Question 11.SP.20: At the instant shown, the length of the boom AB is being dec...
At the instant shown, the length of the boom AB is being decreased at the constant rate of 0.2 m/s, and the boom is being lowered at the constant rate of 0.08 rad/s. Determine (a) the velocity of point B, (b) the acceleration of point B.
STRATEGY: Use polar coordinates, since that is the most natural way to describe the position of point B.
MODELING and ANALYSIS: From the problem statement, you know
\dot{r} = –0.2 m/s \ddot{r} = 0 \dot{\theta} = –0.08 rad/s \ddot{\theta} = 0
a. Velocity of B. Using Eqs.(11.44), you can determine the values of v_{r} and v_{θ} at this instant to be
\begin{aligned}&v_{r}=\dot{r}=-0.2 \mathrm{~m} / \mathrm{s} \\&v_{\theta}=r \dot{\theta}=(6 \mathrm{~m})(-0.08 \mathrm{rad} / \mathrm{s})=-0.48 \mathrm{~m} / \mathrm{s}\end{aligned}Therefore, you can write the velocity vector as
\mathbf{v}=(-0.200 \mathrm{~m} / \mathrm{s}) \mathbf{e}_{r}+(-0.480 \mathrm{~m} / \mathrm{s}) \mathbf{e}_{t}b. Acceleration of B. Using Eqs. (11.45), you find
\begin{aligned}&a_{r}=\ddot{r}-r \dot{\theta}^{2}=0-(6 \mathrm{~m})(-0.08 \mathrm{rad} / \mathrm{s})^{2}=-0.0384 \mathrm{~m} / \mathrm{s}^{2} \\&a_{\theta}=r \ddot{\theta}+2 \dot{r} \dot{\theta}=0+2(-0.02 \mathrm{~m} / \mathrm{s})(-0.08 \mathrm{rad} / \mathrm{s})=0.00320 \mathrm{~m} / \mathrm{s}^{2}\end{aligned}or
\mathbf{a}=\left(-0.0384 \mathrm{~m} / \mathrm{s}^{2}\right) \mathbf{e}_{r}+\left(0.00320 \mathrm{~m} / \mathrm{s}^{2}\right) \mathbf{e}_{\theta}REFLECT and THINK: Once you identify what you are given in the problem statement, this problem is quite straightforward. Sometimes you will be asked to express your answer in terms of a magnitude and direction. The easiest way is to first determine the x and y components and then to find the magnitude and direction. From Fig. 1,
\begin{aligned}&\stackrel{+}{\longrightarrow}:\left(v_{B}\right)_{x}=0.48 \cos 60^{\circ}-0.2 \cos 30^{\circ}=0.06680 \mathrm{~m} / \mathrm{s} \\&+\uparrow:\left(v_{B}\right)_{y}=-0.48 \sin 60^{\circ}-0.2 \sin 30^{\circ}=-0.5157 \mathrm{~m} / \mathrm{s}\end{aligned}So the magnitude and direction are
\begin{aligned}v_{B} &=\sqrt{0.06680^{2}+0.5157^{2}} \\&=0.520 \mathrm{~m} / \mathrm{s} \quad \tan \beta=\frac{0.51569}{0.06680}, \beta=82.6^{\circ}\end{aligned}So, an alternative way of expressing the velocity of B is \mathbf{v}_{B}=0.520 \mathrm{~m} / \mathrm{s} \measuredangle 82.6^{\circ}
You could also find the magnitude and direction of the acceleration if you needed it expressed in this way. It is important to note that no matter what coordinate system we choose, the resultant velocity vector is the same. You can choose to express this vector in whatever coordinate system is most useful. Figure 2 shows the velocity vector \mathbf{v}_{B} resolved into x and y components and r and θ coordinates.
