Integrated Concepts

When kicking a football, the kicker rotates his leg about the hip joint.

(a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity?

(b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s?

(c) Find the maximum range of the football, neglecting air resistance.

Solution:

Part A

From the given problem, we are given the following values: v=35.0\ \text{m/s} and r=1.05\ \text{m}. We are required to solve for the angular velocity \omega.

The linear velocity, v and the angular velocity, \omega are related by the equation

v=r\omega \ \text{or} \ \omega=\frac{v}{r}

If we substitute the given values into the formula, we can directly solve for the value of the angular velocity. That is,

\begin{align*}
\omega & = \frac{v}{r} \\ \\
\omega & = \frac{35.0\ \text{m/s}}{1.05\ \text{m}} \\ \\
\omega & = 33.3333\ \text{rad/sec} \\ \\
\omega & = 33.3 \ \text{rad/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

For this part of the problem, we are going to use Newton’s second law of motion in term of linear momentum which states that the net external force equals the change in momentum of a system divided by the time over which it changes. That is

F_{net} = \frac{\Delta p}{\Delta t} = \frac{m\left( v_f - v_i \right)}{t}

For this problem, we are given the following values: m=0.500\ \text{kg}, t=20.0\times 10^{-3} \ \text{s}, v_{f}=20.0\ \text{m/s}, and v_{i}=0. Substituting all these values into the equation, we can solve directly for the value of the net external force.

\begin{align*}
F_{net} & = \frac{\left( 0.500\ \text{kg} \right)\left( 20.0\ \text{m/s}-0\ \text{m/s} \right)}{20.0\times 10^{-3}\ \text{s}} \\ \\
F_{net} & = 500\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

This is a problem on projectile motion. In this particular case, we are solving for the range of the projectile. The formula for the range of a projectile is

R=\frac{v_{0}^2 \sin 2\theta}{g}

We are asked to solve for the maximum range, and we know that the maximum range happens when the angle \theta is 45^\circ .

\begin{align*}
R & = \frac{\left( 20.0\ \text{m/s} \right)^{2} \sin \left( 2\left( 45^\circ  \right) \right)}{9.81 \ \text{m/s}^2} \\ \\
R & = 40.7747\ \text{m} \\ \\
R & = 40.8 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}