Chapter 15.6, Problem 15.202P

In Prob. 15.201, the speed of point B is known to be constant. For the position shown determine (a) the angular acceleration of the guide arm, (b) the acceleration of point C.

To determine

(a)

Angular acceleration of guide arm

Answer to Problem 15.202P

The angular acceleration $\alpha$ of guide arm is $\left(1.125rad/s^2\right)k$.

Explanation of Solution

Given information:

Rod OA slides in straight inclined slot.

Rod OB slides in slot parallel to z axis.

$v_B=\left(9in/s\right)k$

The vector product of angular velocity $\omega$ and of the position vector r of the particle:

$v=\frac{dr}{dt}=\omega \times r$

The acceleration $a$ of a particle is defined as

$a=\left(\alpha \times r\right)+\left(\omega \times v\right)$

$\alpha$ - Angular acceleration

$r$ - Position vector

$\omega$ - Angular velocity

Calculation:

Rod OA slides in a slope of 1:2.

Hence

$\begin{array}{l}{\left(v_D\right)}_x=-2{\left(v_D\right)}_y\\ {\left(v_A\right)}_x=-2{\left(v_A\right)}_y\end{array}$

Angular velocity $\omega$ of guide arm

$\omega ={\omega }_{x}i+{\omega }_{y}j+{\omega }_{z}k$

The velocity $v_B$ of point B

$\begin{array}{l}{v}_B=\omega \times r_B\\ \left(9in/s\right)k=\left({\omega }_{x}i+{\omega }_{y}j+{\omega }_{z}k\right)\times \left(12in\right)j\\ 9k=12{\omega }_{x}k-12{\omega }_{z}i\end{array}$

Equate components

$\begin{array}{l}i:0=-12{\omega }_z\\ k:9=12{\omega }_x\end{array}$

Therefore

$\begin{array}{l}{\omega }_z=0\\ {\omega }_x=0.75rad/s\end{array}$

Velocity $v_B$ of point A

$\begin{array}{l}{v}_A=\omega \times r_A\\ {\left(v_A\right)}_{x}i+{\left(v_A\right)}_{y}j+{\left(v_A\right)}_{z}k=\left(0.75i+{\omega }_{y}j\right)\times 10k\\ {\left(v_A\right)}_{x}i+{\left(v_A\right)}_{y}j+{\left(v_A\right)}_{z}k=-7.5j+10{\omega }_{y}i\end{array}$

Equate components

$\begin{array}{l}i:{\left(v_A\right)}_x=10{\omega }_y\\ j:{\left(v_A\right)}_y=-7.5\\ k:{\left(v_A\right)}_z=0\end{array}$

But we know that

${\left(v_A\right)}_x=-2{\left(v_A\right)}_y$

Therefore

${\left(v_A\right)}_x=-2\left(-7.5\right)=15in/s$

And

$\begin{array}{l}10{\omega }_y=15in/s\\ {\omega }_y=1.5rad/s\end{array}$

Angular velocity $\omega$ of guide arm

$\omega =\left(0.75rad/s\right)i+\left(1.5rad/s\right)j$

The velocity $v_A$ of point A

$v_A=\left(15in/s\right)i-\left(7.5in/s\right)j$

The position vector $r_C$ of point C

$r_C=5i+4j+2k$

The velocity $v_C$ of point C

$\begin{array}{l}{v}_C=\omega \times r_C\\ =\left|\begin{array}{ccc}i& j& k\\ 0.75& 1.5& 0\\ 5& 4& 2\end{array}\right|\\ =\left(3in/s\right)i-\left(1.5in/s\right)j-\left(4.5in/s\right)k\end{array}$

The acceleration $a_B$ of point B

$\begin{array}{l}{a}_B=\alpha \times r_B+\omega \times v_B\\ =\left|\begin{array}{ccc}i& j& k\\ {\alpha }_x& {\alpha }_y& {\alpha }_z\\ 0& 12& 0\end{array}\right|+\left|\begin{array}{ccc}i& j& k\\ 0.75& 1.5& 0\\ 0& 0& 9\end{array}\right|\\ =-12{\alpha }_{z}i+12{\alpha }_{x}k+13.5i-6.75j\\ a_B=\left(13.5-12{\alpha }_z\right)i-6.75j+12{\alpha }_{x}k\end{array}$

Equate components

$\begin{array}{l}{\left(a_B\right)}_x=13.5-12{\alpha }_z=0\\ {\left(a_B\right)}_y=-6.75\\ {\left(a_B\right)}_z=12{\alpha }_x=0\end{array}$

Therefore

$\begin{array}{l}{\alpha }_z=1.125rad/s^2\\ {\left(a_B\right)}_y=-6.75in/s^2\\ {\alpha }_x=0\end{array}$

The acceleration $a_A$ of point A

$\begin{array}{l}{a}_A=\alpha \times r_A+\omega \times v_A\\ =\left|\begin{array}{ccc}i& j& k\\ 0& {\alpha }_y& 1.125\\ 0& 0& 10\end{array}\right|+\left|\begin{array}{ccc}i& j& k\\ 0.75& 1.5& 0\\ 15& -7.5& 0\end{array}\right|\\ =10{\alpha }_{y}i-28.125k\end{array}$

Equate components

$\begin{array}{l}{\left(a_A\right)}_x=10{\alpha }_y\\ {\left(a_A\right)}_y=0\\ {\left(a_A\right)}_z=-28.125\end{array}$

But we know that

${\left(a_A\right)}_x=-2{\left(a_A\right)}_y$

Therefore

$\begin{array}{l}{\left(a_A\right)}_x=10{\alpha }_y=0\\ {\alpha }_y=0\end{array}$

The angular acceleration $\alpha$ of guide arm

$\begin{array}{l}\alpha ={\alpha }_{x}i+{\alpha }_{y}j+{\alpha }_{z}k\\ =\left(1.125rad/s^2\right)k\end{array}$

Conclusion:

The angular acceleration $\alpha$ of guide armis $\left(1.125rad/s^2\right)k$.

To determine

(b)

Acceleration of point C

Answer to Problem 15.202P

The acceleration $a_C$ of point C is $-\left(11.25in/s^2\right)i+\left(9in/s^2\right)j-\left(5.625in/s^2\right)k$.

Explanation of Solution

Given information:

Rod OA slides in straight inclined slot.

Rod OB slides in slot parallel to z axis.

$v_B=\left(9in/s\right)k$

The vector product of angular velocity $\omega$ and of the position vector r of the particle:

$v=\frac{dr}{dt}=\omega \times r$

The acceleration $a$ of a particle is defined as

$a=\left(\alpha \times r\right)+\left(\omega \times v\right)$

$\alpha$ - Angular acceleration

$r$ - Position vector

$\omega$ - Angular velocity

Calculation:

According to sub part a

$\alpha =\left(1.125rad/s^2\right)k$

The velocity $v_C$ of point C

$v_C=\left(3in/s\right)i-\left(1.5in/s\right)j-\left(4.5in/s\right)k$

Angular velocity $\omega$ of guide arm

$\omega =\left(0.75rad/s\right)i+\left(1.5rad/s\right)j$

The position vector $r_C$ of point C

$r_C=5i+4j+2k$

The acceleration $a_C$ of point C

$\begin{array}{l}{a}_C=\alpha \times r_C+\omega \times v_C\\ =\left|\begin{array}{ccc}i& j& k\\ 0& 0& 1.125\\ 5& 4& 2\end{array}\right|+\left|\begin{array}{ccc}i& j& k\\ 0.75& 1.5& 0\\ 3& -1.5& -4.5\end{array}\right|\\ =-4.5i+5.625j-6.75i+3.375j-5.625k\\ =-\left(11.25in/s^2\right)i+\left(9in/s^2\right)j-\left(5.625in/s^2\right)k\end{array}$

Conclusion:

The acceleration $a_C$ of point Cis $-\left(11.25in/s^2\right)i+\left(9in/s^2\right)j-\left(5.625in/s^2\right)k$.