Question

A skater holds her arms outstretched as she spins at$180\text{rpm}$ . What is the speed of her hands if they are$140\text{cm}$ apart?

Short answer

The speed of the hands of the skater is $13.2\text{m}/\text{s}$

Step-by-step solution

Step 1. Given information

The angular velocity is$\omega =180\text{rpm}$

The diameter is $d=140\text{cm}$.

Step 2. Explanation

The relation between the linear velocity and angular velocity is

$v=r\omega$

Here, v is the linear speed, r is the distance,

Converting the annular speed of the skater from rpm to rad/s.

$\omega =180\text{rpm}=\left(180\text{rpm}\right)\left(\frac{2\pi \text{rad}}{1\text{rev}}\right)\left(\frac{1\text{min}}{60\text{s}}\right)=18.84\text{rad}/\text{s}$

The radius and diameter relation.

$\begin{array}{l}r=\frac{d}{2}\\ r=\frac{140\text{cm}}{2}=70\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)=0.7\text{m}\end{array}$

The speed of the hands of the skater is

$\begin{array}{l}v=r\omega \\ v=(0.7\text{m})(18.84\text{rad}/\text{s})=13.195\text{m}/\text{s}\end{array}$

Therefore, the speed of the hands of the skater is$13.2\text{m}/\text{s}$