# [SOLVED]angular velocity and rotation

#### dwsmith

##### Well-known member
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.
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#### Ackbach

##### Indicium Physicus
Staff member
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?

#### dwsmith

##### Well-known member
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

##### Indicium Physicus
Staff member
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
$$\theta(t)= \theta_{0} +\omega_{0} \,t+ \frac{1}{2} \, \alpha \,t^{2}.$$