Question

Question: The four particles in Figure P10.45 are connected by rigid rods of negligible mass. The origin is at the center of the rectangle. The system rotates in the x y plane about the z axis with an angular speed of 6.00 rad/s. Calculate (a) the moment of inertia of the system about the z axis and (b) the rotational kinetic energy of the system.

Short answer

(a) The moment of inertia of the system, $I_{\text{t}}=143 \text{kg}·{\text{m}}^2$

(b) The rotational kinetic energy of the system, $K_{\text{R}}=2574 \text{J}$.

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.

Calculating distance of all the particles

(a)

Since the system rotates in the x y plane about the z-axis, the distance of all the particles from the origin

$\begin{array}{rcl}r& =& \sqrt{x^2+y^2}\\ & =& \sqrt{2^2+3^2}\\ & =& 3.6 \text{m}\end{array}$

The moment of inertia of a point particle is given by

data-custom-editor="chemistry" $I=m{r}^2$

Hence, the moment of inertia of the four particles is

$$\begin{gathered} I_{\text{t}}=m_1{r}^2+m_2{r}^2+m_3{r}^2+m_4{r}^2 \\ =\left(m_1+m_2+m_3+m_4\right)r^2 \end{gathered}$$

Substituting the numerical values, we get

$\begin{array}{rcl}{I}_{\text{t}}& =& (3+2+4+2)(3.6{)}^2\\ & =& 143 \text{kg}·{\text{m}}^2\end{array}$

So, the moment of inertia is $143 \text{kg}·{\text{m}}^2$.

Substituting numerical values

(b)

The total rotational kinetic energy of the system is given by

$K_{\text{R}}=\frac{1}{2}{I}_{\text{t}}{\omega }^2$

Substituting the numerical value

$\begin{array}{rcl}{K}_{\text{R}}& =& \frac{1}{2}\left(143\right)(6{)}^2\\ & =& 2574 \text{J}\end{array}$

So, the energy is $2574 \text{J}$.