Hypothetical kinetics questions, just for fun

In summary, Spider-Man needs to constantly pull on the web in order to maintain a horizontal path, but if the strand is elastic, he can swing horizontally. The equation describing this is not too difficult to solve.
  • #1
Onus14
2
0
Some imagination is required, so let me dress up the question a bit.

Imagine Spider-Man is on a rooftop, and he shoots a web to the top edge of a nearby building that's taller than the one he's on, in an attempt to swing toward it. Assume that the web is non-elastic and is completely adhered to the face of the building, and if he jumps off the rooftop he will naturally move in a perfect pendulum motion (ignore air resistance) A. His goal, however, is to move toward his target building following a path that is perfectly horizontal to the ground (ignore his reason) B. Keep in mind that he is able to produce a force that draws the web back into his hand, thus giving him the ability to pull himself upward if the end of the web is already attached to a surface.

http://img706.imageshack.us/img706/5889/physics.jpg"

So my question is, how much force will he need to produce from his web-hand in order to maintain the perfect horizontal path? I posit that the force will need to be dynamic/changing at each point of the fall, but can the amount of force necessary to maintain the horizontal path be elegantly described in an equation that accounts for all the variables (like building distance, time taken to travel, etc)? I gave up trying once I guessed that calculus would be involved...

If anyone is up for solving this, go for it; I'd be very interested in the solution and your thought process.


Good to be here,
~ Onus
 
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  • #2
Welcome Onus.

Onus14 said:
Imagine Spider-Man is on a rooftop, and he shoots a web to the top edge of a nearby building that's taller than the one he's on, in an attempt to swing toward it. Assume that the web is non-elastic and is completely adhered to the face of the building, and if he jumps off the rooftop he will naturally move in a perfect pendulum motion (ignore air resistance) A.

It won't be a pendulum motion (unless the strand is attached right ahead & he swings forthright). It will be a superposition of two pendulum motions ;a non-planar trajectory.
If you want the strand to be of constant length , a horizontal path is impossible to maintain. If you allow elasticity, a horizontal path is possible. The calculation is not really something to funk at .
Consider the horizontal plane in which the path lies.Assuming that the strand is taut all the times , the tension T points towards the fixed point of adherence of the strand.
The vertical component of the tension must be mg = T sin a (m =mass of the spider & a is the angle of elevation of the fixed point from his current position).Now, the magnitude of the acceleration must be (T/m)cos a . Use polar coordinates & a simple differential equation follows.
 
  • #3
In a regular pendulum motion(the A path), the tension in the web would depend on the angle of elevation at that point in time. So you end up with T = W sin(a), where T is the tension and W is the weight of spiderman, a is angle of elevation.

Now to travel in a straight path(B path), the total net force vertically must be equal to zero. So we consider all the forces acting vertically, W, and the new tension since spiderman is pulling downwards, the new tension is greater than before(Newton's 3rd law and all), let's call it T'.

Now what's the value to T'? We know T' would simply be the old tension, + the added force which spiderman pulls down with, causing the web to pull back with a greater amount as well. So T' = T + F where F is the force which spiderman uses to retract his web.

Now we have the vertical components which must equal to zero.
So W - T' sin(a) = 0,
=> W - T sin(a) - F sin (a) = 0
=> W - W sin^2(a) - F sin(a) = 0
=> F = W(1 - sin^2(a))/sin(a)
F = W cos^2(a)/sin(a)

So we have the force he must pull his web in at any angle a. Now how do we find a in terms of t(time)? We have to consider the horizontal forces. Now the net horizontal force would be equal to T' cos(a) = F cos(a) + T cos(a).

Now letting the horizontal distance of spiderman from the taller building be x, the difference in height be h, we can find cos(a) in terms of h and x.

Using Pythagoras's Theorem, we find cos(a) = d/root(d^2+h^2), sin(a) = h/root(d^2+h^2)

Substituting all this into the net force yields ma = W(x^3+hx)/(hx^2+h^3) which simplifies to:
d2x/dt^2 = g(x^3+hx)/(hx^2+h^3), where g = 9.81, the acceleration due to Earth's gravity at the surface, a rather evil looking second order differential equation.
Solve it for x in terms of t, plug it into F and you get F in terms of t, h and x.

Oh yes another question you might want to consider is how much length of web must spiderman retract per second to obtain this motion, that is, the rate at which web is retracted.
 
  • #4
Thanks heaps, Repainted, for your easy-to-follow solution. :))
 
  • #5


I find this hypothetical scenario to be quite intriguing. To answer your question, we would need to use the principles of kinetics, specifically the laws of motion and energy conservation. Let's break down the problem and see if we can come up with a solution.

First, we need to determine the initial conditions. Spider-Man is on a rooftop and shoots his web to the top edge of a taller building. Let's assume that he releases the web at a height of 10 meters above ground level and that the distance to the building is 20 meters. We also know that the web is non-elastic and completely adhered to the building, and that Spider-Man is able to produce a force to pull himself upward.

Next, we need to consider the forces acting on Spider-Man during his swing. The only external force acting on him is gravity, which is always directed towards the center of the Earth. However, as he swings, there are also internal forces acting on him - the tension in the web and the force he produces with his web-hand. The key to maintaining a horizontal path is to balance these forces.

Now, let's consider the energy conservation principle. As Spider-Man swings, he is converting potential energy (due to his height) into kinetic energy (due to his motion). At the highest point of his swing, all of his potential energy is converted into kinetic energy. At this point, the tension in the web is equal to his weight, and the horizontal component of this tension is what keeps him moving in a horizontal path.

To maintain a horizontal path, the tension in the web must always be equal to his weight. This means that the force he produces with his web-hand must also be equal to his weight. Using Newton's second law (F=ma), we can calculate the force needed as:

F = m x g

Where m is Spider-Man's mass and g is the acceleration due to gravity (9.8 m/s^2). Since we are assuming that Spider-Man is in a perfect pendulum motion, his mass and the acceleration due to gravity will remain constant throughout the swing.

In conclusion, the force Spider-Man needs to produce from his web-hand to maintain a perfect horizontal path is simply his weight multiplied by the acceleration due to gravity. This force will not need to change throughout the swing as long as the initial conditions and the distance to the building remain constant. I hope this answers your question and sheds some light on the role
 

1. What is the purpose of hypothetical kinetics questions?

Hypothetical kinetics questions are used to test a scientist's understanding of chemical reactions and the factors that affect their rates. These questions often involve hypothetical scenarios that require the application of kinetic principles to solve.

2. How are hypothetical kinetics questions different from standard kinetics questions?

Hypothetical kinetics questions are usually more complex and require more critical thinking than standard kinetics questions. They often involve multiple variables and require a deeper understanding of the underlying concepts.

3. Can hypothetical kinetics questions be solved without any prior knowledge of kinetics?

No, a basic understanding of kinetics principles is necessary to solve hypothetical kinetics questions. These questions often build upon the concepts and principles learned in standard kinetics problems.

4. Are hypothetical kinetics questions relevant to real-world applications?

Yes, hypothetical kinetics questions are designed to test a scientist's ability to apply kinetics principles to real-world scenarios. These questions are often used in chemistry and chemical engineering courses to prepare students for practical applications in the field.

5. How can I improve my skills in solving hypothetical kinetics questions?

To improve your skills in solving hypothetical kinetics questions, it is important to have a strong understanding of kinetics principles and to practice solving a variety of problems. You can also work with a study group or seek help from a tutor to gain a better understanding of the concepts involved.

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