Chapter 15.1, Problem 15.35P

Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B is at rest. A period of slipping follows and disk B makes two revolutions before reaching its final angular velocity. Assuming that the angular acceleration of each disk is constant and inversely proportional to the cube of its radius, determine (a) the angular acceleration of each disk, (b) the time during which the disks slip.

Fig. P15.34 and P15.35

To determine

Find the angular acceleration of disk A and B.

Answer to Problem 15.35P

The angular acceleration of disk A and B are $\underset{\_}{4.34{\text{rad/s}}^2\left(\text{Clockwise}\right)}$ and $\underset{\_}{1.832{\text{rad/s}}^2\left(\text{Clockwise}\right)}$.

Explanation of Solution

Given information:

Consider the initial and final angular velocity of disk A are as follows:${\left({\omega }_A\right)}_0=240\text{rpm}\left(\text{Counterclockwise}\right)$ and ${\left({\omega }_A\right)}_1$.

Consider the initial and final angular velocity of disk B are ${\left({\omega }_B\right)}_0=0\text{rpm}$ and ${\left({\omega }_B\right)}_1$.

Consider the angular acceleration of the disk A and B are denoted by ${\alpha }_A$ and ${\alpha }_B$.

Consider the angular displacement of the disk B is denoted by ${\theta }_B\left(\text{Clockwise}\right)$.

Show the value of the angular displacement ${\theta }_B\left(\text{Clockwise}\right)$ as follows:

$\begin{array}{c}{\theta }_B=2\text{rev}\times \left(\frac{2\pi \text{radians}}{1\text{rev}}\right)\\ =4\pi \text{radians}\end{array}$

The angular acceleration of disk A and B are constant.

The radius of the disk A and B are $r_A=150\text{mm}$ and $r_B=200\text{mm}$.

The angular acceleration of disk A and B are inversely proportional to cube of their radius.

$\left(\frac{{\alpha }_B}{{\alpha }_A}\right)={\left(\frac{r_A}{r_B}\right)}^3$ (1)

Calculation:

Calculate the final angular velocity of disk A using the relation:

${\left({\omega }_A\right)}_1={\left({\omega }_A\right)}_0-{\alpha }_A{t}_1$

Substitute $240\text{rpm}$ for ${\left({\omega }_A\right)}_0$.

$\begin{array}{c}{\left({\omega }_A\right)}_1=240\text{rpm}\times \left(\frac{2\pi \text{rad/s}}{60\text{rpm}}\right)-{\alpha }_A{t}_1\\ =8\pi -{\alpha }_A{t}_1\end{array}$

Calculate the angular displacement of the disk B using the relation:

${\theta }_B={\left({\omega }_B\right)}_0{t}_1+\frac{1}{2}{\alpha }_B{t}_1^2$

Modify above Equation using Equation (1).

${\theta }_B={\left({\omega }_B\right)}_0{t}_1+\frac{1}{2}{\alpha }_A{\left(\frac{r_A}{r_B}\right)}^3{t}_1^2$

Substitute 0 for ${\left({\omega }_B\right)}_0$, $150\text{mm}$ for $r_A$, $200\text{mm}$ for $r_B$, and $4\pi \text{radians}$ for ${\theta }_B$

$\begin{array}{l}4\pi =0+\frac{1}{2}{\alpha }_A{\left(\frac{150}{200}\right)}^3{t}_1^2\\ 8\pi =0.421875t_1^2\\ {\alpha }_A{t}_1^2=\frac{8\pi }{0.421875}\\ {\alpha }_A{t}_1^2=59.574\text{radians}\end{array}$ (2)

Calculate the final angular velocity of the disk B using the relation:

${\left({\omega }_B\right)}_1={\left({\omega }_B\right)}_0+{\alpha }_B{t}_1$

Modify above Equation using Equation (1).

${\left({\omega }_B\right)}_1={\left({\omega }_B\right)}_0+{\alpha }_A{\left(\frac{r_A}{r_B}\right)}^3{t}_1$

Substitute 0 for ${\left({\omega }_B\right)}_0$, $150\text{mm}$ for $r_A$, and $200\text{mm}$ for $r_B$.

$\begin{array}{l}{\left({\omega }_B\right)}_1=0+{\alpha }_A{\left(\frac{150}{200}\right)}^3{t}_1\\ {\left({\omega }_B\right)}_1=0.421875{\alpha }_A{t}_1\end{array}$

Consider the contact point between the disk A and B are denoted by C.

Calculate the velocity at point C using the relation:

$v_C=r_A{\omega }_A$

Substitute $150\text{mm}$ for $r_A$ and $\left(8\pi -{\alpha }_A{t}_1\right)$ for ${\omega }_A$.

$v_C=150\times \left(8\pi -{\alpha }_A{t}_1\right)$ (3)

Calculate the velocity at point C using the relation:

$v_C=r_B{\omega }_B$

Substitute $200\text{mm}$ for $r_B$ and $0.421875{\alpha }_A{t}_1$ for ${\omega }_B$.

$\begin{array}{c}{v}_C=200\times \left(0.421875{\alpha }_A{t}_1\right)\\ =84.375{\alpha }_A{t}_1\end{array}$ (4)

Equate Equation (3) and (4).

$\begin{array}{l}150\times \left(8\pi -{\alpha }_A{t}_1\right)=84.375{\alpha }_A{t}_1\\ 8\pi -{\alpha }_A{t}_1=0.5625{\alpha }_A{t}_1\\ 1.5625{\alpha }_A{t}_1=8\pi \\ {\alpha }_A{t}_1=\left(\frac{8\pi }{1.5625}\right)\end{array}$

$\begin{array}{l}{\alpha }_A{t}_1=\left(\frac{8\pi }{1.5625}\right)\\ {\alpha }_A{t}_1=16.0850\text{rad/s}\end{array}$ (5)

Divide Equation (1) by (5).

$t_1=3.7037\text{s}$ (6)

Calculate the angular acceleration of disk A as follows:

Substitute $150\text{mm}$ for $r_A$, $200\text{mm}$ for $r_B$,

$\begin{array}{c}{\alpha }_A\times \left(3.7037\text{s}\right)=16.0850\text{rad/s}\\ {\alpha }_A=\frac{16.0850\text{rad/s}}{3.7037\text{s}}\\ =4.34{\text{rad/s}}^2\left(\text{Clockwise}\right)\end{array}$

Thus, the angular acceleration of the disk A is $\underset{\_}{4.34{\text{rad/s}}^2\left(\text{Clockwise}\right)}$.

Calculate the angular acceleration of disk B as follows:

Substitute $150\text{mm}$ for $r_A$, $200\text{mm}$ for $r_B$, and $4.34\text{rad/s}$ for ${\alpha }_A$.

$\begin{array}{l}\left(\frac{{\alpha }_B}{4.34\text{rad/s}}\right)={\left(\frac{150}{200}\right)}^3\\ {\alpha }_B=4.34{\left(\frac{150}{200}\right)}^3\\ {\alpha }_B\approx 1.832{\text{rad/s}}^2\end{array}$

Thus, the angular acceleration of the disk B is $\underset{\_}{1.832{\text{rad/s}}^2\left(\text{Clockwise}\right)}$.

To determine

Find the time at which the disk slip.

Answer to Problem 15.35P

The disk slip at time $\underset{\_}{t_1=3.70\text{s}}$.

Explanation of Solution

Given information:

Calculation:

Refer to Part (a).

Refer Equation (6).

The time at which the disk slip is at $t_1=3.70\text{s}$.

Thus, the time at which the disk slip is at $\underset{\_}{t_1=3.70\text{s}}$.