Chapter 16.1, Problem 16.34P

Each of the double pulleys shown has a mass moment of inertia of 15 lb·ft·s² and is initially at rest. The outside radius is 18 in., and the inner radius is 9 in. Determine (a) the angular acceleration of each pulley, (b) the angular velocity of each pulley after point A on the cord has moved 10 ft.

Fig. P16.34

To determine

Find the angular acceleration of the pulley 1 $\left({\alpha }_1\right)$.

Find the angular acceleration of the pulley 2 $\left({\alpha }_2\right)$.

Find the angular acceleration of the pulley 3 $\left({\alpha }_3\right)$.

Find the angular acceleration of the pulley 4 $\left({\alpha }_4\right)$.

Answer to Problem 16.34P

The angular acceleration of the pulley 1 $\left({\alpha }_1\right)$ is $\underset{\_}{8.00\text{rad}/{\text{s}}^{\text{2}}}$.

The angular acceleration of the pulley 2 $\left({\alpha }_2\right)$ is $\underset{\_}{6.74\text{rad}/{\text{s}}^{\text{2}}}$.

The angular acceleration of the pulley 3 $\left({\alpha }_3\right)$ is $\underset{\_}{4.24\text{rad}/{\text{s}}^{\text{2}}}$.

The angular acceleration of the pulley 4 $\left({\alpha }_4\right)$ is $\underset{\_}{5.83\text{rad}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

The mass moment of inertia of the double pulleys $\left(I_O\right)$ is $15\text{lb}\cdot \text{ft}\cdot {\text{s}}^2$.

The outside radius of the pulley $\left(R\right)$ is $18\text{in}\text{.}$.

The inner radius of the pulley $\left(r\right)$ is $9\text{in}\text{.}$.

The finial angular velocity of the pulley $\left(\omega \right)$ is 0.

The load of the pulley 1 $\left(W_1\right)$ is $160\text{lb}$.

The load of the pulley 2 $\left(W_2\right)$ is $160\text{lb}$.

The left side load of the pulley 3 $\left(W_3\right)$ is $460\text{lb}$.

The right side load of the pulley 3 $\left(W_4\right)$ is $300\text{lb}$.

The load of the pulley 4 $\left(W_5\right)$ is $80\text{lb}$.

Calculation:

Consider the acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Case 1:

Convert the unit of the outside radius $\left(R\right)$:

$\begin{array}{l}R=18\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ R=1.5\text{ft}\end{array}$

Convert the unit of the inner radius $\left(r\right)$:

$\begin{array}{l}r=9\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ r=0.75\text{ft}\end{array}$

Show the free body diagram of the double pulley 1 as in Figure 1.

Here, $\alpha$ is the angular acceleration of the pulley.

Refer to Figure 1.

Calculate the angular acceleration of the pulley 1 $\left({\alpha }_1\right)$:

Calculate the moment about point O by applying the equation of equilibrium:

$\begin{array}{l}{\displaystyle \sum M_O}=I_O\alpha \\ \left(160\times 0.75\right)=15{\alpha }_1\\ {\alpha }_1=\frac{160\times 0.75}{15}\\ {\alpha }_1=8\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Hence, the angular acceleration of the pulley 1 $\left({\alpha }_1\right)$ is $\underset{\_}{8.00\text{rad}/{\text{s}}^{\text{2}}}$.

Case 2:

Calculate the mass of the pulley 2 $\left(m_2\right)$:

$m_2=\frac{W_2}{g}$

Substitute $160\text{lb}$ for $W_2$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_2=\frac{160}{32.2}\\ =4.9689\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

Show the free body diagram of the double pulley 2 as in Figure 2.

Refer to Figure 2.

Calculate the moment about point O by applying the equation of equilibrium:

$\begin{array}{l}{\displaystyle \sum M_O}=I_O\alpha \\ \left(160\times 0.75\right)=15{\alpha }_2+\left(m_2\times 0.75{\alpha }_2\times 0.75\right)\\ 120=15{\alpha }_2+0.5625m_2{\alpha }_2\\ 120=\left(15+0.5625m_2\right){\alpha }_2\end{array}$ (1)

Calculate the angular acceleration of the pulley 2 $\left({\alpha }_2\right)$:

Substitute $4.9689\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_2$ in Equation (1).

$\begin{array}{l}120=\left[15+\left(0.5625\times 4.9689\right)\right]{\alpha }_2\\ 17.795{\alpha }_2=120\\ {\alpha }_2=\frac{120}{17.795}\\ {\alpha }_2=6.74\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Hence, the angular acceleration of the pulley 2 $\left({\alpha }_2\right)$ is $\underset{\_}{6.74\text{rad}/{\text{s}}^{\text{2}}}$.

Case 3:

Calculate the left side mass of the pulley 3 $\left(m_3\right)$:

$m_3=\frac{W_3}{g}$

Substitute $460\text{lb}$ for $W_3$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_3=\frac{460}{32.2}\\ =14.2857\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

Calculate the right side mass of the pulley 3 $\left(m_4\right)$:

$m_4=\frac{W_4}{g}$

Substitute $300\text{lb}$ for $W_4$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_4=\frac{300}{32.2}\\ =9.3168\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

Show the free body diagram of the double pulley 3 as in Figure 3.

Refer to Figure 3.

Calculate the moment about point O by applying the equation of equilibrium:

$\begin{array}{l}{\displaystyle \sum M_O}=I_O\alpha \\ \left(460\times 0.75\right)-\left(300\times 0.75\right)=\left[\begin{array}{l}15{\alpha }_3+\left(m_3\times 0.75{\alpha }_3\times 0.75\right)\\ +\left(m_4\times 0.75{\alpha }_3\times 0.75\right)\end{array}\right]\\ 120=15{\alpha }_3+0.5625m_3{\alpha }_3+0.5625m_4{\alpha }_3\\ 120=\left(15+0.5625m_3+0.5625m_4\right){\alpha }_3\end{array}$ (2)

Calculate the angular acceleration of the pulley 3 $\left({\alpha }_3\right)$:

Substitute $14.2857\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_3$ and $9.3168\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_4$ in Equation (2).

$\begin{array}{l}120=\left[15+\left(0.5625\times 14.2857\right)+\left(0.5625\times 9.3168\right)\right]{\alpha }_3\\ 28.2764{\alpha }_3=120\\ {\alpha }_3=\frac{120}{28.2764}\\ {\alpha }_3=4.24\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Hence, the angular acceleration of the pulley 3 $\left({\alpha }_3\right)$ is $\underset{\_}{4.24\text{rad}/{\text{s}}^{\text{2}}}$.

Case 4:

Calculate the left side mass of the pulley 4 $\left(m_5\right)$:

$m_5=\frac{W_5}{g}$

Substitute $80\text{lb}$ for $W_5$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}{m}_5=\frac{80}{32.2}\\ =2.4845\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

Show the free body diagram of the double pulley 4 as in Figure 4.

Refer to Figure 4.

Calculate the moment about point O by applying the equation of equilibrium:

$\begin{array}{l}{\displaystyle \sum M_O}=I_O\alpha \\ \left(80\times 1.5\right)=15{\alpha }_4+\left(m_5\times 1.5{\alpha }_4\times 1.5\right)\\ 120=15{\alpha }_3+2.25m_5{\alpha }_4\\ 120=\left(15+2.25m_5\right){\alpha }_3\end{array}$ (3)

Calculate the angular acceleration of the pulley 4 $\left({\alpha }_4\right)$:

Substitute $2.4845\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_5$ in Equation (3).

$\begin{array}{l}120=\left[15+\left(2.25\times 2.4845\right)\right]{\alpha }_4\\ 20.59{\alpha }_4=120\\ {\alpha }_4=\frac{120}{20.59}\\ {\alpha }_4=5.83\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Hence, the angular acceleration of the pulley 4 $\left({\alpha }_4\right)$ is $\underset{\_}{5.83\text{rad}/{\text{s}}^{\text{2}}}$.

To determine

Find the angular velocity of the pulley 1 $\left({\omega }_1\right)$.

Find the angular velocity of the pulley 2 $\left({\omega }_2\right)$.

Find the angular velocity of the pulley 3 $\left({\omega }_3\right)$.

Find the angular velocity of the pulley 4 $\left({\omega }_4\right)$.

Answer to Problem 16.34P

The angular velocity of the pulley 1 $\left({\omega }_1\right)$ is $\underset{\_}{14.61\text{rad}/\text{s}}$.

The angular velocity of the pulley 2 $\left({\alpha }_2\right)$ is $\underset{\_}{13.41\text{rad}/\text{s}}$.

The angular velocity of the pulley 3 $\left({\alpha }_3\right)$ is $\underset{\_}{10.64\text{rad}/\text{s}}$.

The angular velocity of the pulley 4 $\left({\alpha }_4\right)$ is $\underset{\_}{8.82\text{rad}/\text{s}}$.

Explanation of Solution

The mass moment of inertia of the double pulleys $\left(I_O\right)$ is $15\text{lb}\cdot \text{ft}\cdot {\text{s}}^2$.

The outside radius of the pulley $\left(R\right)$ is $18\text{in}\text{.}$.

The inner radius of the pulley $\left(r\right)$ is $9\text{in}\text{.}$.

The finial angular velocity of the pulley $\left(\omega \right)$ is 0.

The load of the pulley 1 $\left(W_1\right)$ is $160\text{lb}$.

The load of the pulley 2 $\left(W_2\right)$ is $160\text{lb}$.

The left side load of the pulley 3 $\left(W_3\right)$ is $460\text{lb}$.

The right side load of the pulley 3 $\left(W_4\right)$ is $300\text{lb}$.

The load of the pulley 4 $\left(W_5\right)$ is $80\text{lb}$.

The moved distance of the point A (l) is $10\text{ft}$.

Calculation:

Refer part (a).

Case 1:

Calculate the angle of the pulley 1 $\left({\theta }_1\right)$:

${\theta }_1=\frac{l}{r}$

Substitute $10\text{ft}$ for l and $0.75\text{ft}$ for r.

$\begin{array}{c}{\theta }_1=\frac{10}{0.75}\\ =13.333\text{rad}\end{array}$

Calculate the angular velocity of the pulley 1 $\left({\omega }_1\right)$:

${\omega }_1^2=2{\alpha }_1{\theta }_1$

Substitute $8\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_1$ and $13.333\text{rad}$ for ${\theta }_1$

$\begin{array}{l}{\omega }_1^2=2\times 8\times 13.333\\ {\omega }_1=\sqrt{213.328}\\ {\omega }_1=14.61\text{rad}/\text{s}\end{array}$

Hence, the angular velocity of the pulley 1 $\left({\omega }_1\right)$ is $\underset{\_}{14.61\text{rad}/\text{s}}$.

Case 2:

Calculate the angle of the pulley 2 $\left({\theta }_2\right)$:

${\theta }_2=\frac{l}{r}$

Substitute $10\text{ft}$ for l and $0.75\text{ft}$ for r.

$\begin{array}{c}{\theta }_2=\frac{10}{0.75}\\ =13.333\text{rad}\end{array}$

Calculate the angular velocity of the pulley 2 $\left({\omega }_2\right)$:

${\omega }_2^2=2{\alpha }_2{\theta }_2$

Substitute $6.74\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_2$ and $13.333\text{rad}$ for ${\theta }_2$

$\begin{array}{l}{\omega }_2^2=2\times 6.74\times 13.333\\ {\omega }_2=\sqrt{179.72884}\\ {\omega }_2=13.41\text{rad}/\text{s}\end{array}$

Hence, the angular velocity of the pulley 2 $\left({\omega }_2\right)$ is $\underset{\_}{13.41\text{rad}/\text{s}}$.

Case 3:

Calculate the angle of the pulley 3 $\left({\theta }_3\right)$:

${\theta }_3=\frac{l}{r}$

Substitute $10\text{ft}$ for l and $0.75\text{ft}$ for r.

$\begin{array}{c}{\theta }_3=\frac{10}{0.75}\\ =13.333\text{rad}\end{array}$

Calculate the angular velocity of the pulley 3 $\left({\omega }_3\right)$:

${\omega }_3^2=2{\alpha }_3{\theta }_3$

Substitute $4.24\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_3$ and $13.333\text{rad}$ for ${\theta }_3$

$\begin{array}{l}{\omega }_3^2=2\times 4.24\times 13.333\\ {\omega }_3=\sqrt{113.06384}\\ {\omega }_3=10.64\text{rad}/\text{s}\end{array}$

Hence, the angular velocity of the pulley 3 $\left({\omega }_3\right)$ is $\underset{\_}{10.64\text{rad}/\text{s}}$.

Case 4:

Calculate the angle of the pulley 4 $\left({\theta }_4\right)$:

${\theta }_4=\frac{l}{R}$

Substitute $10\text{ft}$ for l and $1.5\text{ft}$ for R.

$\begin{array}{c}{\theta }_4=\frac{10}{1.5}\\ =6.6667\text{rad}\end{array}$

Calculate the angular velocity of the pulley 4 $\left({\omega }_4\right)$:

${\omega }_3^2=2{\alpha }_3{\theta }_3$

Substitute $5.83\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_4$ and $6.6667\text{rad}$ for ${\theta }_4$

$\begin{array}{l}{\omega }_4^2=2\times 5.83\times 6.6667\\ {\omega }_4=\sqrt{77.733722}\\ {\omega }_4=8.82\text{rad}/\text{s}\end{array}$

Hence, the angular velocity of the pulley 4 $\left({\omega }_4\right)$ is $\underset{\_}{8.82\text{rad}/\text{s}}$.