Chapter 16.1, Problem 16.43P

Disk A has a mass m_A = 4 kg, a radius r_A = 300 mm, and an initial angular velocity ω₀ = 300 rpm clockwise. Disk B has a mass m_B = 1.6 kg, a radius r_B = 180 mm, and is at rest when it is brought into contact with disk A. Knowing that μ_k = 0.35 between the disks and neglecting bearing friction, determine (a) the angular acceleration of each disk, (b) the reaction at the support C.

Fig. P16.43 and P16.44

To determine

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

Answer to Problem 16.43P

The angular acceleration of each disk $\left({\alpha }_A\text{and}{\alpha }_B\right)$ are $\underset{\_}{9.16\text{rad}/{\text{s}}^{\text{2}}}$ and $\underset{\_}{38.2\text{rad}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

The mass of the disk A $\left(m_A\right)$ is $4\text{kg}$.

The mass of the disk B $\left(m_B\right)$ is $1.6\text{kg}$.

The initial angular velocity of the disk A $\left({\omega }_0\right)$ is $300\text{rpm}$.

The coefficient of the kinetic friction $\left({\mu }_k\right)$ is $0.35$.

The radius of the disk A $\left(r_A\right)$ is $300\text{mm}$.

The radius of the disk B $\left(r_B\right)$ is $180\text{mm}$.

Calculation:

Consider the acceleration due to gravity (g) as $9.81\text{m}/{\text{s}}^{\text{2}}$.

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

$\begin{array}{c}{r}_A=300\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\\ =0.3\text{m}\end{array}$

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

$\begin{array}{c}{r}_B=180\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\\ =0.18\text{m}\end{array}$

Calculate the mass moment of inertia of the disk A $\left(I_A\right)$:

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

Substitute $4\text{kg}$ for $m_A$ and $0.3\text{m}$ for $r_A$.

$\begin{array}{c}{I}_A=\frac{1}{2}\times 4\times {\left(0.3\right)}^2\\ =0.18\text{kg}\cdot {\text{m}}^2\end{array}$

Calculate the mass moment of inertia of the disk B $\left(I_B\right)$:

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

Substitute $1.6\text{kg}$ for $m_B$ and $0.18\text{m}$ for $r_B$.

$\begin{array}{c}{I}_B=\frac{1}{2}\times 1.6\times {\left(0.18\right)}^2\\ =25.92\times {10}^{-3}\text{kg}\cdot {\text{m}}^2\end{array}$

Calculate the load of the disk A $\left(W_A\right)$:

$W_A=m_{A}g$

Substitute $4\text{kg}$ for $m_A$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$.

$\begin{array}{c}{W}_A=4\times 9.81\\ =39.24\text{N}\end{array}$

Calculate the load of the disk B $\left(W_B\right)$:

$W_B=m_{B}g$

Substitute $1.6\text{kg}$ for $m_B$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$.

$\begin{array}{c}{W}_B=1.6\times 9.81\\ =15.696\text{N}\end{array}$

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

Here, $F$ is the magnitude of the friction force, $R_B$ is the reaction force of the disk B, $N$ is the vertical force of the disk B, and ${\alpha }_B$ is the angular acceleration of the disk B.

Refer to Figure 1.

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ N-W_B=0\\ N=W_B\end{array}$

Substitute $15.696\text{N}$ for $W_B$

$N=15.696\text{N}$

Calculate the magnitude of the friction force $\left(F\right)$:

$F={\mu }_{k}N$

Substitute $15.696\text{N}$ for $N$ and $0.35$ for ${\mu }_k$.

$\begin{array}{c}F=0.35\times 15.696\\ =5.4936\text{N}\end{array}$

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F-R_B=0\\ R_B=F\end{array}$

Substitute $5.4936\text{N}$ for $F$

$R_B=5.4936\text{N}\leftarrow$

Calculate the angular acceleration of the disk B $\left({\alpha }_B\right)$:

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

$\begin{array}{l}{\displaystyle \sum M_B}=I_G\alpha +mad\\ F{r}_B={\overline{I}}_B{\alpha }_B\end{array}$

Substitute $5.4936\text{N}$ for $F$, $0.18\text{m}$ for $r_A$, and $25.92\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_A$.

$\begin{array}{l}\left(5.4936\times 0.18\right)=25.92\times {10}^{-3}\times {\alpha }_B\\ {\alpha }_B=\frac{5.4936\times 0.18}{25.92\times {10}^{-3}}\\ {\alpha }_B=38.2\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Show the free body diagram of the disk A as in Figure 2.

Here, $R_{A_x}$ is the horizontal reaction force of the disk A, $R_{A_y}$ is the vertical reaction force of the disk A, and ${\alpha }_A$ is the angular acceleration of the disk A.

Refer to Figure 2.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ F-R_{A_x}=0\\ R_{A_x}=F\end{array}$

Substitute $5.4936\text{N}$ for $F$

$R_{A_x}=5.4936\text{N}\to$

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_{A_y}-W_A-N=0\\ R_{A_y}=W_A+N\end{array}$

Substitute $39.24\text{N}$ for $W_A$ and $15.696\text{N}$ for $N$.

$\begin{array}{c}{R}_{A_y}=39.24+15.696\\ =54.936\text{N}\uparrow \end{array}$

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

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

$\begin{array}{l}{\displaystyle \sum M_A}=I_G\alpha +mad\\ F{r}_A={\overline{I}}_A{\alpha }_A\end{array}$

Substitute $5.4936\text{N}$ for $F$, $0.3\text{m}$ for $r_A$, and $0.18\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_A$.

$\begin{array}{l}\left(5.4936\times 0.3\right)=0.18\times {\alpha }_A\\ {\alpha }_A=\frac{5.4939\times 0.3}{0.18}\\ {\alpha }_A=9.16\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Hence, the angular acceleration of each disk $\left({\alpha }_A\text{and}{\alpha }_B\right)$ are $\underset{\_}{9.16\text{rad}/{\text{s}}^{\text{2}}}$ and $\underset{\_}{38.2\text{rad}/{\text{s}}^{\text{2}}}$.

To determine

Find the reaction at the support C $\left(C_y\text{and}{M}_C\right)$.

Answer to Problem 16.43P

The reaction at the support C $\left(C_y\text{and}{M}_C\right)$ is $\underset{\_}{54.9\text{N}\uparrow }$ and $\underset{\_}{2.64\text{N}\cdot \text{m}}$.

Explanation of Solution

The mass of the disk A $\left(m_A\right)$ is $4\text{kg}$.

The mass of the disk B $\left(m_B\right)$ is $1.6\text{kg}$.

The initial angular velocity of the disk A $\left({\omega }_0\right)$ is $300\text{rpm}$.

The coefficient of the kinetic friction $\left({\mu }_k\right)$ is $0.35$.

The radius of the disk A $\left(r_A\right)$ is $300\text{mm}$.

The radius of the disk B $\left(r_B\right)$ is $180\text{mm}$.

Calculation:

Refer to part (a).

Show the free body diagram of the support C as in Figure 3.

Here, $C_x$ is the horizontal force of the support C, $C_y$ is the vertical force of the support C, and $M_C$ is the moment of the support C.

Refer to Figure 3.

Calculate the horizontal forces by applying the equation of equilibrium:

Sum of horizontal forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ C_x=0\end{array}$

Calculate the vertical forces by applying the equation of equilibrium:

Sum of vertical forces is equal to 0.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ C_y-54.936=0\\ C_y=54.936\text{N}\uparrow \end{array}$

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

Sum of moments about point C is equal to 0.

$\begin{array}{l}{\displaystyle \sum M_C}=0\\ M_C-\left[5.4936\times \left(0.3+0.18\right)\right]=0\\ M_C-2.64=0\\ M_C=2.64\text{N}\cdot \text{m}\end{array}$

Substitute $5.4936\text{N}$ for $F$, $0.18\text{m}$ for $r_A$, and $25.92\times {10}^{-3}\text{kg}\cdot {\text{m}}^2$ for ${\overline{I}}_A$.

$\begin{array}{l}\left(5.4936\times 0.18\right)=25.92\times {10}^{-3}\times {\alpha }_B\\ {\alpha }_B=\frac{5.4936\times 0.18}{25.92\times {10}^{-3}}\\ {\alpha }_B=38.2\text{rad}/{\text{s}}^{\text{2}}\end{array}$

Calculate the time required for the disk to come to rest (t):

Substitute $0.08\text{m}$ for $r_A$, $0.06\text{m}$ for $r_B$, $12\pi \text{rad}/{\text{s}}^{\text{2}}$ for ${\left({\omega }_B\right)}_0$, $12.5\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$, and $33.3\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_B$ in Equation (1).

$\begin{array}{l}12.5t\times 0.08=\left[12\pi -33.3t\right]\times 0.06\\ 1t=2.2619-2t\\ t+2t=2.2619\end{array}$

$\begin{array}{l}3t=2.2619\\ t=\frac{2.2619}{3}\\ t=0.75397\text{s}\end{array}$

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

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

Substitute $12.5\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_A$, and $0.75397\text{s}$ for $t$.

$\begin{array}{c}{\omega }_A=12.5\times 0.75397\text{s}\\ =9.425\text{rad}/\text{s}\times \frac{\frac{60}{2\pi }\text{rpm}}{1\text{rad}/\text{s}}\\ =90\text{rpm}\end{array}$

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

${\omega }_B={\left({\omega }_B\right)}_0-{\alpha }_{B}t$

Substitute $12\pi \text{rad}/{\text{s}}^{\text{2}}$ for ${\left({\omega }_B\right)}_0$, $33.3\text{rad}/{\text{s}}^{\text{2}}$ for ${\alpha }_B$, and $0.75397\text{s}$ for $t$.

$\begin{array}{c}{\omega }_B=12\pi -\left(33.3\times 0.75397\right)\\ =12.5919\text{rad}/\text{s}\times \frac{\frac{60}{2\pi }\text{rpm}}{1\text{rad}/\text{s}}\\ =120\text{rpm}\end{array}$

Hence, the reaction at the support C $\left(C_y\text{and}{M}_C\right)$ is $\underset{\_}{54.9\text{N}\uparrow }$ and $\underset{\_}{2.64\text{N}\cdot \text{m}}$.