An experimental flywheel, used to store energy and replace an automobile engine, is a solid disk of mass \(200.0 \mathrm{kg}\) and radius $0.40 \mathrm{m} .\( (a) What is its rotational inertia? (b) When driving at \)22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph}),$ the fully energized flywheel is rotating at an angular speed of \(3160 \mathrm{rad} / \mathrm{s} .\) What is the initial rotational kinetic energy of the flywheel? (c) If the total mass of the car is \(1000.0 \mathrm{kg},\) find the ratio of the initial rotational kinetic energy of the flywheel to the translational kinetic energy of the car. (d) If the force of air resistance on the car is \(670.0 \mathrm{N},\) how far can the car travel at a speed of $22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph})$ with the initial stored energy? Ignore losses of mechanical energy due to means other than air resistance.

Step by step solution

01

(a) Calculate the rotational inertia

To calculate the rotational inertia (I) of the solid disk, use the formula \(I = \frac{1}{2} MR^2\), where M is the mass and R is the radius of the disk. Given \(M = 200.0\, \mathrm{kg}\) and \(R = 0.40\, \mathrm{m}\), we can calculate I. \(I = \frac{1}{2}(200.0\, \mathrm{kg})(0.40\, \mathrm{m})^2\)

02

Calculate the value of I

By solving for I, we get: \(I = (100)(0.16) = 16\, \mathrm{kg\,m^2}\) The rotational inertia of the flywheel is \(16\, \mathrm{kg\,m^2}\).

03

(b) Calculate the initial rotational kinetic energy

To calculate the initial rotational kinetic energy of the flywheel, use the formula \(K_E = \frac{1}{2} I\omega^2\), where \(\omega\) is the angular speed and I is the rotational inertia. Given \(\omega = 3160\, \mathrm{rad/s}\) and \(I = 16\, \mathrm{kg\,m^2}\), we can calculate the initial rotational kinetic energy. \(K_E = \frac{1}{2}(16\, \mathrm{kg\,m^2})(3160\, \mathrm{rad/s})^2\)

04

Calculate the value of \(K_E\)

Now calculate the initial rotational kinetic energy: \(K_E = (8)(99849600) = 798796800\, \mathrm{J}\) The initial rotational kinetic energy of the flywheel is \(798796800\, \mathrm{J}\).

05

(c) Calculate the ratio of the initial rotational kinetic energy of the flywheel to the translational kinetic energy of the car

To calculate the ratio, we first need to find the translational kinetic energy of the car using the formula \(K_{Et} = \frac{1}{2} Mv^2\), where \(M = 1000.0\, \mathrm{kg}\) is the car's total mass and \(v = 22.4\, \mathrm{m/s}\) is the car's speed. \(K_{Et} = \frac{1}{2}(1000.0\, \mathrm{kg})(22.4\, \mathrm{m/s})^2\) Now calculate the value of \(K_{Et}\): \(K_{Et} = 500(501.76) = 250880\, \mathrm{J}\) The translational kinetic energy of the car is \(250880\, \mathrm{J}\). We can now find the ratio: \(ratio = \frac{K_E}{K_{Et}} = \frac{798796800\, \mathrm{J}}{250880\, \mathrm{J}}\)

06

Calculate the value of the ratio

By solving for the ratio, we get: \(ratio = 3184/25 = 127.36\) The ratio of the initial rotational kinetic energy of the flywheel to the translational kinetic energy of the car is approximately \(127.36\).

07

(d) Calculate how far the car can travel

Since we are given the force of air resistance (\(F_{air} = 670.0\, \mathrm{N}\)), we can calculate the work done by air resistance using the formula: \(W = F_{air}\,d\), where d is the distance the car travels. The car can travel at a speed of \(22.4\, \mathrm{m/s}\) as long as there is energy stored in the flywheel, so we need to use conservation of mechanical energy in our calculations. We know that the initial rotational kinetic energy of the flywheel is equal to the work done by air resistance on the car: \(K_E = W\). Therefore, we can find the distance d as follows: \(798796800\, \mathrm{J} = 670.0\, \mathrm{N}\,d\)

08

Calculate the value of d

By solving for d, we get: \(d = \frac{798796800\, \mathrm{J}}{670.0\, \mathrm{N}} \approx 1192234\, \mathrm{m}\) The car can travel approximately \(1,192,234\, \mathrm{m}\) at a speed of \(22.4\, \mathrm{m/s}\) with the initial stored energy, ignoring losses of mechanical energy due to means other than air resistance.

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