A bicycle has wheels of radius \(0.32 \mathrm{m}\). Each wheel has a rotational inertia of \(0.080 \mathrm{kg} \cdot \mathrm{m}^{2}\) about its axle. The total mass of the bicycle including the wheels and the rider is 79 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?

Let's denote the translational kinetic energy as \(K_{trans}\). It can be computed using the equation: \(K_{trans} = \frac{1}{2}mv^{2}\), where \(m = 79 \mathrm{kg}\) is the total mass of the bicycle and rider, and \(v\) is the velocity. To calculate the translational kinetic energy, we need the speed, \(v\) of the bicycle. Unfortunately, the problem doesn't give the speed directly, but we know that the bicycle is coasting at a constant speed, so we can consider the speed \(v\) to be the linear speed of the wheel's rim. Knowing the radius \(r = 0.32\mathrm{m}\), we can use the relationship between linear and angular speed: \(v = r\omega\), where \(\omega\) is the angular speed. The linear speed can be eliminated in \(K_{trans}\) expression using the formula \(v = r\omega\). So, \(K_{trans} = \frac{1}{2}m(r\omega)^{2} = \frac{1}{2}mr^{2}\omega^{2}\).

Now let's find the rotational kinetic energy of the wheels. The rotational kinetic energy can be computed using the equation: \(K_{rot} = \frac{1}{2}I\omega^{2}\), where \(I = 0.080\,\mathrm{kg}\cdot\mathrm{m}^{2}\) is the moment of inertia of each wheel and \(\omega\) is the angular speed. As there are two wheels, the total rotational kinetic energy of the wheels is: \(K_{rot_{total}} = 2\left(\frac{1}{2}I\omega^{2}\right) = I\omega^{2}\).

Now we have equations for both the translational kinetic energy \(K_{trans}\) and the total rotational kinetic energy of the wheels \(K_{rot_{total}}\) in terms of \(\omega\). We want to find the fraction of the total kinetic energy that corresponds to the rotational kinetic energy of the wheels, which can be expressed as \(\frac{K_{rot_{total}}}{K_{trans} + K_{rot_{total}}}\). Plugging in the expressions found in steps 1 and 2, we get: \(\frac{I\omega^{2}}{\frac{1}{2}mr^{2}\omega^{2} + I\omega^{2}}\).

We can simplify the fraction by dividing both the numerator and denominator by \(\omega^{2}\): \(\frac{I}{\frac{1}{2}mr^{2} + I}\). Now, we substitute the given values of \(I = 0.080 \,\mathrm{kg}\cdot\mathrm{m}^{2}\) and \(m = 79\,\mathrm{kg}\), and \(r = 0.32\,\mathrm{m}\) into the expression: \(\frac{0.080}{\frac{1}{2}(79)(0.32)^{2} + 0.080}\).

Performing the calculations, we obtain: \(\frac{0.080}{6.400 + 0.080} \approx \frac{0.080}{6.480}\). The fraction of the total kinetic energy of the bicycle (including rider) that is the rotational kinetic energy of the wheels is approximately \(\frac{0.080}{6.480} \approx 0.01235\), or about \(1.235 \%\).