Question

A flywheel, that has most of its mass at its rim, can rotate about an axis that is perpendicular to the wheel at its center. Rotating flywheel provide a means for storing energy in the form of rotational kinetic energy, and are being used in some clectric power plant to store excess electrical energy. How fast would a M - 14-kg flywheel with radius R = 0.35 m, have to rotate (that is, what should be its o in rad/s), to store 2.3 x 10' Joules of rotational kinetic energy? (Rotational Inertia of a flywheel is I = MR?)

Answer 1

To store 2.3 x 10⁸ J of rotational kinetic energy, the M-14 kg flywheel with a radius R=0.35 m would have to rotate at a speed of approximately 146.08 rad/s.

Step-by-step explanation:

To calculate the angular velocity (ω) at which the flywheel should rotate to store a given amount of rotational kinetic energy, we can use the formula:

KE = (1/2)Iω²

Where KE is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.

Given that the flywheel has a mass (M) of 14 kg and a radius (R) of 0.35 m, we can calculate the moment of inertia (I) using the formula:

I = MR²

Substituting the values:

I = (14 kg) x (0.35 m)² = 17.675 kg m²

Now, we can rearrange the equation for kinetic energy and solve for ω:

ω = √(2KE / I)

Substituting the given value of rotational kinetic energy (KE = 2.3 x 10⁸ J) and the calculated moment of inertia (I = 17.675 kg m²):

ω = √(2 x 2.3 x 10⁸ J / 17.675 kg m²) ≈ 146.08 rad/s

Therefore, the flywheel would have to rotate at a speed of approximately 146.08 rad/s to store 2.3 x 10⁸ J of rotational kinetic energy.

Answer 2

A 14-kg flywheel with a radius of 0.35 m needs to rotate at an angular velocity of approximately 517.33 rad/s to store 2.3 x 10⁵ Joules of rotational kinetic energy.

Step-by-step explanation:

The question asks how fast a 14-kg flywheel with a radius of 0.35 m needs to rotate to store 2.3 x 10⁵ Joules of rotational kinetic energy. We can use the equation for rotational kinetic energy, KErot = 0.5 * I * ω2, where I is the moment of inertia and ω is the angular velocity.

The moment of inertia for a flywheel, where most of the mass is at the rim, is given by I = MR2. So, for our case,

I = (14 kg)(0.35 m)² = 1.715 kg·m²

We can solve for ω by rearranging the kinetic energy equation to ω = √(2 * KErot / I). Substituting the given values, we get

ω = √(2 * 2.3 x 10⁵ J / 1.715 kg·m²)

ω = √(2 * 2.3 x 10⁵ / 1.715)

ω = √(267637.896)

ω ≈ 517.33 rad/s

Thus, the flywheel needs to rotate at an angular velocity of approximately 517.33 rad/s to store 2.3 x 10⁵ Joules of energy.