Chapter 10, Problem 25P

To determine

The transmitted torque.

Explanation of Solution

Given:

Thickness of the oil film $\left(h\right)$ is 2 mm.

Dynamic viscosity $\left(\mu \right)$ is $0.38\text{N}\cdot \text{s}/{\text{m}}^2$.

Diameter of the clutch $\left(D\right)$ is 30 cm.

Driving speed of the shaft $\left(n_1\right)$ is 1450 rpm.

Driven speed of the shaft $\left(n_2\right)$ is 1398 rpm.

Calculation:

When one of the plate is stationary, the linear velocity of the other plate is $V=\left({\omega }_1-{\omega }_2\right)r$.

Consider a small area dA on the surface of the disc.

Write the expression of the shear force on the disc on the area dA.

  $\begin{array}{l}dF=\mu \left(dA\right)\left(\frac{V}{h}\right)\\ dF=\mu \left(2\pi rdr\right)\left(\frac{V}{h}\right)\\ dF=\mu \left(2\pi r\right)\left(\frac{\left({\omega }_1-{\omega }_2\right)r}{h}\right)dr\end{array}$

Write the expression of torque on the small area dA.

  $\begin{array}{l}dT=\left(\mu \left(2\pi r\right)\left(\frac{\left({\omega }_1-{\omega }_2\right)r}{h}\right)dr\right)r\\ dT=\frac{2\pi \mu r^3\left({\omega }_1-{\omega }_2\right)dr}{h}\end{array}$

Integrate the above equation.

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{T}{\int }}dT}={\displaystyle \underset{r=0}{\overset{D/2}{\int }}\frac{2\pi \mu r^3\left({\omega }_1-{\omega }_2\right)dr}{h}}\\ {\displaystyle \underset{0}{\overset{T}{\int }}dT}=\frac{2\pi \mu \left({\omega }_1-{\omega }_2\right)}{h}{\displaystyle \underset{r=0}{\overset{D/2}{\int }}{r}^{3}dr}\\ {\left(T\right)}_0^T=\frac{2\pi \mu \left({\omega }_1-{\omega }_2\right)}{h}{\left(\frac{r^4}{4}\right)}_{r=0}^{D/2}\\ T-0=\frac{2\pi \mu \left({\omega }_1-{\omega }_2\right)}{h}\left(\frac{{\left(D/2\right)}^4}{4}-0\right)\end{array}$

  $\begin{array}{l}T=\frac{\pi \mu \left({\omega }_1-{\omega }_2\right)D^4}{32h}\\ T=\frac{\pi \mu \left[2\pi \left(n_1-n_2\right)\right]D^4}{32h}\\ T=\frac{\pi \left(0.38\text{N}\cdot \text{s}/{\text{m}}^2\right)\left[2\pi \left(1450\text{ rpm}-1398\text{ rpm}\right)\left(\frac{1\text{ min}}{60\text{ s}}\right)\right]{\left(30\text{ cm}\left(\frac{1\text{ m}}{{10}^2\text{cm}}\right)\right)}^4}{32\left(\text{2 mm}\left(\frac{1\text{ m}}{{10}^3\text{mm}}\right)\right)}\\ T=0.823\text{ N}\cdot \text{m}\end{array}$

Thus, the transmitted torque is $\underset{\_}{0.823\text{ N}\cdot \text{m}}$.