Minimum velocity required by grasshopper

[Googlu02]

Homework Statement

There is a grasshopper who wants to cross a log(cylindrical) of radius $$R$$.Calculate the minimum velocity required by grasshopper to just scrape the log.

Homework Equations

First of all I have a doubt.So equations are out of question. At least I know that it is a question of projectile motion.

The Attempt at a Solution

I have a doubt.For minimum velocity should the grasshopper scrape once or twice over the log?Please someone clarify it.

 

[berkeman]

Homework Statement

There is a grasshopper who wants to cross a log(cylindrical) of radius $$R$$.Calculate the minimum velocity required by grasshopper to just scrape the log.

Homework Equations

First of all I have a doubt.So equations are out of question. At least I know that it is a question of projectile motion.

The Attempt at a Solution

I have a doubt.For minimum velocity should the grasshopper scrape once or twice over the log?Please someone clarify it.

The Relevant Equations are indeed the projectile motion equations.

Since you probably can ignore air resistance in this simple problem, the grasshopper will follow a parabolic path, correct? Sketch some parabolas to figure out more about whether it barely touches the cylinder once or twice. What are your thoughts on the minimum initial velocity Vo that is needed to clear the log? What equation will you differentiate to find the minimum value?

 

[drvrm]

I have a doubt.For minimum velocity should the grasshopper scrape once or twice over the log?Please someone clarify it.

i think one should specify the initial position of the grasshopper first to ask the question...its very similar to a high jump in atheletics and the choice of initial position determines the possible angle of projection so that he can scrape/clear the height. and the horizontal range and vertical distance gets related through angle of projection...

 

[Googlu02]

The initial position of the grasshopper is at ground.

 

[berkeman]

i think one should specify the initial position of the grasshopper first to ask the question.

The initial position is one of the variables in the optimization equation...

The initial position of the grasshopper is at ground.

Now show us some of the equations for the parabolic path of the grasshopper, and how they depend on Vo...

 

[ehild]

Homework Statement

There is a grasshopper who wants to cross a log(cylindrical) of radius $$R$$.Calculate the minimum velocity required by grasshopper to just scrape the log.

Homework Equations

First of all I have a doubt.So equations are out of question. At least I know that it is a question of projectile motion.

The Attempt at a Solution

I have a doubt.For minimum velocity should the grasshopper scrape once or twice over the log?Please someone clarify it.

"Just scrape the log" should mean that the grasshopper does not rise higher than the log, that is 2R above the ground, as shown in the picture. There are two parameters to choose for minimum initial speed: The distance and angle of the jump.

 

[dykuma]

The velocity needed a reach a certain height (velocity in the y direction) is the same that is reached by an object falling with no air resistance. So just think of the velocity attained by an object falling a height of 2R. But to get the true velocity needed to clear the log (the velocity in the x direction), you need to know the initial distance away from the center of the log. That is the key to solving this problem.

 

[Googlu02]

There is no need of the distance from the log.And the grasshopper will scrape twice through the log.And the angle by which it scrapes the first part is 45 degrees.

 

[drvrm]

There is no need of the distance from the log.And the grasshopper will scrape twice through the log.And the angle by which it scrapes the first part is 45 degrees.

pl. show your path as a projectile...

 

[haruspex]

"Just scrape the log" should mean that the grasshopper does not rise higher than the log, that is 2R above the ground

That is not immediately clear, and this is precisely Googlu02's issue. Perhaps clearing a height of 2R is not enough. And if it is not, 45 degrees might not be the ideal launch angle.
@Googlu02, I suggest you first check whether merely clearing a height of 2R with a 45 degree launch will clear the log. What would be the radius of curvature of the parabola at its apex?

 

[dykuma]

It seems to me then that there are two ways to approach this problem, depending on how technical you want to be.

Many intro physics classes would be content with just having the grasshopper passing though a point that is 2R above the ground. If you know that the initial angle was 45°, the magnitude of velocity can be easily obtained by seeing that: V_y=|V|⋅sin(45°), where V_y is obtained though the method I mentioned earlier.

It seems though that you are talking about a situation where, perhaps the two foci of the parabola scrape the sides of the log, with an initial launch angle of 45°? Can you confirm if this depiction is what you are talking about?

 

[haruspex]

with an initial launch angle of 45°?

There is no basis for assuming that.

 

[dykuma]

There is no basis for assuming that.

It seemed to me that the OP is saying that the initial launch angle is 45° in his most recent post, which is why I assumed that.

 

[haruspex]

the grasshopper will scrape twice through the log

Maybe, but you do not know that yet.

the angle by which it scrapes the first part is 45 degrees.

What do you mean by that? It sounds like you are saying that as it scrapes the log the tangent there will be 45 degrees. Perhaps you mean you can assume the launch angle is 45 degrees. Unless you have left something out of the original problem statemennt, neither is necessarily true.

 

[ehild]

The track of the bug, a parabola, touches the circle, cross-section of the log, where the tangents of the circle and the parabola are the same. It can happen at two points, or at a single point, at the top of the log. In case the common point is at the top of the circle, the second derivative of the curves must be also equal. It is quite easy to find the angle of the jump when the required initial speed is minimum.
Assuming the bug scrapes the log twice, the bug touches the log lower than 2R, but rises higher than 2R. How one can be sure that a jump at 45° scrapes the log? To find the parameters of the jump is not easy.

 

[haruspex]

case the common point is at the top of the circle, the second derivative of the curves must be also equal.

No, it (the radius of curvature of the parabola) must be greater than or equal to that of the log.

 

[ehild]

No, it (the radius of curvature of the parabola) must be greater than or equal to that of the log.

I meant, when the bug just scrapes the log.

 

[haruspex]

I meant, when the bug just scrapes the log.

Sure, but the curvatures do not have to be equal. It is only necessary that i) the parabola and circle are tangential at that point and ii) the curvature of the parabola there is greater than or equal to that of the circle.

 

[ehild]

Sure, but the curvatures do not have to be equal. It is only necessary that i) the parabola and circle are tangential at that point and ii) the curvature of the parabola there is greater than or equal to that of the circle.

The speed needed to reach the top of the circle decreases with increasing angle and it reaches the minimum value at 90°, But the parabola must be over the circle, and touching it at the top. That means a limit for the angle of the jump. We get the maximum angle (and minimum speed) when the second derivatives of the circle and parabola are equal at the top of the curves.

 

[haruspex]

The speed needed to reach the top of the circle decreases with increasing angle and it reaches the minimum value at 90°, But the parabola must be over the circle, and touching it at the top. That means a limit for the angle of the jump. We get the maximum angle (and minimum speed) when the second derivatives of the circle and parabola are equal at the top of the curves.

Quite so, but that is the key argument for solving the problem of minimisation. In the post I objected to (#15) you appeared to be saying that equality of curvatures was a direct consequence of just touching the top of the log and nowhere else. No mention of minimisation there.
In fact, the neatest approach is to consider the jump in reverse from the top of the log. What is the least energy required to jump from there and not hit the log?