Question

Knowing that at the instant shown the angular velocity of rod BE is 4 rad/s counterclockwise, determine (a) the angular velocity of rod AD, (b) the velocity of collar D, (c) the velocity of point A.

Answer 1

Concepts and reason Equations of motion:

Equation of motions refers to the equations which describe the attributes of a moving object in terms of displacement, velocity, and acceleration. Vector:

Vector form is used to represent a quantity in terms of magnitude and direction. This form is expressed as magnitude multiplied by direction in terms of $\mathbf{i}, \mathbf{j}$ and $\mathbf{k}$ representing the three perpendicular axes $x, y$ and $z$ respectively of a quantity. Angular displacement:

Angular displacement is the angle through which an object moves on a circular path. It is denoted by $\boldsymbol{\theta}$ Angular velocity:

The rate of change of angular displacement of a body with respect to time is angular velocity. It is denoted by $\omega$ and its unit is $\mathrm{rad} / \mathrm{s}$

Relative velocity It is defined as the velocity measurement of a moving object with reference as a translating frame or another moving object. The relative velocity between points $\mathrm{B}$ and $\mathrm{D}$ can be used to obtain the angular velocity of the link and the velocity of point $\mathrm{D}$. Then relative velocity between points $\mathrm{B}$ and $\mathrm{A}$ can be used to calculate the velocity of point $\mathrm{A} .$

Fundamentals

$A(x, y, z)$

For a point with coordinates as ${A}(x, y, 2)$ in motion, the position vector with respect to the origin is expressed as $\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}$

Here, the position vector is $\mathbf{r}$.

Write the relation for velocity and angular velocity $(\omega)$

$v=r \omega$

Here, the constant velocity is $v$ and the radius of the circular path is $r$ $\left(v_{B / A}\right)$

The formula to calculate the relative velocity $\left(V_{B / A}\right)$ of point $\mathrm{B}$ with respect to point $\mathrm{A}$ of an object is as follows:

$v_{B / A}=v_{B}-v_{A}$

Here, velocity of object $\mathrm{B}$ is ${v_{B}}$ and velocity of object $\mathrm{A}$ is ${v_A}$. From the angular velocity definition, the expression can also be written as follows:

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

$v_{B}-v_{A}=\omega r_{B / A}$

Here, relative position of $\mathrm{B}$ with respect to $\mathrm{A}$ is $\boldsymbol{r}_{B / A}$. The velocity of B with respect to origin A from the observer measurement can be rewritten in vector form as follows:

$\mathbf{v}_{B}=\mathbf{v}_{A}-\boldsymbol{\omega} \times \mathbf{r}_{B / A}$

Cross product rule of vectors:

$\mathbf{i} \times \mathbf{i}=\mathbf{0}$

$\mathbf{j} \times \mathbf{j}=\mathbf{0}$

$\mathbf{k} \times \mathbf{k}=\mathbf{0}$

$\mathbf{i} \times \mathbf{j}=\mathbf{k}$

$\mathbf{j} \times \mathbf{k}=\mathbf{i}$

$\mathbf{k} \times \mathbf{i}=\mathbf{j}$

$\mathbf{j} \times \mathbf{i}=-\mathbf{k}$

$\mathbf{k} \times \mathbf{j}=-\mathbf{i}$

$\mathbf{i} \times \mathbf{k}=-\mathbf{j}$

Calculate the linear velocity $\left(v_{B}\right)$ of point $\mathrm{B}$.

$v_{B}=r_{B E} \omega_{B E}$

Here, the distance BE is $r_{B E}$ and the angular velocity of $\mathrm{BE}$ is $\omega_{B E}$.

Substitute $0.192 \mathrm{~m}$ for $r_{B E}$ and $4 \mathrm{rad} / \mathrm{s}$ for $\omega_{B E}$.

$v_{B}=0.192 \mathrm{~m} \times 4 \mathrm{rad} / \mathrm{s}$

$v_{B}=0.768 \mathrm{~m} / \mathrm{s}$

Write the velocity of point $\mathrm{B}$ in Cartesian vector form. $\mathbf{v}_{B}=0.768 \mathbf{i} \mathrm{m} / \mathrm{s}$

The velocity of point B acts along only in the horizontal direction. Thus the velocity of point B is calculated as the product of radius and angular velocity of link BE.

(A)

Write the position vector for $\mathrm{BD}\left(\mathbf{r}_{B D}\right)$.

$\mathbf{r}_{B D}=\left(0.36 \cos 30^{\circ} \mathbf{i}-0.36 \sin 30^{\circ} \mathbf{j}\right) \mathrm{m}$

Write the expression for relative velocity between points $B$ and D for link AD.

$\mathbf{v}_{D}=\mathbf{v}_{B}-\omega_{A D} \mathbf{k} \times \mathbf{r}_{B D}$

Here, the velocity of point $\mathrm{D}$ is $v_{D}$, and the angular velocity of link $\mathrm{AD}$ is $\omega_{A D}$.

Substitute $-v_{D} \mathbf{j}_{\text {for }} \mathbf{v}_{D}, 0.768 \mathbf{i} \mathrm{m} / \mathrm{s}$ for $\mathbf{v}_{B}$, and $\left(0.36 \cos 30^{\circ} \mathbf{i}-0.36 \sin 30^{\circ} \mathbf{j}\right) \mathrm{m}_{\text {for }} \mathbf{r}_{B D}$.

$\begin{aligned}-v_{D} \mathbf{j} &=0.768 \mathbf{i} \mathrm{m} / \mathrm{s}-\omega_{A D} \mathbf{k} \times\left(0.36 \cos 30^{\circ} \mathbf{i}-0.36 \sin 30^{\circ} \mathbf{j}\right) \mathrm{m} \\ &=0.768 \mathbf{i}-\left(0.36 \cos 30^{\circ}\right) \omega_{A D} \mathbf{j}-\left(0.36 \sin 30^{\circ}\right) \omega_{A D} \mathbf{i} \\-v_{D} \mathbf{j} &=\left[0.768-\left(0.36 \sin 30^{\circ}\right) \omega_{A D}\right] \mathbf{i}-\left(0.36 \cos 30^{\circ}\right) \omega_{A D} \mathbf{j} \end{aligned}$

Equate the coefficients of $\mathbf{i}$ from Equation (1).

$0.768-\left(0.36 \sin 30^{\circ}\right) \omega_{A D}=0$

$\omega_{A D}=4.267 \mathrm{rad} / \mathrm{s}$

The position vector for the link BD is obtained. Then the expression for the relative velocity between points B and D is used to obtain the equation (1). Then i components are equated on both sides of equation (1) to calculate the angular velocity of the link AD.

(B) Rewrite the Equation (1).

$-v_{D} \mathbf{j}=\left[0.768-\left(0.36 \sin 30^{\circ}\right) \omega_{A D}\right] \mathbf{i}-\left(0.36 \cos 30^{\circ}\right) \omega_{A D} \mathbf{j}$

Equate the components of $\mathrm{j}$

$-v_{D}=-\left(0.36 \cos 30^{\circ}\right) \omega_{A D}$

Substitute $4.267 \mathrm{rad} / \mathrm{s}$ for $\omega_{A D}$.

$v_{D}=0.360 \cos 30^{\circ} \times 4.267$

$v_{D}=1.33 \mathrm{~m} / \mathrm{s}$

Equation (1) obtained for the velocity of D in vector form is considered. Then components of j are equated on both sides of the expression to calculate the velocity of collar D.

(c)

Write the position vector for $\mathrm{AB}\left(\mathbf{r}_{A B}\right)$.

$\mathbf{r}_{A B}=\left(0.24 \cos 30^{\circ} \mathbf{i}-0.24 \sin 30^{\circ} \mathbf{j}\right) \mathrm{m}$

Write the expression for relative velocity between points $A$ and $B$ for link $A B$.

$\mathbf{v}_{A}=\mathbf{v}_{B}+\omega_{A D} \mathbf{k} \times \mathbf{r}_{A B}$

Here, velocity of point $\mathrm{A}$ is $\boldsymbol{v}_{A}$.

Substitute $0.768 \mathbf{i} \mathrm{m} / \mathrm{s}$ for $\mathbf{v}_{B}, 4.267 \mathrm{rad} / \mathrm{s}$ for $\omega_{A D, \text { and }}\left(0.24 \cos 30^{\circ} \mathbf{i}-0.24 \sin 30^{\circ} \mathbf{j}\right) \mathrm{m}_{\text {for }} \mathbf{r}_{A B}$.

$\begin{aligned} \mathbf{v}_{A} &=0.768 \mathrm{i} \mathrm{m} / \mathrm{s}+(4.267 \mathrm{rad} / \mathrm{s}) \mathbf{k} \times\left(0.24 \cos 30^{\circ} \mathbf{i}-0.24 \sin 30^{\circ} \mathbf{j}\right) \mathrm{m} \\ &=(1.28 \mathbf{i}+0.887 \mathbf{j}) \mathrm{m} / \mathrm{s} \end{aligned}$

Calculate the magnitude of velocity of point $A$.

$v_{A}=\sqrt{(1.28)^{2}+(0.887)^{2}}$

$=1.557 \mathrm{~m} / \mathrm{s}$

The position vector for the link AB is obtained. Then the expression for the relative velocity between points B and A is used to obtain the velocity of A in vector form.

Part A

The angular velocity of rod AD $\left(\omega_{A D}\right)$ is $4.267 \mathrm{rad} / \mathrm{s}$.

Part B

The velocity of collar D $\left(v_{D}\right)$ is $1.33 \mathrm{~m} / \mathrm{s}$.

Part C

The velocity of point $A\left(v_{A}\right)$ is $1.557 \mathrm{~m} / \mathrm{s}$