Chapter 15.2, Problem 15.46P

To determine

(a)

The angular velocity of the plate

Answer to Problem 15.46P

The angular velocity of the disk is $4rad/s$ clockwise.

Explanation of Solution

Given information:

$\begin{array}{l}{\left(v_A\right)}_x=12in/s\\ {\left(v_B\right)}_x=-4in/s\\ {\left(v_C\right)}_y=-24in/s\end{array}$

The relative velocity of A with respect to B is defined as

$v_{A/B}=\omega k\times r_{A/B}$

$r_{A/B}$ - Position vector of A with respect to B

$\omega$ - Angular velocity

Calculation:

Velocity of A with respect B is

$v_{A/B}=\omega k\times r_{A/B}=\omega k\times \left(-2i+4j\right)=-4\omega i-2\omega j$

The absolute value of point A

$v_A=v_B+v_{A/B}$

Substitute

$12i+{\left(v_A\right)}_{y}j=-4i+{\left(v_B\right)}_{y}j-4\omega i-2\omega j$

Equate components

$\begin{array}{l}i:12=-4-4\omega \to \left(1\right)\\ j:{\left(v_A\right)}_y={\left(v_B\right)}_y-2\omega \to \left(2\right)\end{array}$

From equation 1

$\omega =\frac{16}{-4}=-4rad/s$

Conclusion:

According to the above explanation, the angular velocity of the disk is $4rad/s$ clockwise.

To determine

(b)

The velocity of point B

Answer to Problem 15.46P

$v_B=-\left(4in/s\right)i$

Explanation of Solution

Given information:

$\begin{array}{l}{\left(v_A\right)}_x=12in/s\\ {\left(v_B\right)}_x=-4in/s\\ {\left(v_C\right)}_y=-24in/s\end{array}$

The relative velocity of C with respect to B is defined as

$v_{C/B}=\omega k\times r_{C/B}$

$r_{C/B}$ - Position vector of C with respect to B

$\omega$ - Angular velocity

Calculation:

According to sub part a

$\omega =-4rad/s$

Velocity of C with respect to B is

$v_{C/B}=\omega k\times r_{C/B}=\omega k\times \left(6i-2j\right)=2\omega i+6\omega j$

The absolute velocity of point C

$v_C=v_B+v_{C/B}$

Substitute

${\left(v_C\right)}_{x}i-24j=-4i+{\left(v_B\right)}_{y}j+2\omega i+6\omega j$

Equate components

$\begin{array}{l}i:{\left(v_C\right)}_x=-4+2\omega \to \left(1\right)\\ j:-24={\left(v_B\right)}_y+6\omega \to \left(2\right)\end{array}$

From equation 2

${\left(v_B\right)}_y=-24-6\left(-4\right)=0$

Therefore

$v_B={\left(v_B\right)}_{x}i+{\left(v_B\right)}_{y}j=-\left(4in/s\right)i$

Conclusion:

According to the above explanation, the velocity of point B is $v_B=-\left(4in/s\right)i$

To determine

(c)

The point of plate with zero velocity

Answer to Problem 15.46P

The point of plate with zero velocity is $x=-3in,y=-2in$

Explanation of Solution

Given information:

$\begin{array}{l}{\left(v_A\right)}_x=12in/s\\ {\left(v_B\right)}_x=-4in/s\\ {\left(v_C\right)}_y=-24in/s\end{array}$

The relative velocity of O with respect to B is defined as

$v_{O/B}=\omega k\times r_{O/B}$

$r_{O/B}$ - Position vector of O with respect to B

$\omega$ - Angular velocity

Calculation:

According to sub part a and b

$\omega =-4rad/s$

And

$v_B=-\left(4in/s\right)i$

Find the velocity of point O with respect to B

$\begin{array}{l}{v}_{O/B}=\omega k\times r_{O/B}\\ =\omega k\times \left(-2i-2j\right)\\ =2\omega i-2\omega j\\ =-8i+8j\end{array}$

The absolute velocity of point O

$v_O=v_B+v_{O/B}=-4i-8i+8j=-12i+8j$

To find the point with zero velocity

$0=v_O+\omega k\times \left(xi+yj\right)$

Substitute

$\begin{array}{l}0=-12i+8j-4k\times \left(xi+yj\right)\\ 0=-12i+8j-4xj+4yj\end{array}$

Equate components

$\begin{array}{l}i:0=-12-4x\\ j:0=8+4y\end{array}$

Solve to find x and y

$\begin{array}{l}x=-3in\\ y=-2in\end{array}$

Conclusion:

The point of plate with zero velocity is $x=-3in,y=-2in$