Answer:
The total kinetic energy of the car can be divided into two parts: translational kinetic energy, due to the car's motion as a whole, and rotational kinetic energy, due to the rotation of the wheels about their axles.
The translational kinetic energy of the car is given by:
KE_translational = (1/2)mv^2
where m is the mass of the car and v is its speed.
The rotational kinetic energy of each wheel can be calculated as:
KE_rotational = (1/2)Iω^2
where I is the moment of inertia of a wheel and ω is its angular velocity.
The moment of inertia of a uniform disk is given by:
I = (1/2)mr^2
where m is the mass of the disk and r is its radius.
The mass of each wheel is given as 12 kg, so the moment of inertia of each wheel is:
I = (1/2)(12 kg)(0.5 m)^2 = 1.5 kg·m^2
Assuming that the wheels are rolling without slipping, the angular velocity of each wheel can be related to the speed of the car by:
v = rω
where r is the radius of the wheel.
Solving for ω, we get:
ω = v/r
Substituting this expression into the equation for KE_rotational, we get:
KE_rotational = (1/2)(1.5 kg·m^2)(v/r)^2
Simplifying, we get:
KE_rotational = (3/8)mv^2
So the rotational kinetic energy of each wheel is (3/8) of the translational kinetic energy of the car.
The total rotational kinetic energy of all four wheels is therefore:
KE_total_rotational = 4 KE_rotational = (3/2)mv^2
The total kinetic energy of the car is:
KE_total = KE_translational + KE_total_rotational = (5/2)mv^2
So the fraction of the total kinetic energy of the car that is due to the rotation of the wheels about their axles is:
KE_total_rotational / KE_total = (3/5) = 0.6
Therefore, 60% of the car's total kinetic energy is due to the rotation of the wheels about their axles.