 

[BvU]

Hello sandman, $\qquad$ $\qquad$ !

Question states angular velocity is equal to zero

Agreed.

But I don't understand where your relevant equation comes from: it doesn't even match dimensionally.

How about $\alpha = {d\omega\over dt }$ ?

$\$

 

[sandmanvgc]

Hello sandman, $\qquad$ $\qquad$ !
Agreed.

But I don't understand where your relevant equation comes from: it doesn't even match dimensionally.

How about $\alpha = {d\omega\over dt }$ ?

$\$

Would I solve for tangential acceleration and use
$a_t=R\alpha$?

 

[BvU]

Of what wrt to what ?

 

[Chestermiller]

In my judgment, this is not such a simple kinematics problem. For the initial geometry, drop a vertical line from point A, and draw a horizontal line to the left from point B. Where these lines cross, take it as the origin of a Cartesian coordinate system. Let the length of the rod be L, and let the angle that the rod makes with the vertical at any time be $\phi$. Then, at time zero, $$x_A=0$$$$y_A=L\cos{\phi_0}$$$$x_B=L\sin{\phi_0}$$and $$y_B=0$$where $\phi_0$ is the initial value of $\phi$.

Let R be the radius of the wheel. Then at time t, the coordinates of points A and B are: $$x_A=R\cos{\omega t}-R$$$$y_A=L\cos{\phi_0}+R\sin{\omega t}$$$$x_B=L\sin{\phi_0}$$$$y_B=y$$where y is next to be determined.

OK so far??

 

[sandmanvgc]

In my judgment, this is not such a simple kinematics problem. For the initial geometry, drop a vertical line from point A, and draw a horizontal line to the left from point B. Where these lines cross, take it as the origin of a Cartesian coordinate system. Let the length of the rod be L, and let the angle that the rod makes with the vertical at any time be $\phi$. Then, at time zero, $$x_A=0$$$$y_A=L\cos{\phi_0}$$$$x_B=L\sin{\phi_0}$$and $$y_B=0$$where $\phi_0$ is the initial value of $\phi$.

Let R be the radius of the wheel. Then at time t, the coordinates of points A and B are: $$x_A=R\cos{\omega t}-R$$$$y_A=L\cos{\phi_0}+R\sin{\omega t}$$$$x_B=L\sin{\phi_0}$$$$y_B=y$$where y is next to be determined.

OK so far??

I'm assuming I overlooked the problem stating the wheel rotates at a constant velocity. Would I calculate the angular velocity of member AB while the wheel is at a different position, then with two angular velocities and the fact that its spinning consistent speed, I would be able to take time from $\phi_0$ to $\phi_t$ and plug into angular acceleration equation from BvU's post?

I'm following so far

 

[Chestermiller]

I'm assuming I overlooked the problem stating the wheel rotates at a constant velocity. Would I calculate the angular velocity of member AB while the wheel is at a different position, then with two angular velocities and the fact that its spinning consistent speed, I would be able to take time from $\phi_0$ to $\phi_t$ and plug into angular acceleration equation from BvU's post?

I'm following so far

I don't quite get what you are saying in the first paragraph, but if you are following my approach so far, I'll continue.

For short times, we can expand $\cos{\omega t}$ and $\sin{\omega t}$ up to quadratic terms in time to obtain:
$$\cos{\omega t}=1-\frac{(\omega t)^2}{2}$$and$$\sin{\omega t}=\omega t$$Substituting these into our previous equations for the coordinates of points A and B at time t yields:
$$x_A=-R\frac{(\omega t)^2}{2}$$$$y_A=L\cos{\phi_0}+R\omega t$$$$x_B=L\sin{\phi_0}$$$$y_B=y$$The length of the rigid rod does not change with time, so the distance between points A and B at time t must remain at the original rod length L. This means, according to the Pythagorean theorem that $$(x_A-x_B)^2+(y_A-y_B)^2=L^2$$or, substituting, that
$$\left(L\sin{\phi_0}+R\frac{(\omega t)^2}{2}\right)^2+(L\cos{\phi_0}-y^*)^2=L^2$$where $$y^*=y-R\omega t$$

Now, please rewrite this equation with the two squared terms in parenthesis expanded out, neglecting the terms that are the squares of $R\frac{(\omega t)^2}{2}$ and $y^*$ (as being higher than quadratic in time t). This will enable you to solve for the parameter $y^*$ as a function of t, up to quadratic terms in t.

After you complete this satisfactorily, we will continue.

 

[sandmanvgc]

Step 1. Expand
$$\left(R\frac{(\omega t)^2}{2}\right)^2 + 2L\sin{\phi_0}\left(R\frac{(\omega t)^2}{2}\right)+L^2\sin^2{\phi_0} + L^2\cos^2{\phi_0}+(y^*)^2 -2Ly^*\cos{\phi_0}=L^2$$
Step 2. Combine like terms
$$\left(R\frac{(\omega t)^2}{2}\right)^2 + 2L\sin{\phi_0}\left(R\frac{(\omega t)^2}{2}\right)+L^2(\sin^2{\phi_0} + \cos^2{\phi_0})+(y^*)^2 -2Ly^*\cos{\phi_0}=L^2$$
Step 3. Sin and Cos identity means L^2 terms cancel
$$\left(R\frac{(\omega t)^2}{2}\right)^2 + 2L\sin{\phi_0}\left(R\frac{(\omega t)^2}{2}\right)+(y^*)^2 -2Ly^*\cos{\phi_0}=0$$

 

[Chestermiller]

Step 1. Expand
$$\left(R\frac{(\omega t)^2}{2}\right)^2 + 2L\sin{\phi_0}\left(R\frac{(\omega t)^2}{2}\right)+L^2\sin^2{\phi_0} + L^2\cos^2{\phi_0}+(y^*)^2 -2Ly^*\cos{\phi_0}=L^2$$
Step 2. Combine like terms
$$\left(R\frac{(\omega t)^2}{2}\right)^2 + 2L\sin{\phi_0}\left(R\frac{(\omega t)^2}{2}\right)+L^2(\sin^2{\phi_0} + \cos^2{\phi_0})+(y^*)^2 -2Ly^*\cos{\phi_0}=L^2$$
Step 3. Sin and Cos identity means L^2 terms cancel
$$\left(R\frac{(\omega t)^2}{2}\right)^2 + 2L\sin{\phi_0}\left(R\frac{(\omega t)^2}{2}\right)+(y^*)^2 -2Ly^*\cos{\phi_0}=0$$

If you drop the two quadratic terms that I indicated you can neglect, you obtain:
$$ 2L\sin{\phi_0}\left(R\frac{(\omega t)^2}{2}\right) -2Ly^*\cos{\phi_0}=0$$
From this, what is your solution for y*, and then, what is your solution for y?

 

[sandmanvgc]

If you drop the two quadratic terms that I indicated you can neglect, you obtain:
$$ 2L\sin{\phi_0}\left(R\frac{(\omega t)^2}{2}\right) -2Ly^*\cos{\phi_0}=0$$
From this, what is your solution for y*, and then, what is your solution for y?

$$ 2(.26)(5/13)\left(R\frac{(\omega t)^2}{2}\right) -2(0.26)y^*(12/13)=0$$

At this point I'm lost what would plug into term $$\left(R\frac{(\omega t)^2}{2}\right)$$

EDIT: Unless looking from earlier posts $$x_a=0=\left(R\frac{(\omega t)^2}{2}\right)$$
y* would be zero and y would be a function of rotation of wheel?

$$y = (6/10)(\omega t)$$
?

 

[Chestermiller]

$$ 2(.26)(5/13)\left(R\frac{(\omega t)^2}{2}\right) -2(0.26)y^*(12/13)=0$$

At this point I'm lost what would plug into term $$\left(R\frac{(\omega t)^2}{2}\right)$$

EDIT: Unless looking from earlier posts $$x_a=0=\left(R\frac{(\omega t)^2}{2}\right)$$?

Let's. continue to keep it algebraic for now. Solving for y* yields: $$y^*=R\frac{(\omega t)^2}{2}\tan{\phi_0}$$So $$y=R\omega t+R\frac{(\omega t)^2}{2}\tan{\phi_0}$$From this, the coordinates of our two points at time t now become: $$x_A=-R\frac{(\omega t)^2}{2}$$$$y_A=L\cos{\phi_0}+R\omega t$$$$x_B=L\sin{\phi_0}$$$$y_B=R\omega t+R\frac{(\omega t)^2}{2}\tan{\phi_0}$$The cosine and sine of the angle $\phi$ (of the rod) at time t are given by:
$$\cos{\phi}=\frac{y_A-y_B}{L}$$$$\sin{\phi}=\frac{x_B-x_A}{L}$$What do you get if you substitute the previous four equations for the coordinates into these equations?

 

[sandmanvgc]

Let's. continue to keep it algebraic for now. Solving for y* yields: $$y^*=R\frac{(\omega t)^2}{2}\tan{\phi_0}$$So $$y=R\omega t+R\frac{(\omega t)^2}{2}\tan{\phi_0}$$From this, the coordinates of our two points at time t now become: $$x_A=-R\frac{(\omega t)^2}{2}$$$$y_A=L\cos{\phi_0}+R\omega t$$$$x_B=L\sin{\phi_0}$$$$y_B=R\omega t+R\frac{(\omega t)^2}{2}\tan{\phi_0}$$The cosine and sine of the angle $\phi$ (of the rod) at time t are given by:
$$\cos{\phi}=\frac{y_A-y_B}{L}$$$$\sin{\phi}=\frac{x_B-x_A}{L}$$What do you get if you substitute the previous four equations for the coordinates into these equations?

$$\sin{\phi_0}=(L\cos{\phi_0} -R\frac{(\omega t)^2}{2}\tan{\phi_0})/L$$

$$\cos{\phi_0}=(L\sin{\phi_0} +R\frac{(\omega t)^2}{2})/L$$

 

[Chestermiller]

$$\sin{\phi_0}=(L\cos{\phi_0} -R\frac{(\omega t)^2}{2}\tan{\phi_0})/L$$

$$\cos{\phi_0}=(L\sin{\phi_0} +R\frac{(\omega t)^2}{2})/L$$

These are not correct. They should read:
$$\cos{\phi}=(L\cos{\phi_0} -R\frac{(\omega t)^2}{2}\tan{\phi_0})/L$$$$\sin{\phi}=(L\sin{\phi_0} +R\frac{(\omega t)^2}{2})/L$$or, equivalently,
$$\cos{\phi}=\cos{\phi_0} -\tan{\phi_0}\frac{R\omega^2}{L} \frac{t^2}{2}$$$$\sin{\phi}=\sin{\phi_0} +\frac{R\omega^2}{L} \frac{t^2}{2}$$
Next, please write down the time derivative of each of these two equations.

 

[sandmanvgc]

These are not correct. They should read:
$$\cos{\phi}=(L\cos{\phi_0} -R\frac{(\omega t)^2}{2}\tan{\phi_0})/L$$$$\sin{\phi}=(L\sin{\phi_0} +R\frac{(\omega t)^2}{2})/L$$or, equivalently,
$$\cos{\phi}=\cos{\phi_0} -\tan{\phi_0}\frac{R\omega^2}{L} \frac{t^2}{2}$$$$\sin{\phi}=\sin{\phi_0} +\frac{R\omega^2}{L} \frac{t^2}{2}$$
Next, please write down the time derivative of each of these two equations.

$$\cos{\phi}dt=\cos{\phi_0}dt -\tan{\phi_0}\frac{R\omega^2}{L}{t}$$

$$\sin{\phi}dt=\sin{\phi_0}dt +\frac{R\omega^2}{L}{t}$$

 

[Chestermiller]

$$\cos{\phi}dt=\cos{\phi_0}dt -\tan{\phi_0}\frac{R\omega^2}{L}{t}$$

$$\sin{\phi}dt=\sin{\phi_0}dt +\frac{R\omega^2}{L}{t}$$

These are differentiated incorrectly. The correct results are:
$$-\sin{\phi}\frac{d\phi}{dt}= -\tan{\phi_0}\frac{R\omega^2}{L}{t}$$or $$\sin{\phi}\frac{d\phi}{dt}= \tan{\phi_0}\frac{R\omega^2}{L}{t}\tag{1}$$and
$$\cos{\phi}\frac{d\phi}{dt}= \frac{R\omega^2}{L}{t}\tag{2}$$

Next, multiply the left hand side of Eqn. 1 by $\sin{\phi}$ and the right hand side by its equivalent, $\sin{\phi_0} +\frac{R\omega^2}{L} \frac{t^2}{2}$. Then do the analogous thing with Eqn. 2 for $\cos{\phi}$. What do you get?

 

[Chestermiller]

I can see that the OP has lost interest in this problem (or lost confidence in the approach we are using), so I will complete this analysis on my own. Following what I asked the OP to do in my previous post, we have: $$\sin^2{\phi}\frac{d\phi}{dt}= \tan{\phi_0}\frac{R\omega^2}{L}{t}\left(\sin{\phi_0} +\frac{R\omega^2}{L} \frac{t^2}{2}\right)=\frac{\sin^2{\phi_0}}{\cos{\phi_0}}\frac{R\omega^2}{L}{t}+\tan{\phi_0}\left(\frac{R\omega^2}{L}\right)^2 \frac{t^3}{2}$$and$$\cos^2{\phi}\frac{d\phi}{dt}= \frac{R\omega^2}{L}{t}\left(\cos{\phi_0} -\tan{\phi_0}\frac{R\omega^2}{L} \frac{t^2}{2}\right)=\cos{\phi_0}\frac{R\omega^2}{L}{t}-\tan{\phi_0}\left(\frac{R\omega^2}{L}\right)^2 \frac{t^3}{2}$$If we add these two equations together, we obtain the angular velocity of the member AB at short times: $$\frac{d\phi}{dt}=\frac{R\omega^2}{L\cos{\phi_0}}{t}$$The angular acceleration of the rod at short times is obtained by taking the time derivative of this equation:$$\alpha=\frac{d^2\phi}{dt^2}=\frac{R\omega^2}{L\cos{\phi_0}}$$
For the parameter values in this problem, this gives an angular acceleration at time zero of 9 rad/sec^2.

 

[sandmanvgc]

Due to the holiday I have been tied up and regret my absence and appreciate Chestermiller's effort to carry out the remainder of the work. My original thought from my earlier post was to use the fact that wheel is rotating consistently to try the following to solve the problem. Attachment below shows diagram in relation to my method.

$$V_Bx=V_Ax+\frac{160}{260}V_\frac{B}{A}$$

$$0=0.36m/s+\frac{160}{260}V_\frac{B}{A}$$

$$V_\frac{B}{A}=0.36m/s\frac{260}{160}$$

$$V_\frac{B}{A}=0.26mω=0.36m/s\frac{260}{160}$$

$$ω=2.25rad/s$$​

Then taking the time it would take point A to move from my drawn point to the point in the original drawing from the problem.

$$\frac{pi/2}{6rad/s}=0.261799sec$$

$$\alpha=\frac{\delta\omega}{\delta t}$$

$$α=\frac{2.25rad/s−0}{0−0.261799s}=8.6\frac{rad}{s^2}counterclockwise$$​

While method above comes close it doesn't get it exactly right as Chestermiller's process does and I was wandering why my method isn't coming as close , if I messed up a calculation, or if it is entirely not practical?

 

[vela]

While method above comes close, it doesn't get it exactly right as Chestermiller's process does, and I was wondering why my method isn't coming as close. If I messed up a calculation, or if it is entirely not practical?

For one thing, the problem is asking for the instantaneous angular acceleration while your method is finding an average angular acceleration.

I found a method that's quite a bit simpler than Chet's. Start with $\vec{r}_\text{A} = \vec{r}_\text{B} + \vec {r}_\text{A/B}$ where $\vec{r}_\text{A}$ is the position of point A, $\vec{r}$ is the position of point B, and $\vec {r}_\text{A/B}$ is the displacement of A relative to B. The vector $\vec {r}_\text{A/B}$ only changes if the rod rotates, hence you can show that
$$\vec a_\text{A} = \vec{a}_\text{B} + \vec \alpha \times \vec r_\text{A/B}$$ when $\vec \omega =0$. Then consider the $x$-component of the equation.

 

[sandmanvgc]

Is this what your referring to?$$\vec a_\text{A} = 0.6rad(6^2)\frac{rad^2}{s^2} = 21.6 i = \vec a_\text{B}j + \vec \alpha_\text{A/B} k \times \vec r[\frac{5}{13}i +\frac{12}{13}j]$$

$$21.6 i = \vec a_\text{B}j + \vec \alpha_\text{A/B} k \times [\frac{130}{1300}i +\frac{312}{1300}j]$$

$$21.6 i = \vec a_\text{B}j -\frac{3\vec \alpha_\text{A/B}}{13}i +\frac{\vec \alpha_\text{A/B}}{10}j$$

x component

$$21.6 i = -\frac{3\vec \alpha_\text{A/B}}{13}i $$

solving this isn't matching the answer

 

[vela]

$$\vec a_\text{A} = 0.6rad(6^2)\frac{rad^2}{s^2} = 21.6 i = \vec a_\text{B}j + \vec a_\text{A/B} k \times \vec r[\frac{5}{13}i +\frac{12}{13}j]$$

$$\underbrace{\vec a_\text{A}}_\text{vector} = \underbrace{0.6rad(6^2)\frac{rad^2}{s^2}}_\text{scalar, units, conversion} = \underbrace{\vec a_\text{B}j}_\text{no idea what this means} + \underbrace{\vec a_\text{A/B} k}_\text{a? k?} \times \underbrace{\vec r[\frac{5}{13}i +\frac{12}{13}j]}_\text{no idea what this means}$$
You've made so many errors here it would be easier if you clean it up first before we proceed.

 

[sandmanvgc]

$$\underbrace{\vec a_\text{A}}_\text{vector} = \underbrace{0.6rad(6^2)\frac{rad^2}{s^2}}_\text{scalar, units, conversion} = \underbrace{\vec a_\text{B}j}_\text{no idea what this means} + \underbrace{\vec a_\text{A/B} k}_\text{a? k?} \times \underbrace{\vec r[\frac{5}{13}i +\frac{12}{13}j]}_\text{no idea what this means}$$

$$\underbrace{\vec a_\text{B}j}_\text{no idea what this means}$$ Due to B being confined to a vertical motion, I designated variable you had above with a j to show it was a y component.

$$\underbrace{\vec a_\text{A/B} k}_\text{a? k?} \times \underbrace{\vec r[\frac{5}{13}i +\frac{12}{13}j]}_\text{no idea what this means}$$
a should've been an alpha, r has been broken up into its x and y coordinates of location A with respect to B?

That was my though process to take the x-component of the equation. If that is incorrect I don't know where to go from where you started above in your original post

Edit: How should I go about cleaning it up? I fixed the alpha in the original problem as well removing the i,j,k components

$$\vec a_\text{A} = 0.6rad(6^2)\frac{rad^2}{s^2} = 21.6 = \vec a_\text{B} + \vec \alpha_\text{A/B} \times \vec r_\text{A/B}$$

 

[vela]

OK, that cleared it up a bit. You're on the right track, but it's important to use precise notation to communicate your ideas clearly.

Due to B being confined to a vertical motion, I designated variable you had above with a j to designate it was a y component.

Typically, you'd write that using notation like $a_{\text{B},y}$, the subscript indicating which component you're referring to. Note that we also drop the arrow over the $a$ since we're not referring to a vector anymore but just one component.

I inferred that $i$, $j$, and $k$ were the unit vectors, so what you seemed to mean was something like $\vec{a}\hat j$, which just mashes two vectors together and doesn't make sense.

You started with the vector equation
$$\vec{a}_\text{A} = \vec{a}_\text{B} + \vec{\alpha} \times \vec r_\text{A/B}.$$ Now if you take just the $x$-component, you get
$$a_{\text{A},x} = a_{\text{B},x} + (\vec{\alpha} \times \vec r_\text{A/B})_x,$$ where $a_{\text{A},x} = -\omega_\text{OA}^2 R$ and $a_{\text{B},x}=0$. The $a$'s, being linear accelerations, should have units of $\rm m/s^2$ (or $\rm mm/s^2$ if you don't want to bother with the conversion to meters). You should find $a_{\text{A},x} = -2.16~\rm m/s^2$. You were off by a factor of 10 and a sign.

The displacement vector is $\vec r_\text{A/B} = (260~{\rm mm})\left[-\frac 5{13} \hat i + \frac{12}{13} \hat j\right]$. You had the direction of the $x$-component backward.

$\vec \alpha$, as you indicated, points along the $z$-axis. You have to choose the sign so that the cross product yields an $x$-component that points in the $-x$ direction.

 

[sandmanvgc]

Looking back at on my scrap paper I see where I made a mistake when I toke the cross product of $\alpha_\text{A/B} \times \vec r_\text{A/B} = (260~{\rm mm})\left[-\frac 5{13} \hat i + \frac{12}{13} \hat j\right]$

I wrote down it right the first time then 25 was a little smudgy and look like a 26 so I wrote 26 down instead of former. Now going back and plugging back into the x component I'm able to get the correct answer. Thank you for the clarification in regards to the vectors.

 

[Chestermiller]

For one thing, the problem is asking for the instantaneous angular acceleration while your method is finding an average angular acceleration.

I found a method that's quite a bit simpler than Chet's. Start with $\vec{r}_\text{A} = \vec{r}_\text{B} + \vec {r}_\text{A/B}$ where $\vec{r}_\text{A}$ is the position of point A, $\vec{r}$ is the position of point B, and $\vec {r}_\text{A/B}$ is the displacement of A relative to B. The vector $\vec {r}_\text{A/B}$ only changes if the rod rotates, hence you can show that
$$\vec a_\text{A} = \vec{a}_\text{B} + \vec \alpha \times \vec r_\text{A/B}$$ when $\vec \omega =0$. Then consider the $x$-component of the equation.

I don't quite follow. Can you please flesh this out a little?

 

[vela]

We have the constraint $\lvert \vec{r}_\text{A/B} \rvert = \lvert \vec{r}_\text{A} - \vec{r}_\text{B} \rvert = L_0$, so $\vec{r}_\text{A/B}$ changes only because its direction changes. So differentiating $\vec{r}_\text{A} = \vec{r}_\text{B} + \vec {r}_\text{A/B}$ with respect to time gives
$$\vec{v}_\text{A} = \vec{v}_\text{B} + \vec{\omega}\times\vec {r}_\text{A/B}.$$ Differentiating again, we get
$$\vec{a}_\text{A} = \vec{a}_\text{B} + \vec{\alpha}\times\vec {r}_\text{A/B} + \vec{\omega}\times \frac{d\vec {r}_\text{A/B}}{dt}.$$ Since $\omega=0$ at the instant in question, the last term vanishes.

 

[sandmanvgc]

We have the constraint $\lvert \vec{r}_\text{A/B} \rvert = \lvert \vec{r}_\text{A} - \vec{r}_\text{B} \rvert = L_0$, so $\vec{r}_\text{A/B}$ changes only because its direction changes. So differentiating $\vec{r}_\text{A} = \vec{r}_\text{B} + \vec {r}_\text{A/B}$ with respect to time gives
$$\vec{v}_\text{A} = \vec{v}_\text{B} + \vec{\omega}\times\vec {r}_\text{A/B}.$$ Differentiating again, we get
$$\vec{a}_\text{A} = \vec{a}_\text{B} + \vec{\alpha}\times\vec {r}_\text{A/B} + \vec{\omega}\times \frac{d\vec {r}_\text{A/B}}{dt}.$$ Since $\omega=0$ at the instant in question, the last term vanishes.

Coming back to this, I'm actually incredibly confused now and am wondering If my answer was just a coincidence. I've been getting mixed up with remembering my drawing and the original one from problem. How would taking x component help here as in diagram both a and b are moving vertically?

 

[vela]

The velocities point upward, but you're interested in the accelerations, which depend only on how the velocities are changing. Don't confuse the direction of velocity with the direction of acceleration.

 

[sandmanvgc]

The velocities point upward, but you're interested in the accelerations, which depend only on how the velocities are changing. Don't confuse the direction of velocity with the direction of acceleration.

So just to clarify, the acceleration we solved for $\vec{a}_A = -r\omega^2$ or in the diagram below $e_r$ is always perpendicular to the motion of the object and for our case that is why A would be accelerating in what we designated as the x plane?

$\vec{a}_B$ is equal to zero only because B is confined to a vertical motion or could the following work?

$\vec{a}_B =\omega_B^2R_\text{B/A} = \omega^2R_\text{B/A} = (0^2)(R_\text{B/A}) = 0?$

 

[vela]

You can't say that a vector is equal
$$\underbrace{\vec{a}_A}_\text{magnitude and direction} \ne \underbrace{-r\omega^2}_\text{just "magnitude"}$$

or in the diagram below $e_r$ is always perpendicular to the motion of the object and for our case that is why A would be accelerating in what we designated as the x plane?

I'm not sure what you're asking. If an object follows a circular path at constant speed, which way does its acceleration point?

$\vec{a}_B$ is equal to zero only because B is confined to a vertical motion or could the following work?

$\vec{a}_\text{B}$ isn't equal to 0.

$\vec{a}_B =\omega_B^2R_\text{B/A} = \omega^2R_\text{B/A} = (0^2)(R_\text{B/A}) = 0?$

Again, this doesn't make sense because you're saying a vector equal to a scalar. It's like trying to say a matrix is equal to a number. Even if you're just referring to magnitudes here, I don't follow at all what you're thinking is. What does $\omega_B$ represent? How is it different from $\omega$?

 

[sandmanvgc]

Again, this doesn't make sense because you're saying a vector equal to a scalar. It's like trying to say a matrix is equal to a number. Even if you're just referring to magnitudes here, I don't follow at all what you're thinking is. What does $\omega_B$ represent? How is it different from $\omega$?

My question was does ${a}_\text{Bx*}=0$ only because the diagram confines it to vertical motion or does the angular velocity of the rod equal angular velocity of B with respect to A and we could use the following below to explain why ${a}_\text{Bx}=0$ has no x component as well.
$a_\text{Bx}=\omega_{B}^2R_\text{B/A} = -R_\text{B/A}\omega_B^2 = -R_\text{B/A}\omega_\text{B/A}^2 = -R_\text{B/A}(0)^2 =0?$