Chapter 15.6, Problem 15.194P

A gun barrel of length $OP=4\text{m}$ is mounted on a turret as shown. To keep the gun aimed at a moving target, the azimuth angle $\beta$ is being increased at the rate $d\beta /dt=30°/s$ and the elevation angle $\gamma$ is being increased at the rate $d\gamma /dt=10°/s$ . For the position $\beta =90°$ and $\gamma =30°$ , determine (a) the angular velocity of the barrel, (b) the angular acceleration of the barrel, (c) the velocity and acceleration of point P.

    

To determine

(a)

The angular velocity of the barrel.

Answer to Problem 15.194P

The angular velocity of the barrel is $-\left(0.1745\text{rad}/\text{s}\right)\text{i}-\left(0.5235\text{rad}/\text{s}\right)\text{j}$.

Explanation of Solution

Given Information:

The length of the gun barrel is $4\text{m}$, the rate of increasing of azimuth angle $\beta$ is $\frac{d\beta }{dt}=30°/\text{s}$, the elevation angle $\gamma$ is increased at the rate of $\frac{d\gamma }{dt}=10°/\text{s}$, the value of azimuth angle is $30°$ and the elevation angle is $10°$.

Write the expression for the angular velocity at the azimuth angle.

${\omega }_1=-\frac{d\beta }{dt}\text{j}$ ...... (I)

Here, the rate of increasing of azimuth angle is $\frac{d\beta }{dt}$.

Write the expression for the angular velocity at the elevation angle.

${\omega }_2=-\frac{d\gamma }{dt}\text{i}$ ...... (II)

Here, the rate of increasing of elevation angle is $\frac{d\gamma }{dt}$.

Write the expression for the angular velocity of the barrel.

$\omega ={\omega }_1+{\omega }_2$ ...... (III)

Calculation:

Substitute $30°/\text{s}$ for $\frac{d\beta }{dt}$ in Equation (I).

$\begin{array}{c}{\omega }_1=-\left(30°/\text{s}\right)\text{j}\\ \text{=}-\left(30°/\text{s}\left(\frac{\pi \text{rad}}{180°}\right)\right)\text{j}\\ \text{=}\left(-\frac{\pi }{6}\text{rad}/\text{s}\right)\text{j}\\ =\left(-0.5235\text{rad}/\text{s}\right)\text{j}\end{array}$

Substitute $10°/\text{s}$ for $\frac{d\gamma }{dt}$ in Equation (II).

$\begin{array}{c}{\omega }_2=-\left(10°/\text{s}\right)\text{i}\\ \text{=}-\left(10°/\text{s}\left(\frac{\pi \text{rad}}{180°}\right)\right)\text{i}\\ \text{=}\left(-\frac{\pi }{18}\text{rad}/\text{s}\right)\text{i}\\ =\left(-0.1745\text{rad}/\text{s}\right)\text{i}\end{array}$

Substitute $\left(-0.5235\text{rad}/\text{s}\right)\text{j}$ for ${\omega }_1$ and $\left(-0.1745\text{rad}/\text{s}\right)\text{i}$ for ${\omega }_2$ in Equation (III)

$\begin{array}{c}\omega =\left(-0.1745\text{rad}/\text{s}\right)\text{i}+\left(-0.5235\text{rad}/\text{s}\right)\text{j}\\ =-\left(0.1745\text{rad}/\text{s}\right)\text{i}-\left(0.5235\text{rad}/\text{s}\right)\text{j}\end{array}$

Conclusion:

The angular velocity of the barrel is $-\left(0.1745\text{rad}/\text{s}\right)\text{i}-\left(0.5235\text{rad}/\text{s}\right)\text{j}$.

To determine

(b)

The angular acceleration of the barrel.

Answer to Problem 15.194P

The angular acceleration of the barrel is $\left(-\text{0}\text{.09135}\text{rad}/{\text{s}}^2\right)\text{k}$.

Explanation of Solution

Write the expression for the angular acceleration of the barrel.

$\alpha ={\omega }_1\times \omega$ ...... (IV)

Calculation:

Substitute $\left(-0.5235\text{rad}/\text{s}\right)\text{j}$ for ${\omega }_1$ and $-\left(0.1745\text{rad}/\text{s}\right)\text{i}-\left(0.5235\text{rad}/\text{s}\right)\text{j}$ for $\omega$ in Equation (IV)

$\begin{array}{c}\alpha =\left(-0.5235\text{rad}/\text{s}\right)\text{j}\left(-\left(0.1745\text{rad}/\text{s}\right)\text{i}-\left(0.5235\text{rad}/\text{s}\right)\text{j}\right)\\ =\left[\left(-0.5235\text{rad}/\text{s}\right)\left(-\left(0.1745\text{rad}/\text{s}\right)\right)\right]\text{k}\\ =\left(-\text{0}\text{.09135}\text{rad}/{\text{s}}^2\right)\text{k}\end{array}$

Conclusion:

The angular acceleration of the barrel is $\left(-\text{0}\text{.09135}\text{rad}/{\text{s}}^2\right)\text{k}$.

To determine

(c)

The velocity of point $P$.

The acceleration of point $P$.

Answer to Problem 15.194P

The velocity of point $P$ is $\left(-1.8138\text{m}/\text{s}\right)\text{i}+\left(0.6046\text{m}/\text{s}\right)\text{j}-\left(0.34906\text{m}/\text{s}\right)\text{k}$.

The acceleration of point $P$ is $\left(0.3656\text{m}/{\text{s}}^{\text{2}}\right)\text{i}-\left(0.0609\text{m}/{\text{s}}^{\text{2}}\right)\text{j}-\left(1.056\text{m}/{\text{s}}^{\text{2}}\right)\text{k}$.

Explanation of Solution

Write the expression for the velocity of point $P$.

$v_P=\omega \times r_P$ ...... (V)

Here, the position vector is $r_P$.

Write the expression for the position vector.

$r_P=\left(l_{OP}\sin \beta \right)\text{j}+\left(l_{OP}\cos \beta \right)\text{k}$ ...... (VI)

Here, the length of the barrel is $l_{OP}$ and the azimuth angle is $\beta$.

Write the expression for the acceleration of point $P$.

$a_P=\left(\alpha \times r_P\right)+\left(\omega \times v_P\right)$ ...... (VII)

Calculation:

Substitute $\text{4}\text{m}$ for $l_{OP}$ and $30°$ for $\beta$ in Equation (VI).

$\begin{array}{c}{r}_P=\left(\left(\text{4}\text{m}\right)\sin 30°\right)\text{j}+\left(\left(\text{4}\text{m}\right)\cos 30°\right)\text{k}\\ =\left(\left(\text{4}\text{m}\right)0.5\right)\text{j}+\left(\left(\text{4}\text{m}\right)0.8660\right)\text{k}\\ =\left(2\text{m}\right)\text{j}+\left(3.4641\text{m}\right)\text{k}\end{array}$

Substitute $\left(2\text{m}\right)\text{j}+\left(3.4641\text{m}\right)\text{k}$ for $r_P$ and $-\left(0.1745\text{rad}/\text{s}\right)\text{i}-\left(0.5235\text{rad}/\text{s}\right)\text{j}$ for $\omega$ in Equation (V)

$\begin{array}{c}{v}_P=\left(\left(2\text{m}\right)\text{j}+\left(3.4641\text{m}\right)\text{k}\right)\left(-\left(0.1745\text{rad}/\text{s}\right)\text{i}-\left(0.5235\text{rad}/\text{s}\right)\text{j}\right)\\ =\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ -0.1745\text{rad}/\text{s}& -0.5236\text{rad}/\text{s}& 0\\ 0& 2\text{m}& 3.4641\text{m}\end{array}\right|\\ =-1.8138\text{m}/\text{s}\text{i}+0.6046\text{m}/\text{s}\text{j}-0.34906\text{m}/\text{s}\text{k}\\ \text{=}\left(-1.8138\text{m}/\text{s}\right)\text{i}+\left(0.6046\text{m}/\text{s}\right)\text{j}-\left(0.34906\text{m}/\text{s}\right)\text{k}\end{array}$

Substitute $\left(-\text{0}\text{.09135}\text{rad}/{\text{s}}^2\right)\text{k}$ for $\alpha$, $\left(-1.8138\text{m}/\text{s}\right)\text{i}+\left(0.6046\text{m}/\text{s}\right)\text{j}-\left(0.34906\text{m}/\text{s}\right)\text{k}$ for $v_P$, $\left(2\text{m}\right)\text{j}+\left(3.4641\text{m}\right)\text{k}$ for $r_P$ and $-\left(0.1745\text{rad}/\text{s}\right)\text{i}-\left(0.5235\text{rad}/\text{s}\right)\text{j}$ for $\omega$ in Equation (VII).

$\begin{array}{c}{a}_P=\left[\begin{array}{l}\left(\left(-\text{0}\text{.09135}\text{rad}/{\text{s}}^2\right)\text{k}\times \left(2\text{m}\right)\text{j}+\left(3.4641\text{m}\right)\text{k}\right)+\\ \left(\begin{array}{l}-\left(0.1745\text{rad}/\text{s}\right)\text{i}-\left(0.5235\text{rad}/\text{s}\right)\text{j}\times \left(-1.8138\text{m}/\text{s}\right)\text{i}+\\ \left(0.6046\text{m}/\text{s}\right)\text{j}-\left(0.34906\text{m}/\text{s}\right)\text{k}\end{array}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(\left(-\text{0}\text{.09135}\text{rad}/{\text{s}}^2\right)\text{k}\times \left(2\text{m}\right)\text{j}+\left(3.4641\text{m}\right)\text{k}\right)+\\ \left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ -0.1745\text{rad}/\text{s}& -0.5236\text{rad}/\text{s}& 0\\ -1.814\text{m}/\text{s}& 0.6046\text{m}/\text{s}& -0.349\text{m}\end{array}\right|\end{array}\right]\\ =\left[\begin{array}{l}0.1828\text{m}/{\text{s}}^{\text{2}}\text{i}+0.1828\text{m}/{\text{s}}^{\text{2}}\text{j}-0.0609\text{m}/{\text{s}}^{\text{2}}\text{j}\\ -0.1056\text{m}/{\text{s}}^{\text{2}}\text{k}-\text{0}\text{.9498}\text{m}/{\text{s}}^{\text{2}}\text{k}\end{array}\right]\\ =\left(0.3656\text{m}/{\text{s}}^{\text{2}}\right)\text{i}-\left(0.0609\text{m}/{\text{s}}^{\text{2}}\right)\text{j}-\left(1.056\text{m}/{\text{s}}^{\text{2}}\right)\text{k}\end{array}$

Conclusion:

The velocity of point $P$ is $\left(-1.8138\text{m}/\text{s}\right)\text{i}+\left(0.6046\text{m}/\text{s}\right)\text{j}-\left(0.34906\text{m}/\text{s}\right)\text{k}$.

The acceleration of point $P$ is $\left(0.3656\text{m}/{\text{s}}^{\text{2}}\right)\text{i}-\left(0.0609\text{m}/{\text{s}}^{\text{2}}\right)\text{j}-\left(1.056\text{m}/{\text{s}}^{\text{2}}\right)\text{k}$.