Chapter 16.1, Problem 16.45P

(a)

To determine

Find the angular acceleration of each cylinder $\left({\alpha }_A,{\alpha }_B\text{and}{\alpha }_C\right)$.

Answer to Problem 16.45P

The angular acceleration of each cylinder $\left({\alpha }_A,{\alpha }_B\text{and}{\alpha }_C\right)$ are $\underset{\_}{102.7\text{rad}/{\text{s}}^{\text{2}}}$, $\underset{\_}{20.5\text{rad}/{\text{s}}^{\text{2}}}$, and $\underset{\_}{10.27\text{rad}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

The weight of the cylinder A $\left(W_A\right)$ is $5\text{lb}$.

The weight of the cylinder B $\left(W_B\right)$ is $5\text{lb}$.

The weight of the cylinder C $\left(W_C\right)$ is $20\text{lb}$.

The initial angular velocity of the cylinder A $\left({\omega }_0\right)$ is $720\text{rpm}$.

The coefficient of the kinetic friction $\left({\mu }_k\right)$ is $0.25$.

The radius of the cylinder A $\left(r_A\right)$ is $4\text{in}\text{.}$.

The radius of the cylinder B $\left(r_B\right)$ is $4\text{in}\text{.}$.

The radius of the cylinder C $\left(r_C\right)$ is $8\text{in}\text{.}$.

Calculation:

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

Convert the unit of the radius of the cylinder A $\left(r_A\right)$:

$\begin{array}{c}{r}_A=4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =\frac{1}{3}\text{ft}\end{array}$

Convert the unit of the radius of the cylinder B $\left(r_B\right)$:

$\begin{array}{c}{r}_B=4\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =\frac{1}{3}\text{ft}\end{array}$

Convert the unit of the radius of the cylinder C $\left(r_C\right)$:

$\begin{array}{c}{r}_C=8\text{in}\text{.}\times \frac{1\text{ft}}{12\text{in}\text{.}}\\ =\frac{2}{3}\text{ft}\end{array}$

Calculate the mass of the cylinder A $\left(m_A\right)$:

$m_A=\frac{W_A}{g}$

Substitute $5\text{lb}$ for $W_A$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$m_A=\frac{5}{32.2}\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$

Calculate the mass of the cylinder B $\left(m_B\right)$:

$m_B=\frac{W_B}{g}$

Substitute $5\text{lb}$ for $W_B$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$m_B=\frac{5}{32.2}\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$

Calculate the mass of the cylinder C $\left(m_C\right)$:

$m_C=\frac{W_C}{g}$

Substitute $20\text{lb}$ for $W_B$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$m_C=\frac{20}{32.2}\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$

Calculate the mass moment of inertia of the cylinder A $\left({\overline{I}}_A\right)$:

${\overline{I}}_A=\frac{1}{2}{m}_A{r}_A^2$

Substitute $\frac{5}{32.2}\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_A$ and $\frac{1}{3}\text{ft}$ for $r_A$.

$\begin{array}{c}{\overline{I}}_A=\frac{1}{2}\times \frac{5}{32.2}\times {\left(\frac{1}{3}\right)}^2\\ =8.6266\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

Calculate the mass moment of inertia of the cylinder B $\left({\overline{I}}_B\right)$:

${\overline{I}}_B=\frac{1}{2}{m}_B{r}_B^2$

Substitute $\frac{5}{32.2}\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_B$ and $\frac{1}{3}\text{ft}$ for $r_B$.

$\begin{array}{c}{\overline{I}}_B=\frac{1}{2}\times \frac{5}{32.2}\times {\left(\frac{1}{3}\right)}^2\\ =8.6266\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

Calculate the mass moment of inertia of the cylinder C $\left({\overline{I}}_C\right)$:

${\overline{I}}_C=\frac{1}{2}{m}_C{r}_C^2$

Substitute $\frac{20}{32.2}\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for $m_C$ and $\frac{2}{3}\text{ft}$ for $r_C$.

$\begin{array}{c}{\overline{I}}_C=\frac{1}{2}\times \frac{20}{32.2}\times {\left(\frac{2}{3}\right)}^2\\ =138.0262\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}\end{array}$

Calculate the tangential acceleration of contact point between cylinder B and C ${\left(a_t\right)}_{BC}$:

$\begin{array}{c}{\left(a_t\right)}_{BC}=r{\alpha }_B\\ =2r{\alpha }_C\\ {\alpha }_B=2{\alpha }_C\end{array}$

Here, ${\alpha }_B$ is the angular acceleration of the cylinder B and ${\alpha }_C$ is the angular acceleration of the cylinder C.

The friction force at the contact point between cylinder A and C $\left(F_{AC}\right)$ as follows:

$F_{AC}={\mu }_k{N}_{AC}$ (1)

Show the free body diagram of the cylinder B as in Figure 1.

Here, $F_{BC}$ is the friction force at the contact point between cylinder B and C, $N_{BC}$ is the normal force at the contact point between cylinder A and C, $B_x$ is the horizontal force of the cylinder B, and $B_y$ is the vertical force of the cylinder B.

Refer to Figure 1.

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

$\begin{array}{l}{\displaystyle \sum M_B}=I_B\alpha \\ F_{BC}{r}_B={\overline{I}}_B{\alpha }_B\end{array}$

Substitute $\frac{1}{3}\text{ft}$ for $r_B$, $2{\alpha }_C$ for ${\alpha }_B$, and $8.6266\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for ${\overline{I}}_B$.

$\begin{array}{l}\left(F_{BC}\times \frac{1}{3}\right)=8.6266\times {10}^{-3}\times 2{\alpha }_C\\ \frac{1}{3}{F}_{BC}=17.2532\times {10}^{-3}{\alpha }_C\\ F_{BC}=17.2532\times {10}^{-3}{\alpha }_C\times 3\\ F_{BC}=51.7596\times {10}^{-3}{\alpha }_C\end{array}$ (2)

Show the free body diagram of the cylinder C as in Figure 2.

Here, $F_{AC}$ is the friction force at the contact point between cylinder A and C and $N_{AC}$ is the normal force at the contact point between cylinder A and C.

Refer to Figure 2.

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

$\begin{array}{l}{\displaystyle \sum M_C}=I_G\alpha +mad\\ F_{AC}{r}_C-F_{BC}{r}_C={\overline{I}}_C{\alpha }_C\end{array}$

Substitute $\frac{2}{3}\text{ft}$ for $r_C$, $51.7596\times {10}^{-3}{\alpha }_C$ for $F_{BC}$, and $138.0262\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for ${\overline{I}}_C$.

$\begin{array}{l}\left(F_{AC}\times \frac{2}{3}\right)-\left(51.7596\times {10}^{-3}{\alpha }_C\times \frac{2}{3}\right)=138.0262\times {10}^{-3}{\alpha }_C\\ \frac{2}{3}{F}_{AC}=34.5064\times {10}^{-3}{\alpha }_C+138.0262\times {10}^{-3}{\alpha }_C\\ F_{AC}=172.5326\times {10}^{-3}{\alpha }_C\times \frac{3}{2}\\ F_{AC}=258.7989\times {10}^{-3}{\alpha }_C\end{array}$ (3)

Calculate the normal force at the contact point between cylinder A and C $\left(N_{AC}\right)$:

Substitute $0.25$ for ${\mu }_k$ and $258.7989\times {10}^{-3}{\alpha }_C$ for $F_{AC}$ in Equation (1).

$\begin{array}{l}258.7989\times {10}^{-3}{\alpha }_C=0.25N_{AC}\\ N_{AC}=\frac{258.7989\times {10}^{-3}{\alpha }_C}{0.25}\\ N_{AC}=1.0352{\alpha }_C\end{array}$ (4)

Consider that the contact point between cylinder A and C is P.

Calculate the components of forces acting along the line CP:

$W_C\sin 30°+F_{BC}-F_{AC}\cos 60°-N_{AC}\sin 60°=0$ (5)

Calculate the angular acceleration of the cylinder C $\left({\alpha }_C\right)$:

Substitute $20\text{lb}$ for $W_C$, $51.7596\times {10}^{-3}{\alpha }_C$ for $F_{BC}$, $258.7989\times {10}^{-3}{\alpha }_C$ for $F_{AC}$, and $1.0352{\alpha }_C$ for $N_{AC}$ in Equation (5).

$\begin{array}{l}\left[\begin{array}{l}20\sin 30°+51.7596\times {10}^{-3}{\alpha }_C-\\ \left(258.7989\times {10}^{-3}{\alpha }_C\times \cos 60°\right)-\\ \left(1.0352{\alpha }_C\times \sin 60°\right)\end{array}\right]=0\\ 10-974.1493\times {10}^{-3}{\alpha }_C=0\\ 974.1493\times {10}^{-3}{\alpha }_C=10\end{array}$

$\begin{array}{l}{\alpha }_C=\frac{10}{974.1493\times {10}^{-3}}\\ {\alpha }_C=10.2654\text{rad}/{\text{s}}^{\text{2}}\\ {\alpha }_C\approx 10.27\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Calculate the friction force at the contact point between cylinder A and C $\left(F_{AC}\right)$:

Substitute $10.2654\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_C$ in Equation (3).

$\begin{array}{c}{F}_{AC}=258.7989\times {10}^{-3}\times 10.2654\\ =2.6567\text{lb}\end{array}$

Calculate the friction force at the contact point between cylinder B and C $\left(F_{BC}\right)$:

Substitute $10.2654\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_C$ in Equation (2).

$\begin{array}{c}{F}_{BC}=51.7596\times {10}^{-3}\times 10.2654\\ =0.53134\text{lb}\end{array}$

Calculate the normal force at the contact point between cylinder A and C $\left(N_{AC}\right)$:

Substitute $10.2654\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_C$ in Equation (4).

$\begin{array}{c}{N}_{AC}=1.0352\times 10.2654\\ =10.6267\text{lb}\end{array}$

Show the free body diagram of the cylinder A as in Figure 3.

Here, $A_x$ is the horizontal force of the cylinder A, and $A_y$ is the vertical force of the cylinder A.

Refer to Figure 3.

Consider that the contact point between cylinder B and C is Q.

Calculate the components of forces acting along the line CQ:

$N_{BC}-W_C\cos 30°+N_{AC}\cos 60°-F_{AC}\sin 60°=0$ (6)

Calculate the normal force at the contact point between cylinder B and C $\left(N_{BC}\right)$:

Substitute $20\text{lb}$ for $W_C$, $10.6267\text{lb}$ for $N_{AC}$, and $2.6567\text{lb}$ for $F_{AC}$ in Equation (6).

$\begin{array}{l}\left[\begin{array}{l}{N}_{BC}-20\cos 30°+\left(10.6267\times \cos 60°\right)-\\ \left(2.6567\times \sin 60°\right)\end{array}\right]=0\\ N_{BC}-14.3079=0\\ N_{BC}=14.3079\text{lb}\end{array}$

Calculate the angular acceleration of the cylinder A $\left({\alpha }_A\right)$:

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

$\begin{array}{l}{\displaystyle \sum M_A}=I_A\alpha \\ F_{AC}{r}_A={\overline{I}}_A{\alpha }_A\end{array}$

Substitute $\frac{1}{3}\text{ft}$ for $r_A$, $2.6567\text{lb}$ for $F_{AC}$, and $8.6266\times {10}^{-3}\text{lb}\cdot {\text{s}}^2\cdot \text{ft}$ for ${\overline{I}}_A$.

$\begin{array}{l}\left(2.6567\times \frac{1}{3}\right)=8.6266\times {10}^{-3}{\alpha }_A\\ \frac{2.6567}{3}=8.6266\times {10}^{-3}{\alpha }_A\\ {\alpha }_A=\frac{2.6567}{3\times 8.6266\times {10}^{-3}}\\ {\alpha }_A=102.7\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Calculate the angular acceleration of the cylinder A $\left({\alpha }_A\right)$:

${\alpha }_B=2{\alpha }_C$

Substitute $10.2654\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_C$ Equation (1).

$\begin{array}{c}{\alpha }_B=2\times 10.2654\\ =20.5\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Hence, the angular acceleration of each cylinder $\left({\alpha }_A,{\alpha }_B\text{and}{\alpha }_C\right)$ are $\underset{\_}{102.7\text{rad}/{\text{s}}^{\text{2}}}$, $\underset{\_}{20.5\text{rad}/{\text{s}}^{\text{2}}}$, and $\underset{\_}{10.27\text{rad}/{\text{s}}^{\text{2}}}$.

(b)

To determine

Find the final angular velocity of each disk $\left({\omega }_A,{\omega }_B\text{and}{\omega }_C\right)$.

Answer to Problem 16.45P

The final angular velocity of each disk $\left({\omega }_A,{\omega }_B\text{and}{\omega }_C\right)$ are $\underset{\_}{120\text{rpm}}$, $\underset{\_}{120\text{rpm}}$, and $\underset{\_}{60\text{rpm}}$.

Explanation of Solution

The weight of the cylinder A $\left(W_A\right)$ is $5\text{lb}$.

The weight of the cylinder B $\left(W_B\right)$ is $5\text{lb}$.

The weight of the cylinder C $\left(W_C\right)$ is $20\text{lb}$.

The initial angular velocity of the cylinder A $\left({\omega }_0\right)$ is $720\text{rpm}$.

The coefficient of the kinetic friction $\left({\mu }_k\right)$ is $0.25$.

The radius of the cylinder A $\left(r_A\right)$ is $4\text{in}\text{.}$.

The radius of the cylinder B $\left(r_B\right)$ is $4\text{in}\text{.}$.

The radius of the cylinder C $\left(r_C\right)$ is $8\text{in}\text{.}$.

Calculation:

Refer to part (a).

The convert the unit of the initial angular velocity of the disk A $\left({\omega }_0\right)$:

$\begin{array}{c}\left({\omega }_0\right)=720\text{rpm}\times \frac{\frac{2\pi }{60}\text{rad}/{\text{s}}^{\text{2}}}{1\text{rpm}}\\ =24\pi \text{rad}/{\text{s}}^{\text{2}}\end{array}$

Calculate the angular velocity of cylinder A $\left({\omega }_A\right)$:

${\omega }_A={\omega }_0-{\alpha }_{A}t$

Substitute $24\pi \text{rad}/{\text{s}}^{\text{2}}$ for ${\omega }_0$ and $102.7\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$.

${\omega }_A=24\pi -102.7t$ (7)

Calculate the tangential velocity of cylinder A ${\left(v_t\right)}_{AC}$:

${\left(v_t\right)}_{AC}=r_A{\omega }_A$

Substitute $\left(24\pi -102.7t\right)$ for ${\omega }_A$ and $\frac{1}{3}\text{ft}$ for $r_A$.

$\begin{array}{c}{\left(v_t\right)}_{AC}=\frac{1}{3}\times \left(24\pi -102.7t\right)\\ =8\pi -34.23t\end{array}$

Calculate the angular velocity of cylinder C $\left({\omega }_C\right)$:

${\omega }_C={\alpha }_{C}t$

Substitute $10.27\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_C$.

${\omega }_C=10.27t$ (8)

Calculate the tangential velocity of cylinder C ${\left(v_t\right)}_{CA}$:

${\left(v_t\right)}_{CA}=2r_A{\omega }_C$

Substitute $10.27t$ for ${\omega }_C$ and $\frac{1}{3}\text{ft}$ for $r_A$.

$\begin{array}{c}{\left(v_t\right)}_{CA}=2\times \frac{1}{3}\times 10.27t\\ =6.846t\end{array}$

Calculate the time taken when tangential velocities are equal:

${\left(v_t\right)}_{AC}={\left(v_t\right)}_{CA}$

Substitute $\left(8\pi -34.23t\right)$ for ${\left(v_t\right)}_{AC}$ and $6.846t$ for ${\left(v_t\right)}_{CA}$.

$\begin{array}{l}8\pi -34.23t=6.846t\\ 6.846t+34.23t=8\pi \\ 41.076t=8\pi \end{array}$

$\begin{array}{l}t=\frac{8\pi }{41.076}\\ t=0.6119\text{s}\end{array}$

Calculate the final angular velocity of the disk A $\left({\omega }_A\right)$:

Substitute $0.6119\text{s}$ for $t$ in Equation (7).

$\begin{array}{c}{\omega }_A=24\pi -\left(102.7\times 0.6119\right)\\ =12.56\text{rad}/\text{s}\times \frac{\frac{60}{2\pi }\text{rpm}}{1\text{rad}/\text{s}}\\ =120\text{rpm}\end{array}$

Calculate the final angular velocity of the disk C $\left({\omega }_C\right)$:

Substitute $0.6119\text{s}$ for $t$ in Equation (8).

$\begin{array}{c}{\omega }_C=\left(10.27\times 0.6119\right)\\ =6.28\text{rad}/\text{s}\times \frac{\frac{60}{2\pi }\text{rpm}}{1\text{rad}/\text{s}}\\ =60\text{rpm}\end{array}$

Calculate the final angular velocity of the disk B $\left({\omega }_B\right)$:

${\omega }_B={\alpha }_{B}t$

Substitute $20.5\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_B$, and $0.6119\text{s}$ for $t$.

$\begin{array}{c}{\omega }_B=20.5\times 0.6119\\ =12.54\times \frac{\frac{60}{2\pi }\text{rpm}}{1\text{rad}/\text{s}}\\ =120\text{rpm}\end{array}$

Hence, the final angular velocity of each disk $\left({\omega }_A,{\omega }_B\text{and}{\omega }_C\right)$ are $\underset{\_}{120\text{rpm}}$, $\underset{\_}{120\text{rpm}}$, and $\underset{\_}{60\text{rpm}}$.