Some special vehicles have spinning disks (flywheels) to store energy while they roll downhill. They use that stored energy to lift themselves uphill later on. Their flywheels have relatively small rotational masses but spin at enormous angular speeds. How would a flywheel’s kinetic energy change if its rotational mass were 7 times larger but its angular speed were 7 times smaller?

Question

Some special vehicles have spinning disks (flywheels) to store energy while they roll downhill. They use that stored energy to lift themselves uphill later on. Their flywheels have relatively small rotational masses but spin at enormous angular speeds. How would a flywheel’s kinetic energy change if its rotational mass were 7 times larger but its angular speed were 7 times smaller?

KE2/KE1 =

Answer 1

Ke = (1/2) I w^2

KEmodified = (1/2)(7I)(w^2/49) = original Ke/7

Answer 2

To determine how the kinetic energy (KE) of a flywheel would change if its rotational mass were 7 times larger and its angular speed were 7 times smaller, we can use the formula for kinetic energy of a rotating object:

KE = (1/2) I ω²

Where:
KE is the kinetic energy
I is the moment of inertia (rotational mass)
ω is the angular speed (rotational velocity)

To compare the kinetic energy, let's assign subscripts to the initial (1) and final (2) states of the flywheel.

Given that the initial rotational mass (I₁) is 7 times smaller, we can represent the final rotational mass (I₂) as:

I₂ = 7I₁

Also, the final angular speed (ω₂) is 7 times smaller, meaning:

ω₂ = ω₁/7

Now we can substitute these values into the kinetic energy formula:

KE₂ = (1/2) I₂ ω₂²
= (1/2) (7I₁) (ω₁/7)²
= (1/2) (7I₁) (ω₁²/49)
= (7/98) I₁ ω₁²

Therefore, the ratio of the final kinetic energy (KE₂) to the initial kinetic energy (KE₁) is:

KE₂/KE₁ = ((7/98) I₁ ω₁²) / ( (1/2) I₁ ω₁²)
= (7/98) / (1/2)
= (7/98) * (2/1)
= 7/49

So, if the flywheel's rotational mass were 7 times larger and its angular speed were 7 times smaller, the kinetic energy would be reduced to 7/49 (approximately 0.14) of its initial value.

Answer 3

To determine how the kinetic energy (KE) of the flywheel changes, we can use the formula for rotational kinetic energy:

KE = (1/2)Iω^2

Where:
- KE is the kinetic energy
- I is the moment of inertia (rotational mass)
- ω is the angular speed

Let's denote KE1 as the initial kinetic energy when the flywheel has a certain rotational mass (I1) and angular speed (ω1). Similarly, let's denote KE2 as the final kinetic energy when the flywheel has a larger rotational mass (I2) and smaller angular speed (ω2).

We can write the ratio of KE2 to KE1 as:

KE2/KE1 = [(1/2)I2ω2^2] / [(1/2)I1ω1^2]

Simplifying the equation, we have:

KE2/KE1 = (I2ω2^2) / (I1ω1^2)

Given that the rotational mass (moment of inertia) is 7 times larger (I2 = 7I1) and the angular speed is 7 times smaller (ω2 = ω1/7), we can substitute these values into the equation:

KE2/KE1 = (7I1)(ω1/7)^2 / (I1ω1^2)

Simplifying further:

KE2/KE1 = (7I1)(1/49)(1/I1) = 7/49 = 1/7

Therefore, the ratio of KE2 to KE1 is 1/7, which means that the final kinetic energy is 1/7th of the initial kinetic energy.