In Figure, two skaters, each of mass $\mathbf{50}\mathbf{}\mathbf{kg}$, approach each other along parallel paths separated by$\mathbf{3}\mathbf{.0}\mathbf{}\mathbf{m}$. They have opposite velocities ofeach $\mathbf{1.4}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}$. One skater carries one end of a long pole of negligible mass, and the other skater grabs the other end as she passes. The skaters then rotate around the centre of the pole. Assume that the friction between skates and ice is negligible.

(a) What is the radius of the circle?

(b) What are the angular speeds of the skaters?

(c) What is the kinetic energy of the two-skater system? Next, the skaters pull along the pole until they are separated by$\mathbf{1.0}\mathbf{}\mathbf{m}$.

(d) What then are their angular speed?

(e) What then are the kinetic energy of the system?

(f) What provided the energy for the increased kinetic energy?

1. The mass of each skater is,$\text{m}=50\text{kg}.$
2. The distance between two skaters is$\text{d}=3.0\text{m}.$
3. The magnitude of the velocity of each skater is,$\text{v}=1.4\frac{\text{m}}{\text{s}}.$

Find the radius of the circular motion from the distance between two stars. Then by using this and the relation between $\mathbf{\text{v}}\mathbf{,}\mathbf{\text{randω}}$we can find angular speed $\left(\text{ω}\right)$. By putting it in the formula for rotational K.E, findthe kinetic energy of the two-skater system. Follow the same procedure to find the $\mathbf{\text{ωandK}}\mathbf{.}\mathbf{\text{E}}$after coming closer.

Formulae:

$\begin{array}{l}\mathbf{v}\mathbf{=}\mathbf{r\omega }\\ \mathbf{Kinetic}\mathbf{}\mathbf{Energy}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}\end{array}$

The law of conservation of momentum, localid="1660977085911" ${\mathbf{I}}_{\mathbf{i}}{\mathbf{\omega }}_{\mathbf{i}}\mathbf{}\mathbf{=}{\mathbf{I}}_{\mathbf{f}}{\mathbf{\omega }}_{\mathbf{f}}$

(a)

As the mass of both skaters are the same and their velocities are equal and opposite their total linear momentum is zero. Hence, the center of mass of the system remains fixed at the center of the pole.

The distance between them is $3\mathrm{m}$.and is not changing; it will be the diameter of the circular motion performed by them

So the radius of the circular motion is

$\mathrm{r}=\frac{\mathrm{d}}{2}=\frac{3}{2}$

$\Rightarrow \text{r}=1.5\text{m}.$

(b)

The angular speedof the skaters is

$\begin{array}{l}\mathrm{\omega }=\frac{\mathrm{v}}{\mathrm{r}}\\ \mathrm{\omega }=\frac{1.4}{1.5}\\ \mathrm{\omega }=0.93\mathrm{rad}/\mathrm{s}\end{array}$

(c)

The moment of inertia of the two-skater system is

$\begin{array}{c}\text{I}=\sum \left({\text{mr}}^2\right)\\ \Rightarrow \text{I}=2\left(50\right){\left(1.5\right)}^2\\ \Rightarrow \text{I}=225{\text{kgm}}^2\end{array}$

The kinetic energy of the two-skater system is

$\begin{array}{c}\text{K}.\text{E}=\frac{1}{2}{\text{Iω}}^2\\ \Rightarrow \text{K}.\text{E}=\frac{1}{2}\left(225\right){\left(0.93\right)}^2\\ \Rightarrow \mathrm{K}.\mathrm{E}.=98\mathrm{J}\end{array}$

(d)

If the skaters pull along the pole until they are separated by , the radius of the rotational motion would be

${\text{r}}_{\text{f}}=\frac{1}{2}=0.5\text{m}$

Then, the M.I of the two-skater system would be

$\begin{array}{c}{\text{I}}_{\text{f}}=2\left(50\right){\left(0.5\right)}^2\\ \Rightarrow {\text{I}}_{\text{f}}=25{\text{kgm}}^2\end{array}$

According to the law of conservation of angular momentum,

${\text{I}}_{\text{i}}{\text{ω}}_{\text{i}}={\text{I}}_{\text{f}}{\text{ω}}_{\text{f}}$

So, the angular speed of the skaters is,

$\begin{array}{c}{\text{ω}}_{\text{f}}=\frac{{\text{I}}_{\text{i}}{\text{ω}}_{\text{i}}}{{\text{I}}_{\text{f}}}\\ \Rightarrow {\text{ω}}_{\text{f}}=\frac{225\times 0.93}{25}\\ \Rightarrow {\mathrm{\omega }}_{\mathrm{f}}=8.4\mathrm{rad}/\mathrm{s}\end{array}$

(e)

The kinetic energy of the two-skater system is,

$\text{K}.\text{E}=\frac{1}{2}{\text{Iω}}^2$

$\begin{array}{l}\Rightarrow \text{K}.{\text{E}}_{\text{f}}=\frac{1}{2}\left(25\right){\left(8.4\right)}^2\\ \Rightarrow \mathrm{K}.\mathrm{E}.=8.8\times {10}^2\mathrm{J}\end{array}$

(f)

The work done by skaters while coming closer is converted into the internal energy of the system. This provides the energy for increased kinetic energy.