# How Long Does It Take for a Boomerang to Return?

• caters
In summary: The y direction would be the component of the force that is perpendicular to the x direction. So if we take the x component of the force, it would be 10N * sin(30°) = 5N. But since the y component is perpendicular to the x direction, it would be 10N * cos(30°) = 15N.The velocity in both the x and y directions also is split in half so each velocity initially is ##2.5\frac{m}{s}##So the x velocity is 2.5 m/s and the y velocity is .5 m/s.Now, in order to not fall and to actually come back to its original position
caters

## Homework Statement

A boomerang is thrown with an initial linear velocity of 5 m/s at an angle of 30 degrees vertically. The initial angular velocity is ##2\frac{revolutions}{s}## At its peak, it has a displacement about the z axis of 2 meters and about the x-axis of 10 meters. The force applied on the boomerang during the throw is 10N with no force in the z direction. It has a mass of 2 kg and a length of .2 m. How long does it take for the boomerang to return?

##F=ma##
##ω=ω_0+αt##

## The Attempt at a Solution

Okay so first off let's assume it starts at 0 on both the x and z axes. Since it goes further along the x-axis than the z axis, it must have a higher velocity in the x direction for most of its path. Next up we need to take all the forces into account. There are 2 forces imparted on the boomerang. There is the initial throwing force of 10N which becomes the torque of the boomerang. There is also the force of gravity.

The force of gravity is ##F_g=mg## according to Newton's second law. This is ##F_g=2 kg*9.8\frac{m}{s^2}## which multiplies out to give us ##F_g=19.6N##

The initial force needs to be separated into its x and y components. The x component is going to be ##F_x=10N * sin(30°) = 10N * .5 = 5N## This means that ##F_y=5N## The velocity in both the x and y directions also is split in half so each velocity initially is ##2.5\frac{m}{s}##

Now, in order to not fall and to actually come back to its original position, the boomerang has to accelerate in the direction of its motion since gravity starts out being the predominant force. It must have a rotational acceleration such that at its peak, the torque balances out with gravity. In other words as time passes, the acceleration must increase which means that the angular velocity has to change, even if the linear velocity stays constant for some time ##t##.

And this is where I'm stuck. I should have enough information here to figure out the angular acceleration and thus the time it takes for the boomerang to return but I can't seem to figure it out. Can somebody help me with this? Also I have no idea why the inline latex isn't showing up properly.

EDIT: I fixed the error with the Latex

Last edited:
After the boomerang is thrown, how many forces act on it if you ignore air resistance?
This problem is simpler than it looks. Just think. What determines the time of flight? If the boomerang cast a shadow on a vertical wall, what path would the shadow trace on the wall?

Click on the LaTeX link, bottom left to see how it is done here. You need  or ## brackets.

Well of course gravity is acting on it because gravity acts on all masses. Wait a second, are you suggesting that after the throw, there is no torque causing it to rotate faster and slower and that the only force acting on it is gravity and that the reason it comes back instead of simply falling has to do with that initial force of 10N causing it to rotate?

This is a very poorly stated question.
caters said:
at an angle of 30 degrees vertically.
What does this mean? That its velocity is 30 degrees to the horizontal, or 30 degrees to the vertical, or that its plane of rotation is at 30 degrees to the vertical?
(In fact, they are thrown in a vertical plane, or nearly, "elbow" pointing backwards.)
caters said:
initial angular velocity
Hmm, implying the angular velocity changes. Yes, it would decrease because of drag, but since we have no way to estimate that it strongly suggests the rotation info is irrelevant.
caters said:
The axes have not been defined. Is this vertical or normal to the plane of the boomerang?
caters said:
The force applied on the boomerang during the throw is 10N
Since we are not told the duration for which the force is applied this cannot be useful.
caters said:
no force in the z direction.
I guess we can deduce that z is vertical and this just means it is thrown in a horizontal direction. But in that case I do not see why it would climb. Maybe the author thinks they are thrown flat, more like a Frisbee.
caters said:
throwing force of 10N which becomes the torque of the boomerang.
How does a force "become" a torque? A force can have a torque about a specified axis. But throwing a boomerang cannot be just a question of applying a linear force - it would not spin very well. It is thrown with an overarm rotation, applying some extra spin with the wrist.
caters said:
initial force needs to be separated into its x and y components.
Eh? Where did the y direction come into it? How are you interpreting the 30 degree angle info?
caters said:
the boomerang has to accelerate in the direction of its motion
I see no way it could nor any need for it to do so.

haruspex said:
This is a very poorly stated question.

What does this mean? That its velocity is 30 degrees to the horizontal, or 30 degrees to the vertical, or that its plane of rotation is at 30 degrees to the vertical?
(In fact, they are thrown in a vertical plane, or nearly, "elbow" pointing backwards.)

Hmm, implying the angular velocity changes. Yes, it would decrease because of drag, but since we have no way to estimate that it strongly suggests the rotation info is irrelevant.

The axes have not been defined. Is this vertical or normal to the plane of the boomerang?

Since we are not told the duration for which the force is applied this cannot be useful.

I guess we can deduce that z is vertical and this just means it is thrown in a horizontal direction. But in that case I do not see why it would climb. Maybe the author thinks they are thrown flat, more like a Frisbee.

How does a force "become" a torque? A force can have a torque about a specified axis. But throwing a boomerang cannot be just a question of applying a linear force - it would not spin very well. It is thrown with an overarm rotation, applying some extra spin with the wrist.

Eh? Where did the y direction come into it? How are you interpreting the 30 degree angle info?

I see no way it could nor any need for it to do so.

I am interpreting the 30 degree angle as being 30 degrees to the horizontal because that would be the way to interpret going upwards at a 30 degree angle. And I have the axes oriented such that the y-axis is the vertical axis and the z axis is normal to the x-axis like in this picture:

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caters said:
I am interpreting the 30 degree angle as being 30 degrees to the horizontal because that would be the way to interpret going upwards at a 30 degree angle. And I have the axes oriented such that the y-axis is the vertical axis and the z axis is normal to the x-axis like in this picture:
View attachment 239900
That would be unusual. The standard is that z is the vertical axis. With your arrangement, how do you interpret this:
caters said:
At its peak, it has a displacement about the z axis of 2 meters
?
(I guess it means along the z axis.)

The question doesn't say it rises at a 30 degree angle. That's just your interpretation. Indeed, if I am right that z is vertical then it does not maintain a 30 degree angle.
However, I do agree the author probably means that the initial 5m/s is at 30 degrees to the horizontal.

The author appears to have no understanding of how boomerangs work, and has deliberately supplied a lot of data that cannot help. We need to locate just a few facts that suggest an answer, but it’s hard without guessing exactly what misconceptions he is under.
I think a key question is how the horizontal component of velocity changes over time. If the flight had been consistently at 30 degrees to the horizontal then we could have treated it as a block sliding up and down a frictionless slope, but given that it seems to level out somewhat I am left with no obvious way to model it.

## 1. What is the boomerang problem?

The boomerang problem is a mathematical puzzle about finding the minimum force required to throw a boomerang in order for it to return to its starting point.

## 2. How does one solve the boomerang problem?

The boomerang problem can be solved using mathematical equations and principles of physics, such as conservation of energy and angular momentum.

## 3. What factors affect the solution to the boomerang problem?

The solution to the boomerang problem is affected by factors such as the shape and weight distribution of the boomerang, air resistance, and the strength and angle of the throw.

## 4. Why is the boomerang problem important?

The boomerang problem is important because it helps us understand the principles of physics and can also be applied to real-life situations, such as designing and improving boomerangs for recreational or competitive purposes.

## 5. Are there any real-life applications of the boomerang problem?

Yes, the boomerang problem has been used in the development of boomerangs for recreational and competitive use, as well as in the design of flying objects such as drones and airplanes.

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