In: Physics

# Knowing that at the instant shown the angular velocity of rod BE is 4 rad/s counterclockwise

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.

## Solutions

##### Expert Solution

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