# Minimum velocity required by grasshopper

1. Sep 23, 2016

### Googlu02

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.
3. 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.

2. Sep 23, 2016

### Staff: Mentor

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?

3. Sep 23, 2016

### drvrm

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...

4. Sep 23, 2016

### Googlu02

The initial position of the grasshopper is at ground.

5. Sep 23, 2016

### Staff: Mentor

The initial position is one of the variables in the optimization equation...
Now show us some of the equations for the parabolic path of the grasshopper, and how they depend on Vo...

6. Sep 24, 2016

### ehild

"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.

7. Sep 24, 2016

### 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.

8. Sep 26, 2016

### 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.

9. Sep 26, 2016

### drvrm

pl. show your path as a projectile....

10. Sep 26, 2016

### haruspex

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?

11. Sep 26, 2016

### 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: Vy=|V|⋅sin(45°), where Vy 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?

Last edited: Sep 26, 2016
12. Sep 26, 2016

### haruspex

There is no basis for assuming that.

13. Sep 26, 2016

### dykuma

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.

14. Sep 26, 2016

### haruspex

Maybe, but you do not know that yet.
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.

15. Sep 27, 2016

### 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.

Last edited: Sep 27, 2016
16. Sep 27, 2016

### haruspex

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

17. Sep 27, 2016

### ehild

I meant, when the bug just scrapes the log.

18. Sep 27, 2016

### haruspex

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.

19. Sep 27, 2016

### ehild

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.

20. Sep 27, 2016

### haruspex

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?