Arrow wave problem

[Tokspor]

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In flight, an arrow has two nodes equidistant from the two ends. The less amount of flex an arrow has the less kinetic energy it loses. Keeping the overall mass the same, which of the following will not optimize the flight in an arrow?

A. Concentrate the mass of the arrow in the center of the arrow
B. Increasing its diameter
C. Concentrate the mass of the arrow at the ends
D. Decreasing its length

The answer is C, because "concentrating the mass at the ends actually decreases the stiffness of the arrow and increases the amplitude of the anti-node."

I'm lost on this one. Which equations are being used to determine this? I can understand like this: if we take this "decrease of stiffness" to be an increase in elasticity, I can intuitively understand how the amplitude might increase since there is more potential energy. But are there any specific formulas being referenced here? How does this not optimize the arrow?

 

[matth87]

In flight, an arrow has two nodes equidistant from the two ends. The less amount of flex an arrow has the less kinetic energy it loses. Keeping the overall mass the same, which of the following will not optimize the flight in an arrow?

A. Concentrate the mass of the arrow in the center of the arrow
B. Increasing its diameter
C. Concentrate the mass of the arrow at the ends
D. Decreasing its length

The answer is C, because "concentrating the mass at the ends actually decreases the stiffness of the arrow and increases the amplitude of the anti-node."

I'm lost on this one. Which equations are being used to determine this? I can understand like this: if we take this "decrease of stiffness" to be an increase in elasticity, I can intuitively understand how the amplitude might increase since there is more potential energy. But are there any specific formulas being referenced here? How does this not optimize the arrow?

This problem is dealing with Torque. Torque due to the weight of the arrow at it ends will cause the most flex. If you concentrate the mass at the ends of the arrows then the torques at the end increase and the arrow will bend alot.

Torque = rFsin(theta)

 

[Tokspor]

This problem is dealing with Torque. Torque due to the weight of the arrow at it ends will cause the most flex. If you concentrate the mass at the ends of the arrows then the torques at the end increase and the arrow will bend alot.

Torque = rFsin(theta)

The question asks, though, about how this would not change the flight of the arrow. How would the arrow fly the same as before?

 

[matth87]

The question asks, though, about how this would not change the flight of the arrow. How would the arrow fly the same as before?

Keeping the overall mass the same, which of the following will not optimize the flight in an arrow?

In other words: which will create the most flex.

Having the greatest torques on the ends will create the most flex.

 

[capn jazz]

Also, problems like these can easily be POE'd to either A or C because they're mutually exclusive. Which one sounds more streamlined? An arrow with, say, a lead ball in the middle or an arrow with a lead ball on each end?

 

[Tokspor]

Keeping the overall mass the same, which of the following will not optimize the flight in an arrow?

In other words: which will create the most flex.

Having the greatest torques on the ends will create the most flex.

This must seem very obvious to everyone else, but I'm lost on what optimizing the flight entails and why creating more torque does not do anything to the arrow in this respect.

 

[matth87]

This must seem very obvious to everyone else, but I'm lost on what optimizing the flight entails and why creating more torque does not do anything to the arrow in this respect.

To optimize means "to provide the most favoring conditions"

Optimizing the flight would be providing conditions which will make the flight the best.

"The less amount of flex an arrow has the less kinetic energy it loses"
The more amount of flex the more kinetic energy it loses the less optimized its flight is.

Torques due to the masses at the end create the most flex and make the flight the least optimized.

 

[Tokspor]

To optimize means "to provide the most favoring conditions"

Optimizing the flight would be providing conditions which will make the flight the best.

"The less amount of flex an arrow has the less kinetic energy it loses"
The more amount of flex the more kinetic energy it loses the less optimized its flight is.

Torques due to the masses at the end create the most flex and make the flight the least optimized.

Ah, got it. Thank you.

 

[Dr Gerrard]

I would assume that the greater the kinetic energy the arrow has, the more optimized is this. What is that all about??

Also, what is r and what is F in the torque equation?

If the mass was concentrated in the center, then wouldn't the force on the center be the greatest, thus maximizing torque?

EDIT: Oh yes, I get it. A greater mass will create a greater resistive force. This force goes against the force you apply. Since this force goes against the force you apply, the torque would be lower, because the F in the equation would be as well.

Still, my first question does stand though.

 

[matth87]

I would assume that the greater the kinetic energy the arrow has, the more optimized is this. What is that all about??

Also, what is r and what is F in the torque equation?

If the mass was concentrated in the center, then wouldn't the force on the center be the greatest, thus maximizing torque?

EDIT: Oh yes, I get it. A greater mass will create a greater resistive force. This force goes against the force you apply. Since this force goes against the force you apply, the torque would be lower, because the F in the equation would be as well.

Still, my first question does stand though.

Right the greater kinetic energy, the more optimized it is.
It loses kinetic energy with greater flex, becoming less optimized.

r = the radius vector, or distance from the fulcrum or pivot point to where the force is applied
Fsin(theta) = perpendicular force to radius vector

If the mass was at the center there would be no torque, consider the pivot point or fulcrum the center of the arrow. If all the mass is at the center r = 0. and Torque = 0. The more distributed the mass is the greater r is the great the opposing Clockwise and Counterclockwise torques, the greater the flex.

You are not changing the mass in any of the answer choice, you are simply redistributing it throughout the object. And "you" are not applying any force, the forces applied that create the torque are due to the Force Gravity aka the Weight of the arrow.

 

[Tokspor]

I would assume that the greater the kinetic energy the arrow has, the more optimized is this. What is that all about??

You are right. The question says "the less amount of flex an arrow has the less kinetic energy it loses." So if you create a lot of flex, you dissipate more kinetic energy, making conditions less optimal. The latter is what the question is asking you to look for.

 

[Tokspor]

You are not changing the mass in any of the answer choice, you are simply redistributing it throughout the object. And "you" are not applying any force, the forces applied that create the torque are due to the Force Gravity aka the Weight of the arrow.

Are we assuming that the masses distributed to both ends are uneven? If they are equal masses, wouldn't the arrow still remain balanced?

 

[matth87]

Are we assuming that the masses distributed to both ends are uneven? If they are equal masses, wouldn't the arrow still remain balanced?

If the arrow mass = 10kg, then you would be moving 5kg to each end of the arrow. The weight of the arrow is the same. But since you now have a radius from the pivot point, the center of the arrow, you have torques. The opposing torques, CCW and CW, create a flex in the arrow.

The answer choice says : Concentrate the mass of the arrow at the ends <-- key word, so you concentrate it at both ends equally

Torques are not forces, the merely create rotational acceleration, and are due to forces acting at some distance from a pivot point.