Question

Many machines employ cams for various purposes, such as opening and closing valves. In Figure P10.46, the cam is a circular disk of radius R with a hole of diameter R cut through it. As shown in the figure, the hole does not pass through the center of the disk. The cam with the hole cut out has mass M. The cam is mounted on a uniform, solid, cylindrical shaft of diameter R and also of mass M. What is the kinetic energy of the cam–shaft combination when it is rotating with angular speed v about the shaft’s axis?

Short answer

Kinetic energy of the cam-shaft, $K=\frac{13}{24}M{R}^2{\omega }^2$

Step-by-step solution

Definition of moment of inertia:

Moment of inertia about a given axis of rotation resists a change in its rotational motion; it can be regarded as a measure of rotational inertia of the body

Calculation for the disk:

Let m be the mass of the hole (small disk)

Let M be the mass of the cam with the hole cut.

The diagram is shown below:

The area of the small disk is given by:

$$\begin{gathered} A_{smalldisk}=\pi {\left(\frac{R}{2}\right)}^2 \\ =\frac{1}{4}\pi R^2 \end{gathered}$$

The area of the big disk is given by:

$A_{bigdisk}=\pi R^2$

Hence, the area of the cam is

$$\begin{gathered} A_{cam}=\pi R^2-\frac{1}{4}\pi R^2 \\ =\frac{3}{4}\pi R^2 \end{gathered}$$

Because the mass is uniformly distributed, the mass is proportional to the area

$$\begin{gathered} \frac{m}{M}=\frac{\frac{1}{4}\pi R^2}{\frac{3}{4}\pi R^2} \\ =\frac{1}{3} \end{gathered}$$

Therefore,

$m=\frac{M}{3} ........ \left(1\right)$

The moment of inertia of big disc

The moment of inertia of the big disc without the hole cut about the center of mass (A)

$I_A=\frac{1}{2}(m+M)R^2$

Substitute for (m) from Equation (1)

$$\begin{gathered} I_A=\frac{1}{2}\left(m+\frac{M}{3}\right)R^2 \\ =\frac{2}{3}M{R}^2 .......... \left(2\right) \end{gathered}$$

According to parallel-axis theorem, the moment of inertia of the big disc without the hole cut about the axis (O) which is a parallel axis to (A) with an offset distance is

$I_{mo}=I_A+\left(M+\frac{M}{3}\right){\left(\frac{R}{2}\right)}^2$

Substitute forfrom Equation (2)

$$\begin{gathered} I_0=\frac{2}{3}M{R}^2+\frac{1}{3}M{R}^2 \\ =M{R}^2 ........... \left(3\right) \end{gathered}$$

The rotating axis (O) is

$I_{mo}=\frac{1}{2}m{\left(\frac{R}{2}\right)}^2$

Substitute for (m) from Equation (1)

$$\begin{gathered} I_{mo}=\frac{1}{2}\left(\frac{M}{3}\right)\left(\frac{R^2}{4}\right) \\ =\frac{1}{24}M{R}^2 .......... \left(4\right) \end{gathered}$$

The big disk about the axis of rotation

The moment of inertia of the big disk about the axis of rotation is equal to the moment of inertia of the cam $\left(I_{Mo}\right)$and the cut hole $\left(I_{Mo}\right)$about the axis of rotation

$I_0=I_{Mo}+I_{mo}$

Rearrange fordata-custom-editor="chemistry" $\left(I_{Mo}\right)$

$I_0=I_{Mo}-I_{mo}$

Substitute for from Equation (3) and for $\left(I_{Mo}\right)$from Equation (4).

$$\begin{gathered} I_{Mo}=m{R}^2-\frac{1}{24}m{R}^2 \\ =\frac{23}{24}M{R}^2 ............. \left(5\right) \end{gathered}$$

The moment of inertia of the shaft which is a cylinder of mass (M) and radius is

$$\begin{gathered} I_{shaft}=\frac{1}{2}M{\left(\frac{R}{2}\right)}^2 \\ =\frac{1}{8}M{R}^2 \end{gathered}$$

Step: 6 The total moment of inertia:

Hence, the total moment of inertia of the cam-shaft system about the axis of rotation is just the moment of inertia of the cam and the moment of inertia of the shaft about the axis of rotation

$I=I_{Mo}+I_{shaft}$

Substitute for $\left(I_{Mo}\right)$from Equation (5) and for$\left(I_{shaft}\right)$ from Equation (6)

$$\begin{gathered} I=\frac{23}{24}M{R}^2+\frac{1}{8}M{R}^2 \\ =\frac{13}{12}M{R}^2 \end{gathered}$$

And the kinetic energy of the cam-shaft combination when it is rotating with angular speedabout the shaft’s axis is

$$\begin{gathered} K=\frac{1}{2}I{\omega }^2 \\ =\frac{1}{2}\left(\frac{13}{12}M{R}^2\right){\omega }^2 \\ =\frac{13}{24}M{R}^2{\omega }^2 \end{gathered}$$

The solution is $K=\frac{13}{24}M{R}^2{\omega }^2$