Chapter 15.1, Problem 15.23P

(a)

To determine

The angular velocity and angular acceleration of the point C on the output belt.

Answer to Problem 15.23P

The angular velocity $\left(v_C\right)$ and angular acceleration $\left({\alpha }_C\right)$ of the point C on the output belt is $\underset{\_}{0.5\text{ft/s}(\to )}$ and $\underset{\_}{1.50{\text{ft/s}}^2(\leftarrow )}$ respectively.

Explanation of Solution

Given information:

The inner radius of left pulley $\left(r_1\right)$ is 2 in..

The outer radius of left pulley $\left(r_2\right)$ is 4 in..

The velocity of point A $\left(v_A\right)$ to the right is 2 ft/s.

The tangential component of acceleration of point A ${\left(a_A\right)}_t$ is $-6{\text{ft/s}}^{\text{2}}$.

The inner radius of right pulley $\left(r_3\right)$ is 2 in..

The outer radius of right pulley $\left(r_4\right)$ is 4 in..

Calculation:

Left pulley:

Determine the angular velocity of the left pulley ${\omega }_1$ using the relation.

${\omega }_1=\frac{v_A}{r_2}$

Substitute 2 ft/s for $v_A$ and 4 in. for $r_2$.

$\begin{array}{c}{\omega }_1=\frac{2}{4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}}\\ =6\text{rad/s}\left(\text{Clockwise}\right)\end{array}$

Draw the free body diagram of the left pulley as in Figure (1).

Determine the angular acceleration of the left pulley ${\alpha }_1$ using the relation.

${\alpha }_1=\frac{{\left(a_A\right)}_t}{r_2}$

Substitute $-6{\text{ft/s}}^{\text{2}}$ for ${\left(a_A\right)}_t$ and 4 in. for $r_2$.

$\begin{array}{c}{\alpha }_1=\frac{-6}{4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}}\\ =-18{\text{rad/s}}^{\text{2}}\\ =18{\text{rad/s}}^{\text{2}}\left(\text{Anticlockwise}\right)\end{array}$

Intermediate belt:

Determine the velocity of the intermediate belt $v_1$ using the relation.

$v_1=r_1{\omega }_1$

Substitute 2 in. for $r_1$ and $6\text{rad/s}$ for ${\omega }_1$.

$\begin{array}{c}{v}_1=2\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\times 6\\ =1\text{ft/s}\end{array}$

Determine the tangential component of acceleration of the intermediate belt ${\left(a_1\right)}_t$ using the relation.

${\left(a_1\right)}_t=r_1{\alpha }_1$

Substitute 2 in. for $r_1$ and $-18{\text{rad/s}}^{\text{2}}$ for ${\alpha }_1$.

$\begin{array}{c}{\left(a_1\right)}_t=2\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\times -18\\ =-3{\text{ft/s}}^{\text{2}}\end{array}$

Right pulley:

Draw the free body diagram of the right pulley as in Figure (2).

Determine the angular velocity of the right pulley ${\omega }_2$ using the relation.

${\omega }_2=\frac{v_1}{r_4}$

Substitute 1 ft/s for $v_1$ and 4 in. for $r_4$.

$\begin{array}{c}{\omega }_2=\frac{1}{4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}}\\ =3\text{rad/s}\left(\text{Clockwise}\right)\end{array}$

Determine the angular acceleration of the right pulley ${\alpha }_2$ using the relation.

${\alpha }_2=\frac{{\left(a_1\right)}_t}{r_4}$

Substitute $3{\text{ft/s}}^{\text{2}}$ for ${\left(a_1\right)}_t$ and 4 in. for $r_4$.

$\begin{array}{c}{\alpha }_2=\frac{3}{4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}}\\ =9{\text{rad/s}}^{\text{2}}\left(\text{Anticlockwise}\right)\end{array}$

Determine the velocity of point C.

$v_C=r_3{\omega }_2$

Substitute 2 in. for $r_3$ and $3\text{rad/s}$ for ${\omega }_2$.

$\begin{array}{c}{v}_C=2\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\times 3\\ =0.5\text{ft/s}(\to )\end{array}$

Determine the tangential component of acceleration of point C ${\left(a_C\right)}_t$ using the relation.

${\left(a_C\right)}_t=r_3{\alpha }_2$

Substitute 2 in. for $r_3$ and $-9{\text{rad/s}}^{\text{2}}$ for ${\alpha }_2$.

$\begin{array}{c}{\left(a_C\right)}_t=2\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\times -9\\ =-1.5{\text{ft/s}}^{\text{2}}\\ =1.5{\text{ft/s}}^{\text{2}}(\leftarrow )\end{array}$

Therefore, the angular velocity $\left(v_C\right)$ and angular acceleration $\left({\alpha }_C\right)$ of the point C on the output belt is $\underset{\_}{0.5\text{ft/s}(\to )}$ and $\underset{\_}{1.5{\text{ft/s}}^2(\leftarrow )}$ respectively.

(b)

To determine

The acceleration of the point B on the output pulley.

Answer to Problem 15.23P

The acceleration of point B on the output pulley is $\underset{\_}{4.24{\text{ft/s}}^2}$ in the direction of $\underset{\_}{45°\left(\text{IV-Quadrant}\right)}$.

Explanation of Solution

Given information:

The inner radius of left pulley $\left(r_1\right)$ is 2 in..

The outer radius of left pulley $\left(r_2\right)$ is 4 in..

The velocity of point A $\left(v_A\right)$ to the right is 2 ft/s.

The tangential component of acceleration of point A ${\left(a_A\right)}_t$ is $-6{\text{ft/s}}^{\text{2}}$.

The inner radius of right pulley $\left(r_3\right)$ is 2 in..

The outer radius of right pulley $\left(r_4\right)$ is 4 in..

Calculation:

Determine the normal component of acceleration of point B ${\left(a_B\right)}_n$ using the relation.

${\left(a_B\right)}_n=r_4{\omega }_2^2$

Substitute 4 in. for $r_4$ and $3\text{rad/s}$ for ${\omega }_2$.

$\begin{array}{c}{\left(a_B\right)}_n=4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\times 3^2\\ =3{\text{ft/s}}^{\text{2}}(\to )\end{array}$

Determine the tangential component of acceleration of point B ${\left(a_B\right)}_t$ using the relation.

${\left(a_B\right)}_t=r_4{\alpha }_2^2$

Substitute 4 in. for $r_4$ and $-9{\text{rad/s}}^{\text{2}}$ for ${\alpha }_2$.

$\begin{array}{c}{\left(a_B\right)}_t=4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\times \left(-9\right)\\ =-3{\text{ft/s}}^{\text{2}}\\ =3{\text{ft/s}}^{\text{2}}(\downarrow )\end{array}$

Determine the magnitude of the total acceleration of the machine component at B.

${\text{a}}_B=\sqrt{{\left(a_B\right)}_t^2+{\left(a_B\right)}_n^2}$

Substitute $-3{\text{ft/s}}^{\text{2}}$ for ${\left(a_B\right)}_t$ and $3{\text{ft/s}}^{\text{2}}$ for ${\left(a_B\right)}_n$.

$\begin{array}{c}{\text{a}}_B=\sqrt{{\left(-3\right)}^2+3^2}\\ =4.24{\text{ft/s}}^{\text{2}}\end{array}$

Determine the direction of the acceleration of point B $\left(\beta \right)$ using the relation.

$\tan \beta =\frac{{\left(a_B\right)}_n}{{\left(a_B\right)}_t}$

Substitute $3{\text{ft/s}}^{\text{2}}$ for ${\left(a_B\right)}_n$ and $3{\text{ft/s}}^{\text{2}}$ for ${\left(a_B\right)}_t$.

$\begin{array}{l}\tan \beta =\frac{2}{3}\\ \beta ={\tan }^{-1}\left(1\right)\\ \beta =45°\left(\text{IV-Quadrant}\right)\end{array}$

Therefore, the acceleration of point B on the output pulley is $\underset{\_}{4.24{\text{ft/s}}^2}$ in the direction of $\underset{\_}{45°\left(\text{IV-Quadrant}\right)}$.