Olympic ice skaters are able to spin at about 5.00 rev/s.

(a) What is their angular velocity in radians per second?

(b) What is the centripetal acceleration of the skater’s nose if it is 0.120 m from the axis of rotation?

(c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9.00 rev/s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius?

(d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins.

Solution:

We are given an angular velocity, \omega = 5 \ \text{rev/sec}

Part A

For this part, we are asked to convert the angular velocity to units of radians per second.

\begin{align*}
\omega & = \frac{5.00\ \text{rev}}{\text{sec}}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \\ \\
\omega & = 31.4159 \ \text{rad/sec} \\ \\
\omega & = 31.4 \ \text{rad/sec}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

For this part, we are asked to solve for the centripetal acceleration. We are going to use the formula a_{c} = r \omega ^2 given r=0.120\ \text{m} and \omega = 31.4159 \ \text{rad/s} .

\begin{align*}
a_{c} & = r \omega ^2 \\ \\ 
a_{c} & = \left( 0.120 \ \text{m} \right) \left( 31.4159 \ \text{rad/s} \right)^2 \\ \\
a_{c} & = 118.4350 \ \text{m/s}^2 \\ \\
a_{c} & = 118 \ \text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

For this part, we are going to directly solve the centripetal acceleration.

\begin{align*}
a_{c} & = r \omega ^2 \\ \\ 
a_{c} & = \left( 0.120 \ \text{m} \right)\left( \frac{9\ \text{rev}}{\text{s}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}}\right)^2 \\ \\
a_{c} & = 383.7302 \ \text{m/s}^2 \\ \\
a_{c} & = 384 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part D

The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That is quite a lot of acceleration in itself. The centripetal acceleration felt by Button’s nose was 39.2 times larger than the acceleration due to gravity! It is no wonder that he ruptured small blood vessels in his spins.

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If the runner completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of their centripetal acceleration as they run the curved portion of the track?

Solution:

Centripetal acceleration a_{c} is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity v and has the magnitude

a_{c}=\frac{v^{2}}{r}

We can solve for the constant speed of the runner using the formula

v=\frac{\Delta x}{\Delta t}

We are given the distance \Delta x = 200 \ \text{m} , and the total time \Delta t = 23.2\ \text{s} . Therefore, the velocity is

\begin{align*}
v & =\frac{\Delta x}{\Delta t} \\ \\ 
v & = \frac{200\ \text{m}}{23.2\ \text{s}} \\ \\
v & = 8.6207\ \text{m/s}
\end{align*}

From the given problem, we are given the following values: r=30\ \text{m} . We now have the details to solve for the centripetal acceleration.

\begin{align*}
a_{c} & = \frac{v^{2}}{r} \\ \\
a_{c} & = \frac{\left( 8.6207\ \text{m/s} \right)^2}{30\ \text{m}} \\ \\
a_{c} & = 2.4772\ \text{m/s}^{2} \\ \\
a_{c} & = 2.5\  \text{m/s}^{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}