Chapter 16.1, Problem 16.44P

Disk B is at rest when it is brought into contact with disk A, which has an initial angular velocity ω₀. (a) Show that the final angular velocities of the disks are independent of the coefficient of friction μ_k between the disks as long as μ_k ≠ 0. (b) Express the final angular velocity of disk A in terms of ω₀ and the ratio of the masses of the two disks, m_A/m_B.

Fig. P16.43 and P16.44

To determine

Show that the final velocities of disk A and B are independent of coefficient of kinetic friction ${\mu }_k$.

Explanation of Solution

The mass of the disk A is $m_A$.

The mass of the disk B is $m_B$.

The initial angular velocity of the disk A is ${\omega }_0$.

The coefficient of the kinetic friction is ${\mu }_k$.

The radius of the disk A is $r_A$.

The radius of the disk B is $r_B$.

The acceleration due to gravity is g.

The time required for the disk to come to rest is t.

Calculation:

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

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

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

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

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

$W_A=m_{A}g$

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

$W_B=m_{B}g$

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 $m_{B}g$ for $W_B$

$N=m_{B}g$

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

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

Substitute $m_{B}g$ for $N$.

$F={\mu }_k{m}_{B}g$

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 ${\mu }_k{m}_{B}g$ for $F$

$R_B={\mu }_k{m}_{B}g$

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 ${\mu }_k{m}_{B}g$ for $F$ and $\frac{1}{2}{m}_B{r}_B^2$ for ${\overline{I}}_B$.

$\begin{array}{l}{\mu }_k{m}_{B}g{r}_B=\frac{1}{2}{m}_B{r}_B^2{\alpha }_B\\ {\alpha }_B=\frac{2{\mu }_k{m}_{B}g{r}_B}{m_B{r}_B^2}\\ {\alpha }_B=\frac{2{\mu }_{k}g}{r_B}\end{array}$ (1)

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 ${\mu }_k{m}_{B}g$ for $F$

$R_{A_x}={\mu }_k{m}_{B}g$

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 $m_{A}g$ for $W_A$ and $m_{B}g$ for $N$.

$\begin{array}{c}{R}_{A_y}=m_{A}g+m_{B}g\\ =\left(m_A+m_B\right)g\end{array}$

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 ${\mu }_k{m}_{B}g$ for $F$ and $\frac{1}{2}{m}_A{r}_A^2$ for ${\overline{I}}_A$.

$\begin{array}{l}{\mu }_k{m}_{B}g{r}_A=\frac{1}{2}{m}_A{r}_A^2{\alpha }_A\\ {\alpha }_A=\frac{2{\mu }_k{m}_{B}g{r}_A}{m_A{r}_A^2}\\ {\alpha }_A=\frac{2{\mu }_k{m}_{B}g}{m_A{r}_A}\end{array}$ (2)

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

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

Substitute $\frac{2{\mu }_k{m}_{B}g}{m_A{r}_A}$ for ${\alpha }_A$.

${\omega }_A={\omega }_0-\frac{2{\mu }_k{m}_{B}gt}{m_A{r}_A}$ (3)

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

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

Substitute $\frac{2{\mu }_{k}g}{r_B}$ for ${\alpha }_B$.

${\omega }_B=\frac{2{\mu }_{k}gt}{r_B}$ (4)

While there is no slipping between disk A and B, their velocity ratio is same.

Show the free body diagram of the system as in Figure 3.

Refer to Figure 3.

The velocity of pinion $\left(v_P\right)$ for both disk.

$\begin{array}{l}{v}_p={\omega }_B{r}_B\\ {\omega }_A{r}_A={\omega }_B{r}_B\end{array}$

Substitute $\left({\omega }_0-\frac{2{\mu }_k{m}_{B}gt}{m_A{r}_A}\right)$ for ${\omega }_A$ and $\frac{2{\mu }_{k}gt}{r_B}$ for ${\omega }_B$.

$\begin{array}{l}\left({\omega }_0-\frac{2{\mu }_k{m}_{B}gt}{m_A{r}_A}\right)r_A=\frac{2{\mu }_{k}g}{r_B}t\times r_B\\ {\omega }_0{r}_A-\frac{2{\mu }_k{m}_{B}gt{r}_A}{m_A{r}_A}=\frac{2{\mu }_{k}gt{r}_B}{r_B}\\ {\omega }_0{r}_A-\frac{2{\mu }_k{m}_{B}gt}{m_A}=2{\mu }_{k}gt\\ 2{\mu }_{k}gt+\frac{2{\mu }_k{m}_{B}gt}{m_A}={\omega }_0{r}_A\end{array}$

$\begin{array}{l}2{\mu }_{k}gt\left(1+\frac{m_B}{m_A}\right)={\omega }_0{r}_A\\ t\left(\frac{m_A+m_B}{m_A}\right)=\frac{{\omega }_0{r}_A}{2{\mu }_{k}g}\\ t=\frac{{\omega }_0{r}_A{m}_A}{2{\mu }_{k}g\left(m_A+m_B\right)}\end{array}$

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

Substitute $\frac{{\omega }_0{r}_A{m}_A}{2{\mu }_{k}g\left(m_A+m_B\right)}$ for t Equation (3).

$\begin{array}{c}{\omega }_A={\omega }_0-\left[\frac{2{\mu }_k{m}_{B}g}{m_A{r}_A}\times \frac{{\omega }_0{r}_A{m}_A}{2{\mu }_{k}g\left(m_A+m_B\right)}\right]\\ ={\omega }_0-\frac{{\omega }_0{m}_B}{m_A+m_B}\\ ={\omega }_0\left(1-\frac{m_B}{m_A+m_B}\right)\end{array}$

$\begin{array}{c}={\omega }_0\left(\frac{m_A+m_B-m_B}{m_A+m_B}\right)\\ =\frac{{\omega }_0{m}_A}{m_A+m_B}\end{array}$ (5)

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

Substitute $\frac{{\omega }_0{r}_A{m}_A}{2{\mu }_{k}g\left(m_A+m_B\right)}$ for t Equation (4).

$\begin{array}{c}{\omega }_B=\frac{2{\mu }_{k}g}{r_B}\times \frac{{\omega }_0{r}_A{m}_A}{2{\mu }_{k}g\left(m_A+m_B\right)}\\ =\frac{{\omega }_0{r}_A{m}_A}{r_A\left(m_A+m_B\right)}\end{array}$ (6)

From Equation (5) and (6), it is clear that the final velocities of disk A and B are independent of coefficient of kinetic friction ${\mu }_k$.

To determine

Express the final angular velocity of disk A in terms of ${\omega }_0$ and the ratio of the masses of the two disks $\left(\frac{m_A}{m_B}\right)$.

Answer to Problem 16.44P

The final angular velocity of disk A in terms of ${\omega }_0$ and the ratio of the masses of the two disks $\left(\frac{m_A}{m_B}\right)$ is $\underset{\_}{\frac{{\omega }_0}{\left(1+\frac{m_B}{m_A}\right)}}$.

Explanation of Solution

The mass of the disk A is $m_A$.

The mass of the disk B is $m_B$.

The initial angular velocity of the disk A is ${\omega }_0$.

The coefficient of the kinetic friction is ${\mu }_k$.

The radius of the disk A is $r_A$.

The radius of the disk B is $r_B$.

The acceleration due to gravity is g.

The time required for the disk to come to rest is t.

Calculation:

Refer to part (a).

Calculate the final angular velocity of disk A in terms of ${\omega }_0$ and the ratio of the masses of the two disks $\left(\frac{m_A}{m_B}\right)$:

Refer Equation (5).

$\begin{array}{c}{\omega }_A=\frac{{\omega }_0{m}_A}{m_A+m_B}\\ =\frac{{\omega }_0}{\frac{m_A+m_B}{m_A}}\\ =\frac{{\omega }_0}{\frac{m_A}{m_A}+\frac{m_B}{m_A}}\\ =\frac{{\omega }_0}{\left(1+\frac{m_B}{m_A}\right)}\end{array}$

Hence, the final angular velocity of disk A in terms of ${\omega }_0$ and the ratio of the masses of the two disks $\left(\frac{m_A}{m_B}\right)$ is $\underset{\_}{\frac{{\omega }_0}{\left(1+\frac{m_B}{m_A}\right)}}$.