How Does Angular Velocity Affect Crane Boom Dynamics?

[Dustinsfl]

The telescopic boom of a crane rotates with the angular velocity and rotation as indicated about point $A$.
At the same instant, the boom is extending with a constant speed of 0.5ft/s, measured relative to the boom.
Determine the magnitude and acceleration of the absolute acceleration of point $B$ at this instant.
I am not familiar with working with angular velocity.
[URL="http://http://img23.imageshack.us/img23/8895/telescope.png

 

[Ackbach]

You have three equations that are relevant:
\begin{align*}
s &=\theta r\\
v_{\tan} &= \omega r\\
a_{\tan} &= \alpha r.
\end{align*}
Here $s$ is arc length, and $\theta$ is measured in radians. So you can see that the pattern here is "tangential variable is corresponding angular variable times the radius". Now the absolute acceleration of point B is going to be
$$a=\sqrt{a_{\tan}^{2}+a_{c}^{2}},$$
where $a_{\tan}$ is the tangential acceleration, and $a_{c}$ is the centripetal acceleration; the centripetal acceleration is given by
$$a_{c}=\frac{v_{\tan}^{2}}{r}=\omega^{2}\,r.$$

Also note that the angular velocity is a measure of how fast the angle $\theta$ is changing. That is,
$$\omega=\frac{d\theta}{dt}.$$
Also,
$$\alpha=\frac{d\omega}{dt}.$$
These are angular analogues to the linear variable case.

Is that enough to get you started?

 

[Dustinsfl]

You have three equations that are relevant:
\begin{align*}
s &=\theta r\\
v_{\tan} &= \omega r\\
a_{\tan} &= \alpha r.
\end{align*}
Here $s$ is arc length, and $\theta$ is measured in radians. So you can see that the pattern here is "tangential variable is corresponding angular variable times the radius". Now the absolute acceleration of point B is going to be
$$a=\sqrt{a_{\tan}^{2}+a_{c}^{2}},$$
where $a_{\tan}$ is the tangential acceleration, and $a_{c}$ is the centripetal acceleration; the centripetal acceleration is given by
$$a_{c}=\frac{v_{\tan}^{2}}{r}=\omega^{2}\,r.$$

Also note that the angular velocity is a measure of how fast the angle $\theta$ is changing. That is,
$$\omega=\frac{d\theta}{dt}.$$
Also,
$$\alpha=\frac{d\omega}{dt}.$$
These are angular analogues to the linear variable case.

Is that enough to get you started?

I have done it this way:
Is this correct? How would you do it your way?
Let the $y$ axis be along the line of $AB$, the $z$ axis come out of the page at $A$, and the $x$ axis run perpendicular to $y$ and $z$ from $A$.
That is, $A$ is the origin.
With this coordinate system, $\mathbf{v}_A = \mathbf{a}_A = 0$, $\omega_{AB} = -0.02\mathbf{k}\text{ rad}/\text{s}$, and $\dot{\omega}_{AB} = -0.01\mathbf{k}\text{ rad}/\text{s}^2$.
The position vector for $B$ is $\mathbf{r}_B = 60\mathbf{j}$, the relative velocity at $B$ is $\mathbf{v}_{\text{rel}} = 0.5\mathbf{j}\text{ ft}/\text{s}$, and the relative acceleration is $\mathbf{a}_{\text{rel}} = 0$.
The the velocity and magnitude at $B$ is
\begin{alignat*}{3}
\mathbf{v}_B & = & \mathbf{v}_A + \omega_{AB}\times\mathbf{r}_B + \mathbf{v}_{\text{rel}}\\
& = & 0 + -0.02\mathbf{k}\times 60\mathbf{j} + 0.5\mathbf{j}\\
& = & (1.2\mathbf{i} + 0.5\mathbf{j})\text{ ft}/\text{s}\\
\lVert\mathbf{v}_B\rVert & = & \sqrt{1.2^2 + 0.5^2}\\
& = & 1.3\text{ ft}/\text{s}
\end{alignat*}
The acceleration at $B$ is
\begin{alignat*}{3}
\mathbf{a}_B & = & \mathbf{a}_A + \dot{\omega}_{AB}\times\mathbf{r}_B + \omega_{AB}\times(\omega_{AB}\times\mathbf{r}_B) + 2\omega_{AB}\times\mathbf{v}_{\text{rel}} + \mathbf{a}_{\text{rel}}\\
& = & 0 + -0.01\mathbf{k}\times 60\mathbf{j} + -0.02\mathbf{k}\times(-0.02\mathbf{k}\times 60\mathbf{j}) + -0.04\mathbf{k}\times 0.5\mathbf{j} + 0\\
& = & 0.6\mathbf{i} - 0.024\mathbf{j} + 0.02\mathbf{i}\\
& = & 0.62\mathbf{i} - 0.024\mathbf{j}
\end{alignat*}

 

[Ackbach]

I think you're making this more complicated than you need - although it is more complicated than I initially thought. You can assume that the base of the crane is an inertial reference frame - hence you do not need the equations for a non-inertial reference frame such as you used. However, you do need to account for the fact that $r$ is changing at a constant speed. You can simply write down $\mathbf{r}$ by using the following:
$$ r(t)=60+0.5t$$
and
$$ \mathbf{r}(t)=r(t) \langle \cos( \theta (t)), \sin( \theta (t)) \rangle.$$
Then simply write it all in and differentiate accordingly. You have
$$ \theta(t)= \theta_{0} +\omega_{0} \,t+ \frac{1}{2} \, \alpha \,t^{2}.$$

 

[CellCoree]

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I would first clarify some terminology to ensure a clear understanding of the problem. Angular velocity refers to the rate of change of angular displacement, measured in radians per unit of time. Rotation can refer to the act of rotating or the state of being rotated. In this context, it likely refers to the movement of the crane's telescopic boom around point A.

To solve for the magnitude and acceleration of the absolute acceleration of point B, we can use the formula for absolute acceleration, which is the sum of tangential acceleration and centripetal acceleration. Tangential acceleration is the rate of change of tangential velocity, while centripetal acceleration is the acceleration towards the center of rotation.

First, we can calculate the tangential velocity of point B using the given information. Since the boom is extending at a constant speed of 0.5 ft/s relative to the boom, we can use the Pythagorean theorem to find the tangential velocity of point B, which is the hypotenuse of the triangle formed by the boom and the extension. This gives us a tangential velocity of 0.707 ft/s.

Next, we can calculate the centripetal acceleration of point B using the formula a = v^2/r, where v is the tangential velocity and r is the distance from point B to the center of rotation, which is point A. Plugging in the values, we get a centripetal acceleration of 0.353 ft/s^2.

Finally, we can calculate the magnitude of the absolute acceleration using the formula a = √(at^2 + ac^2), where at is the tangential acceleration and ac is the centripetal acceleration. Plugging in the values, we get a magnitude of 0.795 ft/s^2.

In conclusion, at this instant, the magnitude of the absolute acceleration of point B is 0.795 ft/s^2, with a tangential acceleration of 0.5 ft/s^2 and a centripetal acceleration of 0.353 ft/s^2. This information can be useful for understanding the dynamics of the crane's movement and ensuring safe operation in construction or other applications.