Question

On an ice rink two skaters of equal mass grab hands and spin in a mutual circle once every 2.5 s. If we assume their arms are each 0.80 m long and their individual masses are 55.0 kg, how hard are they pulling on one another?

Short answer

The skaters are pulling on one another with a force of $2.8\times {10}^2\mathrm{N}$.

Step-by-step solution

Step 1. Understanding the uniform circular motion

When an object of mass m moves in a circular path of radius r with a constant speed v, then the motion of that object is referred as the uniform circular motion.

A net force which is required to keep moving the object in circular motion acts on the object towards the circle’s center and is given as,

$F_{\mathrm{R}}=\frac{m{v}^2}{r}$

Here, skaters are moving in uniform circular motion.

Step 2. Identification of the given information

• The time period of revolution of the skaters is, $T=2.5\mathrm{s}$.
• The radius of the circle made by skater’s motion is, $r=0.80\mathrm{m}$.
• The mass of each skater is, $m=55.0\mathrm{kg}$.

Step 3. Evaluation of velocity of each skater

The speed of each skater while moving along the circle is given as,

$v=\frac{2\pi r}{T}$

Substitute the values in the above equation.

$\begin{array}{c}v=\frac{2\times 3.14\times 0.80\mathrm{m}}{2.5\mathrm{s}}\\ =2.0\mathrm{m}/\mathrm{s}\end{array}$

Step 4. Determination of the force with which skaters pull one another

When the skaters move in a circular path, then they exert radial force on each other along their hands. This force is calculated as,

$F_{\mathrm{R}}=\frac{m{v}^2}{r}$

Substitute the values in the above equation.

$\begin{array}{c}{F}_R=\frac{\left(55.0\mathrm{kg}\right)\times {\left(2.0\mathrm{m}/\mathrm{s}\right)}^2}{0.80\mathrm{m}}\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =275\mathrm{N}\\ \approx 2.8\times {10}^2\mathrm{N}\end{array}$

Thus, the skaters are pulling one another with the force of $2.8\times {10}^2\mathrm{N}$.