You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to...

Question:

You are designing a rotating metal flywheel that will be used to store energy. The flywheel is to be a uniform disk with radius {eq}29.0{/eq} {eq}cm{/eq}. Starting from rest at {eq}t=0{/eq}. The flywheel rotates with constant angular acceleration {eq}3.00{/eq} {eq}rad/s^2{/eq} about an axis perpendicular to the flywheel at its center.

1. If the flywheel density (mass per unit volume) of {eq}8600{/eq} {eq}kg/m^3{/eq}, what thickness must it have to store {eq}800{/eq} {eq}J{/eq} of kinetic energy at {eq}t=8.00{/eq} {eq}s{/eq} ?

Rotational Kinetic Energy:

The energy stored in a rotation flywheel is in the form of rotational kinetic energy given by the formula:

$$\begin{align*} &E_K = \frac{1}{2} I\omega^2 & \text{[Rotational Kinetic Energy of a Flywheel]]} \end{align*} $$

where {eq}\omega{/eq} is the angular velocity of the flywheel.

The moment of inertia {eq}I{/eq} of the flywheel about its axis of symmetry is given by the equation:

$$\begin{align*} &I = \frac{1}{2}mr^2 & \text{[Moment of Inertia of a Flywheel]} \end{align*} $$

where {eq}m{/eq} is the mass of the flywheel and {eq}r{/eq} is its radius.

Answer and Explanation:

Given a rotating metal flywheel with radius {eq}r = 29 \,cm = 0.29\,m{/eq} and density {eq}\rho = 8600\,\frac{kg}{m^3}{/eq}.

The flywheel's mass {eq}M{/eq} and moment of inertia about its center {eq}I{/eq} expressed in terms of its thickness {eq}t{/eq} in meters are:

{eq}\begin{align*} M &= \rho V \\ &= \rho (\pi r^2t) \\ &= \pi (8600)(0.29)^2t \\ &= 723.26\pi t & \text{[Mass of the Flywheel]} \\ \\ I &= \frac{1}{2}mr^2 \\ &= \frac{1}{2} (723.26\pi t)(0.29)^2 \\ &= 30.413083\pi t & \text{[Moment of Inertia of the Flywheel]} \end{align*} {/eq}

The flywheel started spinning from rest at {eq}t = 0{/eq} with an angular acceleration of {eq}\alpha = 3 \,\frac{rad}{s^2}{/eq}. After {eq}t = 8\,s{/eq}, the angular velocity of the flywheel is:

{eq}\begin{align*} \omega &= \alpha t \\ &= (3)(8) \\ &= 24\, \frac{rad}{s} & \text{[Angular Velocity at } t=8 \text{ s]} \end{align*} {/eq}

Finally, to store {eq}800\,J{/eq} of kinetic energy, the flywheel must have a thickness {eq}t{/eq} such that:

{eq}\begin{align*} E_K &= \frac{1}{2} I \omega^2 & \text{[Substitute the known values]} \\ 800 &= \frac{1}{2} (30.413083\pi t) (24)^2 & \text{[Solve for } t \text{]} \\ \\ t &= \frac{2(800)}{(30.413083\pi)(24)^2} \\ &= \frac{25}{9(30.413083\pi)} \\ t &= \boxed{ 0.0291 \,m = 2.91\,cm } & \text{[Required Thickness of the Flywheel to Store } 800 \text{ J of Energy]} \end{align*} {/eq}


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Kinetic Energy of Rotation

from UExcel Physics: Study Guide & Test Prep

Chapter 7 / Lesson 7
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