To drive a typical car at 40 mph on a level road for one hour requires about 3.2 × 10^7 J of energy. Suppose one tried to store this much energy in a spinning solid cylindrical flywheel which was then coupled to the wheels of the car. A large flywheel cannot be spun too fast or it will fracture. If one used a flywheel of radius 0.60 m and mass 400 kg, what angular speed would be required to store 3.2x10^7(Incidentally, 2500 rpm is about the maximum feasible rate of revolution with present materials technology for such a flywheel
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The angular speed required to store 3.2x10^7 J of energy in a flywheel of radius 0.60 m and mass 400 kg is approximately 8.2 rad/s.
To find the angular speed required for the flywheel to store 3.2 × 10^7 J of energy, we can use the formula for rotational kinetic energy:
KE = (1/2) * I * ω^2
Where:
KE is the kinetic energy of the flywheel,
I is the moment of inertia of the flywheel,
ω is the angular speed of the flywheel.
The moment of inertia of a solid cylindrical flywheel can be calculated using the formula:
I = (1/2) * m * r^2
Where:
m is the mass of the flywheel,
r is the radius of the flywheel.
Given:
m = 400 kg,
r = 0.60 m,
KE = 3.2 × 10^7 J.
First, calculate the moment of inertia (I):
I = (1/2) * m * r^2
I = (1/2) * 400 kg * (0.60 m)^2
I = 72 kg·m^2
Substitute the values of KE and I into the formula for rotational kinetic energy:
KE = (1/2) * I * ω^2
3.2 × 10^7 J = (1/2) * 72 kg·m^2 * ω^2
Rearrange the equation and solve for ω:
ω^2 = (2 * KE) / I
ω^2 = (2 * 3.2 × 10^7 J) / 72 kg·m^2
ω^2 = 8.89 × 10^5 rad^2/s^2
Take the square root of both sides to find ω:
ω = √(8.89 × 10^5 rad^2/s^2)
ω ≈ 942 rad/s
The angular speed required to store 3.2 × 10^7 J of energy in the flywheel is approximately 942 rad/s.
To determine the angular speed required to store 3.2 × 10^7 J of energy in the flywheel, we need to use the concept of rotational kinetic energy.
The formula for rotational kinetic energy is given by:
KE = (1/2) I ω^2
Where:
- KE is the rotational kinetic energy
- I is the moment of inertia of the flywheel
- ω is the angular speed of the flywheel
In this case, we are given the value of KE (3.2 × 10^7 J) and the dimensions of the flywheel.
The moment of inertia of a solid cylinder rotating about its central axis is given by:
I = (1/2) m r^2
Where:
- m is the mass of the flywheel
- r is the radius of the flywheel
Again, we are given the values of m (400 kg) and r (0.60 m).
Let's substitute these values into the equations and solve for ω:
KE = (1/2) I ω^2
3.2 × 10^7 J = (1/2) * (1/2) * 400 kg * (0.60 m)^2 * ω^2
To simplify the equation, let's calculate (1/2) * (1/2) * 400 kg * (0.60 m)^2:
(1/2) * (1/2) * 400 kg * (0.60 m)^2 = 36 kg * m^2
Now, let's rearrange the equation to solve for ω:
ω^2 = (2 * 3.2 × 10^7 J) / (36 kg * m^2)
ω^2 = 177.8 × 10^4 J / (36 kg * m^2)
ω^2 = 494,444.44 J / (kg * m^2)
Taking the square root of both sides gives us:
ω = sqrt(494,444.44 J / (kg * m^2))
Now, let's calculate the value of ω:
ω = sqrt(494,444.44 J / (kg * m^2))
ω ≈ 222.22 rad/s
Therefore, the angular speed required to store 3.2 × 10^7 J of energy in the flywheel would be approximately 222.22 rad/s.