Moment of inertia of circular flywheel

In summary: You could get it by subtracting areas of two circles, but it's simpler than that. (And if you do it by subtraction, you get pi (a+dr)2 - pi a2 = pi (2adr + dr2).)In summary, a new member of a forum is seeking help with finding the moment of inertia of a circular flywheel with radius a and mass m. They mention the area density of the flywheel and their process of finding the moment of inertia by using polar coordinates. However, they are having trouble with the integration process and are seeking clarification on their variables.
  • #1
KiNGGeexD
317
1
Hi I'm new to this forum so still getting used to it!

I have to find the moment of inertia of a circular flywheel which has radius, a and mass m! But also had area density (mass per unit area) of ρ0,

Where ρ0=ρa(1+r/a)

I know moment of inertia is the integral of a^2 dm and in order to get mass m I just multiply the above expression by area of the flywheel which I think is pi*a^2

I end up with a rather foolish answer when I have competed the intergratiom!
Any help would be great

Thanks guys
 
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  • #2
Is the answer (9∏[itex]\rho[/itex]a5 /10) ?
If its' correct pls. reply so that i would post my solution.
 
  • #3
KiNGGeexD said:
I know moment of inertia is the integral of a^2 dm and in order to get mass m I just multiply the above expression by area of the flywheel which I think is pi*a^2
Your process is not clear from that description. Please post your actual working.
 
  • #4
Vineeth T said:
Is the answer (9∏[itex]\rho[/itex]a5 /10) ?
If its' correct pls. reply so that i would post my solution.
That's not how this forum works. We don't just post solutions. The idea is to provides hints and guide the thread originator to the solution.
 
  • #5
Oh! Sorry then.
The idea is to take a differential strip(a ring of thickness dr) at a general distance r, find its' M.I and then integrate it to get M.I of the whole disc(a flywheel).
 
  • #6
Yea I know sorry!
Well what I have done is used polar coordinates and defined my moment of inertia as of course

I= integral of r^2 dm

And my element of mass is just ρ0+[1+r/α] *2pi a^2 but I am having trouble with integrating the expression! Are my boundaries a -0 or r-0?
 
  • #7
KiNGGeexD said:
Yea I know sorry!
Well what I have done is used polar coordinates and defined my moment of inertia as of course

I= integral of r^2 dm

And my element of mass is just ρ0+[1+r/α] *2pi a^2 but I am having trouble with integrating the expression! Are my boundaries a -0 or r-0?

Seems to me there's some confusion between variables and constants here.
In the OP you had r as a variable from 0 to a (constant); ρ0 as a variable from ρ a to 2 ρ a, ρ being a constant. That's very unusual - I would have expected ρ and ρ0 to have reversed roles. Assuming you had them crossed over, I'll write ρ(r) = ρ0a(1+r/a). (Maybe you meant ρa, not ρ a, in the OP, but that's just a change to the constant ρ0.)
You want to integrate wrt r. At radius r the density is ρ(r). To get the element of mass you need to multiply that by the element of area, i.e. the area of an annulus between radius r and radius r+dr. In your above post, you seem to have that as 2pi a^2, which is obviously wrong. Also, is "ρ0+[1+r/α]" a typo? Should it be ρ0[1+r/α], or ρ0a[1+r/α]?
 
  • #8
Why is 2pi a^2 obviously wrong? It is the area of the flywheel is it not?
 
  • #9
KiNGGeexD said:
Why is 2pi a^2 obviously wrong? It is the area of the flywheel is it not?
Well, no, the area of the whole flywheel would be pi a2, not 2pi a2. But for the purposes of the integrand we want the area of an element between radius r and radius r+dr, not the whole flywheel, right? That is independent of a.
 

Related to Moment of inertia of circular flywheel

What is the moment of inertia of a circular flywheel?

The moment of inertia of a circular flywheel is a measure of its resistance to changes in rotational motion. It is a property of the flywheel that depends on its mass, shape, and distribution of mass.

How do you calculate the moment of inertia of a circular flywheel?

The moment of inertia of a circular flywheel can be calculated using the formula I = 1/2 * m * r^2, where I is the moment of inertia, m is the mass of the flywheel, and r is the radius of the flywheel. It can also be calculated using the parallel axis theorem, which takes into account the distribution of mass in the flywheel.

What factors affect the moment of inertia of a circular flywheel?

The moment of inertia of a circular flywheel is affected by its mass, shape, and distribution of mass. A heavier flywheel will have a larger moment of inertia, while a flywheel with mass concentrated near its center will have a smaller moment of inertia than one with mass spread out towards the edges.

Why is the moment of inertia important for a circular flywheel?

The moment of inertia is important for a circular flywheel because it determines how much torque is required to accelerate or decelerate the flywheel. A larger moment of inertia means it will take more torque to change the rotational motion of the flywheel, making it more stable and resistant to changes in speed.

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

The moment of inertia of a circular flywheel can affect its performance in various ways. A larger moment of inertia can provide more stability and smoother operation, while a smaller moment of inertia can result in quicker acceleration and deceleration. It is an important factor to consider when designing or selecting a flywheel for a specific application.

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