A car is fitted with an energy-conserving flywheel, which in operation is geared to the driveshaft so that it rotates at 237 rev/s when the car is traveling at \(86.5 \mathrm{~km} / \mathrm{h}\). The total mass of the car is \(822 \mathrm{~kg}\), the flywheel weighs \(194 \mathrm{~N}\), and it is a uniform disk \(1.08 \mathrm{~m}\) in diameter. The car descends a \(1500-\mathrm{m}\) long, \(5.00^{\circ}\) slope, from rest, with the flywheel engaged and no power supplied from the motor. Neglecting friction and the rotational inertia of the wheels, find \((a)\) the speed of the car at the bottom of the slope, \((b)\) the angular acceleration of the flywheel at the bottom of the slope, and \((c)\) the power being absorbed by the rotation of the flywheel at the bottom of the slope.

First let’s find the linear speed the flywheel translates to. Using the relationship \(V = rω\), where V is linear speed, r is the radius of disk and ω is angular speed, we find linear speed = \((0.54 \mathrm{~m})(237 \, (2\pi) \mathrm{~rads/s})\). Convert this speed to kilometers per hour for comparison with the car’s speed.

The car's initial potential energy as it starts to descend is \(mgh\). Here, m is the total mass of the car, g is the acceleration due to gravity and h is the height of hill which is \(1500 \mathrm{m} \cdot \sin(5^{\circ})\). As the car moves down, potential energy is converted to kinetic energy. Assuming no energy loss due to friction or air resistance, conservation of energy allows us to set the potential energy equal to the final kinetic energy, half of which goes to the translational motion of car, and the other half goes to the rotational motion of the flywheel. This gives us the equation \((1/2)mv^2 + (1/2)I\omega^2 = mgh\). Here, \(I\) is the moment of inertia of the flywheel which equals to \(1/2 \cdot (weight of flywheel/g) \cdot (diameter/2)^2 = 1/2 \cdot (194 \mathrm{N}/9.8 \mathrm{m/s^2})(0.54 \mathrm{m})^2).\)

Solve the equation from previous step to find v, the speed of the car at the bottom of the hill.

Angular acceleration (α) or how quickly the flywheel speeds up, can be derived from the relationship between linear and angular speeds, i.e. \(v=aω\). Differentiating both sides with respect to time gives \(dv/dt= a(dω/dt) + αω\). Setting \(dv/dt = g \sin θ\), we get \(g \sin θ = a(dω/dt) + αω\). Solving for \(α\) gives us the angular acceleration at the bottom of the slope.

The power absorbed by the rotation of the flywheel can be found using the formula \(P = Iαω\). Substitute the values from previous steps to calculate the power absorbed.