A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end just before it hits the floor, assuming that the end on the floor does not slip.

1. A meter stick is allowed to fall with its one end on the floor.
2. Hint: Consider the stick to be a thin rod and use the conservation of energy principle.

Find the M.I of the stick from the M.I of the rod about an axis passing through its one end. Then applying the law of conservation of energy to the given system, find an expression forspeed of the other end of the stick just before it hits the floor.Then, inserting M.I and length of the stick,findits value.

Formula:

The M.I of the rod about an axis passing through it’s one end is

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

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

Let m and l be the mass and length of the meter stick respectively. Then, its center of mass will be situated at $l/2$. The angular speed and M.I of the stick is$\omega \text{and\hspace{0.17em}}\text{I}$ respectively.

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

$\begin{array}{c}\mathrm{I}=\frac{1}{3}{\mathrm{ml}}^2\\ =\frac{1}{3}\mathrm{m}{\left(1\text{\hspace{0.17em}m}\right)}^2\\ =\frac{1}{3}\mathrm{m}\end{array}$

According to the conservation of energy,

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

$\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}+\mathrm{P}.{\mathrm{E}}_{\mathrm{i}}=\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}+\mathrm{P}.{\mathrm{E}}_{\mathrm{f}}$

In this case,

$\begin{array}{c}0+\frac{\mathrm{mgl}}{2}=\frac{1}{2}{\mathrm{I\omega }}^2+0\\ {\mathrm{\omega }}^2=\frac{\mathrm{mgl}}{\mathrm{I}}\\ \mathrm{\omega }=\sqrt{\frac{\mathrm{mgl}}{\mathrm{I}}}\end{array}$

The speed of the stick is,

$\text{v}=\text{l}\mathrm{\omega }$

$\begin{array}{c}\mathrm{v}=\mathrm{l\omega }\\ \mathrm{v}=\mathrm{l}\sqrt{\frac{\mathrm{mgl}}{\mathrm{I}}}\end{array}$

But $l=1$and $\mathrm{l}=\frac{1}{3}\mathrm{m}$,

$\begin{array}{c}v=\sqrt{\frac{mgl}{\frac{1}{3}m}}\\ v=\sqrt{3gl}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\text{v}=\sqrt{3\left(9.8{\text{m}/\text{s}}^2\right)\times 1\text{\hspace{0.17em}m}}\\ \text{v}=5.42\text{\hspace{0.17em}m}/\text{s}\end{array}$

Therefore, the speed of the other end of the stick just before it hits the floor is $5.42\text{\hspace{0.17em}m}/\text{s}$.