Jump Over Log with Least Effort: Analyzing Velocity vs Angle

[tempneff]

1.A lazy flea approaches a stationary log with radius R. At what distance from the log must the flea be to jump over the log with the least amount of effort. Hint: Plot Velocity versus the angle.



2.(Not sure how I should type these) I plan to use constant acceleration projectile motion and it seems that work/energy might play a part.



3. To make use of the hint I want try and plot velocity and /theta. I am having trouble understanding how to plot this without values. I guess I can start at an angle of pi halves and work down from there. However, I don't understand how this will yield me any definite values.

 

[willem2]

It will be easier to have the flea jump off the log horizontally, so it just doesn't intersect the log, and then see where it hits the ground.

 

[zgozvrm]

I'm pretty sure the intent of this problem is to assume that the flea's trajectory won't intersect the circumference of the log. If that's so, perhaps it would be better to think of the obstruction as a small "wall" instead.

 

[zgozvrm]

It will be easier to have the flea jump off the log horizontally, so it just doesn't intersect the log, and then see where it hits the ground.

I don't know how this would help, since the flea can jump off the log horizontally at any speed.

 

[willem2]

I'm pretty sure the intent of this problem is to assume that the flea's trajectory won't intersect the circumference of the log. If that's so, perhaps it would be better to think of the obstruction as a small "wall" instead.

But in that case, the flea should jump up from right below the lop to use the least amount of energy from the jump, and the problem would be trivial.

 

[zgozvrm]

But in that case, the flea should jump up from right below the lop to use the least amount of energy from the jump, and the problem would be trivial.

This is a trajectory problem...
If the flea jumps straight up, he comes straight down (unless he's lucky enough to catch a wind current).

 

[zgozvrm]

After thinking it through, I retract my statement referring to the log being thought of as a "wall"

If that were so, it would be almost trivial. The angle of trajectory resulting in the least effort by the flea would then be just less than [itex]90^\circ[/tex] but it would have to be sufficient enough to clear the thickness of the wall.

So, the problem is complicated by the curvature of the log which restricts the range of angles the flea can jump in order to clear it (of course, we are to assume the log's cross-section is perfectly circular).

 

[tempneff]

Can anyone offer advice as to how to solve this problem via graphing velocity vs angle??

 

[sungod]

I don't know if this helps, but this seems to be a trajectory problem with a parabolic trajectory whose vertex is the top of the log. In this case, any trajectory whose highest point is 2R would work.

If it takes off at an angle theta with a velocity v and it takes time t for the entire trajectory (from take-off to landing), then:

0 = v sin(theta)t - (1/2)gt^2 => t = 2 v sin(theta)/g

So the time to reachthe highest point is

t<highest point> = v sin(theta)/g

Thus,
2R = v^2 (sin(theta))^2/g - (1/2)gt^2

However, the point of take-off can be as far away from the log as the flea chooses. If the point of take-off from the log is s, then:
s = v^2 cos(theta) sin(theta)/g = v^2 sin(2 theta)/2g

But, the vertical impulse of the flea on take-off (the change in momentum upwards) is
m v sin(theta). So, presumably for the "least effort", you need to minimise theta. The smaller \theta is, the lower is the impulse.

As for the graph, you can always draw a whole bunch of parabolas such that the vertices are the top of the log. Without any further information, I can't see which is the one with the "least effort". It seems that you have to minimise v sin(theta) while maximising v cos(theta)

 

[zgozvrm]

I don't know if this helps, but this seems to be a trajectory problem with a parabolic trajectory whose vertex is the top of the log. In this case, any trajectory whose highest point is 2R would work.

I haven't proven it yet, but I'm pretty sure that any parabolic trajectory having a vertex of 2R would result in the flea "hitting" the log, unless he jumps from a very long distance. I believe he will have to jump in a (parabolic) path such that the vertex of the trajectory is greater than 2R in order to 1) avoid hitting the log, and 2) minimize the effort of his jump.

 

[sungod]

I see what you mean.

In that case, perhaps there is only one parabola whose vertex is the top of the log. Any other parabola is either too high or too low. So, (looking at it sideways), you draw the line at x = 2R, you draw the circle (x-R)^2 + y^2 = R^2. The y-axis is one tangent to the circle and the line x=2R is another . Then you find a parabola with the origin as vertex, with the y-axis as a common tangent to the circle and cutting the line x=R at two points +/- s. This parabola is unique and the distance away from the log would be s.

I don't think the top of the trajectory has to be greater than 2R since the flea must j-u-u-st clear it. 2R + epsilon would be the minimum clearance, with epsilon being the distance between it's centre of gravity and it's longest leg

 

[tempneff]

Now that we established the unique distance s above 2R. I need to plot (per the assignment) velocity vs angle to determine which angle, and subsequent distance between flea and log, will yield the smallest velocity. I have completely blown a brain gasket and cannot understand how one might develop such a graph.

 

[zgozvrm]

...there is only one parabola whose vertex is the top of the log.

This is incorrect...

There are an infinite number of parabolas that can be found that reach a maximum height of 2R (the top of the log), all of which begin at a distance greater than, or equal to 2R units from the center of the log. Any trajectory that reaches a maximum height of 2R units leaving the ground a distance of less than 2R units from the log will result in the flea hitting the log instead of going over.

I don't think the top of the trajectory has to be greater than 2R since the flea must j-u-u-st clear it.

Again, incorrect...

There is no requirement that the flea must just clear the log. The requirement is that the flea clears the log with the least amount of effort. This means that we are looking for the lowest initial velocity the flea can take, and still clear the log. This does, in fact, mean that the flea will be jumping higher than the height of the log.
An example (see attachment):
A log of radius 4 meters (this is a large log) can be just cleared by the flea taking off from a distance of 8 meters from the center of the log at a velocity of 14 m/sec at approximately 63.435 degrees. (Assuming, of course, that the flea can reach that velocity).
With these values, the path of the flea will intersect the circumference of the log only at the vertex (peak height) of his trajectory. I believe it can be assumed that a height of 8 meters is enough for the flea to "clear" the log.

But, if the flea leaps from only a distance of 6.6 meters away, he can successfully clear the log by jumping with a velocity of 13.846 m/sec at 68.782 degrees. So, it actually takes less energy for the flea to jump higher than necessary, in this case.
Note that this is only an example; 6.6 meters does NOT necessarily result in the least effort required by the flea to clear a log with a 4 meter radius.

 

[zgozvrm]

Now that we established the unique distance s above 2R. I need to plot (per the assignment) velocity vs angle to determine which angle, and subsequent distance between flea and log, will yield the smallest velocity. I have completely blown a brain gasket and cannot understand how one might develop such a graph.

I'm with you on this. I'm not sure how graphing alone will help you find the distance and angle of trajectory.

 

[tempneff]

I have received more guidance on this problem but still require assistance.

Because of symmetry only half of the motion need be considered.

One can solve for the angle at which the projectile becomes tangent to the circle by using constant acceleration equations and the fact that at the top of the motion velocity in the y direction is zero.

I don't know why I am so clueless on this problem, but I can't seem to wrap my mind around it. Maybe it has to do with my phobia of parasites?!

 

[zgozvrm]

At this point, I'm going to assume that the question is mis-stated; that we are looking for the minimum effort by the flea to "just clear" the log. (As I've shown already, the flea can use even less effort if he jumps from a closer distance, higher than necessary to clear the log.)

Even so, this is a fairly difficult problem, but it can be solved utilizing the kinematic equations (along with the formula of a circle).

As far as graphing to solve the problem ... is this a requirement? I can only assume that you are to graph "theoretical" values to help you visualize the problem. (It's not necessary to get to the solution).

 

[tempneff]

You are correct in that the graphing idea was only a tool. So i am no longer concerned with it. But the question remains the same. What is the least effort (min velocity) needed to get over the log, regardless of how. Again, you are correct that a jump that does not try to become tangent to the top of the circle has a lower energy. But, the least effort trajectory will likely still result in 2 tangent points on either side of the circle at some degree theta. If that degree is determined than I believe we can determine the height of that tangent point in relation to radius R. Next if we can determine the x distance of the initial distance of the flea, then I think I can use tan theta = y\x to determine the origin angle of the projectile and in turn the minimum velocity. I think...

 

[zgozvrm]

What level of math have you had (or are currently taking)?

 

[tempneff]

I am in my first Calculus class - single variable.

 

[zgozvrm]

Okay, so first we need to determine the minimum distance the flea can be from the (center) of the log and just clear it. This would be a trajectory with a maximum height of R. Call this distance D.

Obviously, the flea would have to be farther than R away from the log. If he jumps from R, he will be jumping straight up (at an angle of 90 degrees) and would come straight back down. At any distance less than R, the flea would have to jump backwards so as not to hit the log, and therefore wouldn't be able to jump over it.

So next, we need to determine the formula of any trajectory necessary to clear the log for all distances M, such that R < M <= D
These trajectories would, as you stated, be tangent to the log at 2 points (of equal height).

Finally, since each of those trajectories describes a parabolic arc, we need to determine which path has the smallest area beneath it and, hence, require the least amount of effort.


(I'm thinking the flea is going to be too mentally exhausted to even care about getting over this log by now...)

 

[tempneff]

Okay so here I am so far (attatched)

I think i can now find the derivative of eq.(5) then set it equal to zero to get a minimum which will be equal to the min. velocity. but i need help.

 

[Jack the Stri]

Have you tried the following:

Create a parametric equation of a parabola describing a trajectory that 'touches' the circle at just two points (that may or not be the same point, if it just skims the top of the log), but with its maximum being at least 2r. Using this, take the second derivative of the equation to find where the initial acceleration is minimal. That should allow you to plug a into the equation and find theta, if I'm not mistaken.

This is just off the top of my head and I'm not sure if it'll work, but it seems like the most logical (no pun intended) approach. Also, not sure if it's been posted already since my PDF reader isn't working properly.

edit: note to self: read page 2 before posting.

 

[dmriser]

Would it be wrong to set the equation which describes position in the y direction to 2r and then solve for t. This gives a time that you can plug into the position in the x direction equation which leads to a general equation which includes parameters such as v initial, theta, and gravity. Of course... This assumes that a jump of 2r will just clear the log.

This is what I got as a general case..

$$Distance = \frac{V_i^2 sin \theta cos \theta - 2rV_i cos\theta}{g}$$

 

[zgozvrm]

Would it be wrong to set the equation which describes position in the y direction to 2r and then solve for t.

Please read through the thread ... you'll see that it has already been determined that the trajectory requiring the least amount of effort by the flea (i.e. the lowest initial velocity) describes a parabola which more than "just clears" the log (that is, a height greater than 2r).

 

[tempneff]

SOLVED I attached my solution. Funny thing is for literally weeks I have been trying to solve for a variable which I was allowed to leave in the answer (in fact it is impossible otherwise) Guess I got what I deserved for not getting clarification.