Q. 29 - Precalculus Enhanced with Graphing Utilities

Question

Hungarian Adrian Annus won the gold medal for the hammer throw at the $2004$ Olympics in Athens with a winning distance of $83.19$ meters.* The event consists of swinging a $16$-pound weight attached to a wire $190$ centimeters long in a circle and then releasing it. Assuming his release is at a $45°$angle to the ground, the hammer will travel a distance of $\frac{{v_0}^2}{g}$ meters, where $g=9.8$ meters/second² and $v_0$ is the linear speed of the hammer when released. At what rate (rpm) was he swinging the hammer upon release? 

Step-by-Step Solution

Answer

He was swinging the hammer at $143.52$rpm upon release.

Step 1. Given Information

The winning distance for the hammer throw is $83.19$ meters.

Weight of hammer is $16$pounds.

Length of the wire is $190$ centimeters.

We need to find the rate at which the hammer was winging upon release.

Step 2. Finding the value of v 0

The hammer covers a distance of $\frac{{v_0}^2}{g}$meters.

Substituting the value of $g=9.8$ and equating it with $83.19$ meters we get,

$$\begin{gathered} \frac{{v_0}^2}{g}=83.19 \\ \Rightarrow \frac{{v_0}^2}{9.8}=83.19 \\ \Rightarrow {v_0}^2=83.19\times 9.8 \\ \Rightarrow {v_0}^2=815.262 \\ \Rightarrow v_0=\sqrt{815.262} \end{gathered}$$

$\Rightarrow v_0=28.55$m/sec

Step 3. Finding the angular speed

Angular speed $\omega$ is given by $v=r\omega$ where $v$ is the linear velocity and $r$ is the radius of the circle.

Substituting $r=1.9$ meters and $v=28.55$m/sec we get,

$$\begin{gathered} \omega =\frac{v}{r} \\ \Rightarrow \omega =\frac{28.55}{1.9} \end{gathered}$$

$\Rightarrow \omega =15.03$rad/sec

Converting rad/sec to rpm:

$2\mathrm{\pi }$ radians = $1$ revolution

$\therefore 15.03$rad/sec = $\frac{15.03}{2\mathrm{\pi }}\times 60$rpm

$=143.52$rpm