Chapter 15.2, Problem 15.45P

The disk shown moves in the xy p1ane. Knowing that ${\left(v_A\right)}_A =-7\text{ m/s},{\left(v_A\right)}_A=\text{ -}7.4\text{m/s},$ and ${\left(v_3\right)}_3=-1.4\text{m/s},$ determine (a) the ve1ocity of point O, (b) the point of the disk with zero ve1ocity.

Fig. P15.44 and P15.45

To determine

(a)

The velocity of point O

Answer to Problem 15.45P

$v_O=-\left(1.4m/s\right)i-\left(1m/s\right)j$

Explanation of Solution

Given information:

$\begin{array}{l}{\left(v_A\right)}_y=-7m/s\\ {\left(v_B\right)}_x=-7.4m/s\\ {\left(v_C\right)}_x=-1.4m/s\end{array}$

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

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

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

$\omega$ - Angular velocity

Calculation:

Velocity of B with respect A is

$v_{B/A}=\omega k\times r_{B/A}=\omega k\times \left(0.6i+0.6j\right)=-0.6\omega i+0.6\omega j$

Velocity of C with respect to A is

$v_{C/A}=\omega k\times r_{C/A}=\omega k\times \left(1.2i\right)=1.2\omega j$

The absolute value of point B

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

Substitute

$-7.4i+{\left(v_B\right)}_{y}j={\left(v_A\right)}_{x}i-7j-0.6\omega i+0.6\omega j$

Equate components

$\begin{array}{l}i:-7.4+{\left(v_A\right)}_x-0.6\omega \to \left(1\right)\\ j:{\left(v_B\right)}_y=-7+0.6\omega \to \left(2\right)\end{array}$

Similarly for point C

$v_C=v_A+v_{C/A}$

Substitute

$-1.4i+{\left(v_C\right)}_{y}j={\left(v_A\right)}_{x}i-7j+1.2\omega j$

Equate components

$\begin{array}{l}i:-1.4={\left(v_A\right)}_x\to \left(3\right)\\ j:{\left(v_C\right)}_y=-7+1.2\omega \to \left(4\right)\end{array}$

From equation 3

${\left(v_A\right)}_x=-1.4m/s$

Therefore

$v_A=-1.4i-7j$

To find the angular velocity

From equation 1

$\omega =\frac{{\left(v_A\right)}_x+7.4}{0.6}=\frac{-1.4+7.4}{0.6}=10rad/s$

Find the velocity of point O with respect to A

$\begin{array}{l}{v}_{O/A}=\omega k\times r_{O/A}\\ =10k\times \left(0.6i\right)\\ =6j\end{array}$

Find the absolute velocity of point O

$\begin{array}{l}{v}_O=v_A+v_{O/A}\\ =-1.4i-7j+6j\\ =-\left(1.4m/s\right)i-\left(1m/s\right)j\end{array}$

Conclusion:

According to the above explanation, the velocity of point O is equal to

$v_O=-\left(1.4m/s\right)i-\left(1m/s\right)j$

To determine

(b)

The point of the disk with zero velocity

Answer to Problem 15.45P

The point of disk with zero velocity is $x=100mm,y=-140mm$

Explanation of Solution

Given information:

$\begin{array}{l}{\left(v_A\right)}_y=-7m/s\\ {\left(v_B\right)}_x=-7.4m/s\\ {\left(v_C\right)}_x=-1.4m/s\end{array}$

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

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

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

$\omega$ - Angular velocity

Calculation:

According to sub part a

$\omega =10rad/s$

And

$v_O=-\left(1.4m/s\right)i-\left(1m/s\right)j$

To find the point with zero velocity

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

Substitute

$\begin{array}{l}0=-1.4i-j+10k\times \left(xi+yj\right)\\ 0=-1.4i-j-10yi+10xj\end{array}$

Equate components

$\begin{array}{l}i:0=-1.4-10y\\ j:0=-1+10x\end{array}$

Solve to find x and y

$\begin{array}{l}x=100mm\\ y=-140mm\end{array}$

Conclusion:

The point of disk with zero velocity is $x=100mm,y=-140mm$