Moment of inertia of a flywheel

In summary, to solve the homework equation, you need to know the change in angular kinetic energy and the initial and final radii.
  • #1
Arejang
32
0
[SOLVED] moment of inertia

Homework Statement


The flywheel of a gasoline engine is required to give up 750 J of kinetic energy while its angular velocity decreases from 870 rev/min to 410 rev/min.

What moment of inertia is required?

Homework Equations



[tex]I=\frac{2K}{\omega^{2}}[/tex]

The Attempt at a Solution



I'm not sure how to derive the rad/sec used in this equation. I've converted both 870 rev/min and 410 rev/min to radians before attempting the problem (91.1 rad/sec and 42.9 rad/sec respectively). I've tried taking the difference between the two, then attempting the average, then just using the final radial velocity and ignoring the initial. I'm not sure what to do at the moment. Can anyone help?
 
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  • #2
Well they're saying "give up" 750 joules of KE. So you don't actually know any absolute kinetic energies. You know there is a Ki, and that Kf=Ki-750J, you know Ki corresponds with 870rpm and Kf corresponds with 410 rpm, and that is all you're given to work with
 
  • #3
You started on the right track but got off on the wrong foot.

[tex]
\begin{aligned}
K_i &= \frac 1 2 I \omega_i^2 \\[4pt]
K_f &= \frac 1 2 I \omega_f^2 \\
&\Rightarrow \\
K_i-K_f &= \frac 1 2 I (\omega_i^2 - \omega_f^2)
\end{aligned}[/tex]

You can take it from here.
 
  • #4
I did something, but I'm still not sure I understand it, but it seemed to work.

If I assume [tex]K_{f}[/tex] is 0, then I can set up the problem like so:

[tex]K_{i}=\frac{1}{2}I(\omega_{i}^{2}-\omega_{f}^{2})[/tex]

Solving for I, I get[tex]\frac{2K_{i}}{(\omega_{i}^{2}-\omega_{f}^{2})}=I[/tex]

Plug and chug and got .232 kg*m^2, which is the correct answer.

What I don't understand is why [tex]K_{f}[/tex] is 0.
 
  • #5
You cannot assume [itex]K_f[/itex] is zero. You are given the change in angular kinetic energy. The problem statement says exactly what [itex]K_i-K_f[/itex] is: 750 joules.
 
  • #6
So in other words. What K initial and K final are is not important, because the change in kinetic energy is already given in the problem?
 
  • #7
yeah
 
  • #8
^_^; man I feel stupid now. Well, anyway thanks!
 

Related to Moment of inertia of a flywheel

1. What is moment of inertia?

The moment of inertia of an object is a measure of its resistance to changes in rotational motion, similar to how mass is a measure of an object's resistance to changes in linear motion.

2. How is moment of inertia calculated?

Moment of inertia is calculated by multiplying the mass of the object by the square of its distance from the axis of rotation.

3. Why is moment of inertia important in a flywheel?

In a flywheel, moment of inertia determines how much energy can be stored and how smoothly the flywheel can maintain its rotational motion.

4. How does the shape of a flywheel affect its moment of inertia?

The shape of a flywheel can significantly affect its moment of inertia. A flywheel with a larger diameter or more mass concentrated near the edges will have a higher moment of inertia compared to a flywheel with a smaller diameter or more mass concentrated near the center.

5. How does the moment of inertia of a flywheel affect its performance?

A flywheel with a higher moment of inertia will have a slower rate of acceleration and deceleration, but it will also be able to store more energy and maintain its rotational motion for a longer period of time. A flywheel with a lower moment of inertia will have a faster rate of acceleration and deceleration, but it will not be able to store as much energy or maintain its motion for as long.

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