MHB How Does Angular Velocity Affect Crane Boom Dynamics?

AI Thread Summary
The discussion focuses on the relationship between angular velocity and crane boom dynamics, particularly in calculating the absolute acceleration of point B on a telescopic crane boom. Key equations are provided to relate tangential and centripetal acceleration to angular variables, emphasizing the importance of angular velocity and its derivatives. A coordinate system is established to simplify calculations, with specific values for angular velocity and acceleration given. The participant shares their approach to calculating velocity and acceleration at point B, suggesting that the problem may be more complex than initially anticipated but can be simplified by treating the crane's base as an inertial reference frame. Overall, the conversation highlights the mathematical principles necessary for understanding crane boom dynamics in relation to angular motion.
Dustinsfl
Messages
2,217
Reaction score
5
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
 
Mathematics news on Phys.org
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?
 
Ackbach said:
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*}
 
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
$$ \theta(t)= \theta_{0} +\omega_{0} \,t+ \frac{1}{2} \, \alpha \,t^{2}.$$
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top