# Physics of Boomerangs: Learn Moment of Inertia Calculation

• RicardoMarques
In summary, Ricardo is a new member of a Physics forum who joined to get information for his paper on "The Physics of Boomerangs." He found a topic on the forum where someone was explaining how to calculate the moment of inertia for a boomerang model. He had some difficulty understanding the last part and asked for clarification, which was provided. He also asked about calculating angular momentum and proving that a boomerang returns, but it was suggested that this would be a more complicated problem involving aerodynamics and other factors. A link to a paper on the topic was also provided.
RicardoMarques
Member advised to use the homework template for posts in the homework sections of PF.
Im new at this Physics forum and i don't quite know anyone here.
I came here because I'm doing a paper about "The Physics of Boomerangs" and so i found a topic about it where you explained a guy how to calculate the moment of inertia on a simple model of a boomerang.

https://www.physicsforums.com/threa...rtia-for-wierdly-shaped-objects.503476/page-2

However, I didnt understand the last part, about the end of the calculus. Gneill said I new = I + M.r^2 . So I use that formula 1/12 x M(a^2+b^2) and then add M x r^2 ? And after that you multiply by 3 ? In that case scenario r = 5.4 ?

Greetings from Portugal

Hi Ricardo.

In future please use the formatting template that is provided in the edit window when a new thread is started in a homework forum. Don't delete it!

The model boomerang in question consisted of three identical rectangular vanes that were arranged at 120° angle separation, and overlapped at the center (presumably glued together). Here's the picture provided by the original poster of the thread in question:

RicardoMarques said:
However, I didnt understand the last part, about the end of the calculus. Gneill said I new = I + M.r^2 . So I use that formula 1/12 x M(a^2+b^2) and then add M x r^2 ? And after that you multiply by 3 ? In that case scenario r = 5.4 ?

Yes that's the idea. The r had units of cm, and was, I believe, an approximation based on the described dimensions of the model boomerang. The parallel axis theorem was applied to find the moment of inertia of one vane about the center of of rotation of the boomerang as a whole. The result was multiplied by three to account for the three vanes.

The method relies on the fact that the vanes are in fact whole rectangles. If the vanes weren't overlapped and the boomerang was cut from a single sheet of material then another method would have to be found.

RicardoMarques
I'll have that in mind next time I create a post

Thank you so much !

One more thing, adding to the past one...

I calculated the moment of inertia right ? Now I want to know the angular momentum, i only have to multiply this for angular velocity and its done ?

Also, If I want to prove that the boomerang returns using this calculus what would you counsel me to do ? Adding torque calculus would help ?

RicardoMarques said:
One more thing, adding to the past one...

I calculated the moment of inertia right ? Now I want to know the angular momentum, i only have to multiply this for angular velocity and its done ?
Yes.
Also, If I want to prove that the boomerang returns using this calculus what would you counsel me to do ? Adding torque calculus would help ?
I fear that proving that a given boomerang returns, or simply determining any part of its trajectory would be a much more difficult problem involving aerodynamics. Angular momentum would be just a small factor in a much larger problem.

Edit:
Have a look at the Hyperphysics entry on the boomerang for an overview.

What if I was allowed to consider only angular momentum ? Would I be able to prove it ? And how ?

This is a small paper, I think I can assume that other stuff don't matter, or simply not consider them in my calculus

See my Edit update in my previous post.

## What is the physics behind a boomerang's flight?

The physics behind a boomerang's flight is based on the principles of aerodynamics and angular momentum. When the boomerang is thrown, it experiences lift and drag forces due to its curved shape and spin. These forces cause the boomerang to curve in flight and return to the thrower.

## How is the moment of inertia calculated for a boomerang?

The moment of inertia for a boomerang can be calculated by using the formula I = mr², where I is the moment of inertia, m is the mass of the boomerang, and r is the distance from the axis of rotation to the center of mass.

## What factors affect a boomerang's moment of inertia?

The moment of inertia for a boomerang can be affected by its shape, size, and weight distribution. A boomerang with a larger mass or a larger distance from the axis of rotation will have a higher moment of inertia.

A boomerang returns to the thrower due to its unique shape and aerodynamic properties. As the boomerang spins, it creates lift and drag forces that cause it to curve in flight and return to the thrower. The moment of inertia also plays a role in the boomerang's return, as it helps stabilize the flight path.

## Can the moment of inertia be manipulated to improve a boomerang's flight?

Yes, the moment of inertia can be manipulated by adjusting the weight distribution or shape of the boomerang. By changing these factors, the boomerang's flight path can be altered and potentially improve its return to the thrower.

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