Question

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 \(1.557 \mathrm{~m} / \mathrm{s}\)

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