A thin rod of length $\mathbf{0}\mathbf{.75}\mathbf{\text{\hspace{0.17em}m}}$ and mass$\mathbf{\text{0.42kg}}$ is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed$\mathbf{4.0}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}\mathbf{\text{s}}$ . Neglecting friction and air resistance, find (a) the rod’s kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

1. Length of rod is,$\mathrm{l}=0.75\mathrm{m}$.
2. Mass of the rod is,$\mathrm{m}=0.42\text{\hspace{0.17em}kg}$
3. Angular speed of the rod is, $\mathrm{\omega }=4.0\text{\hspace{0.17em}rad}/\text{s}$.

You can find thekinetic energy of the rod at its lowest position using the formula for rotational K. E. Using the law of conservation of energy, you can find the position of the center of mass oftherod above its lowest position. The formulas used are given below.

$\begin{array}{c}\mathbf{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}{\mathbf{ml}}^{\mathbf{2}}\\ \mathbf{K}\mathbf{.}\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}\mathbf{}\end{array}$

The M.I of the rod about the axis passing through one end of the rod is

$\mathrm{I}=\frac{1}{3}{\mathrm{ml}}^2$

Rotational kinetic energy of the rod is

$\begin{array}{c}K.E=\frac{1}{2}I{\omega }^2\\ =\frac{1}{2}\left(\frac{1}{3}m{l}^2\right){\omega }^2\\ =\frac{1}{2}\left(\frac{1}{3}\left(0.42\text{\hspace{0.17em}kg}\right){\left(0.75\text{\hspace{0.17em}m}\right)}^2\right){(4\text{\hspace{0.17em}rad}/\text{s})}^2\\ =0.63\text{\hspace{0.17em}J}\end{array}$

Therefore, kinetic energy of the rod at its lowest position is $0.63\text{\hspace{0.17em}J}$.

Let’s assume that center of mass is at h above the lowest position of the rod.

According to the conservation of energy,

${\mathrm{E}}_{\mathrm{i}}={\mathrm{E}}_{\mathrm{f}}$

In this case,

$\begin{array}{c}\mathrm{K}.\mathrm{E}=\mathrm{P}.\mathrm{E}\\ \mathrm{K}.\mathrm{E}=\mathrm{mgh}\\ \mathrm{h}=\frac{\mathrm{K}.\mathrm{E}}{\mathrm{mg}}\\ \mathrm{h}=\frac{0.63\text{\hspace{0.17em}J}}{\left(0.42\text{\hspace{0.17em}kg}\right)(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})}\\ \mathrm{h}=0.153\text{\hspace{0.17em}m}\\ \simeq 0.15\text{\hspace{0.17em}m}\end{array}$

Therefore, the position of the center of mass of the rod above its lowest position is $0.15\text{\hspace{0.17em}m}$.