Can't Solve This Rigid Solid Problem? Let's Figure It Out!

In summary: Thanks for the clarification!I think that ideal rigid bodies in pendulum motion follow a path that is a straight line because the center of mass of the system is located at the center of the path.
  • #1
Feynmanfan
129
0
I don't know how to solve this rigid solid problem

Let there be a rod hanging of a string fixed to a certain point. Why is not possible for both the string and the rod to be in the same direction all the time. (it is common sense that it's impossible but how can I prove it?)

I've tried to figure it out thinking that it is possible and look for a contradiction.

Thanks a lot!
 
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  • #2
Can you make your question clearer?
 
  • #3
I'm sorry for not explaining myself clearer. the string is attached to the rod at the center of one of its ends, no external forces are acting on it other then gravity. The thing is that the rod will start oscillating and I want to prove that it is impossible that both the rod and the string stay in the same direction (as if it were a single rod).

What properties of the rigid solid should I use to prove it is impossible?

Thanks again
 
  • #4
Well, it all depends on how you start it oscillating, of course (what is the initial position of the rod and string and so forth). But one thing that might be more intuitive to think about is this - how about if you had the rigid rod _above_ the string, what would the end of the string do as the rod swung to and fro?

A further hint: the tangential acceleration at every point along a pendulum system is the same (ma=mgsinx, so the ms cancel out). Rigid bodies swing all in one line because of internal attractive forces among their constituent atoms that "drag" each other along each other - what if the forces were weaker? If every point in the string-rod system was accelerating the same, would the system remain a straight line?

Actually, in what you're proposing, the observable effect would be miniscule because tension would work to keep the system fairly straight (assuming the rod was significantly heavier than the string). You _would_ observe a "kink" if you looked hard enough, though.

Btw, it'd probably be useful to think some more about why ideal rigid bodies fall as a straight line in pendulum motion - that is, why the center of mass concept works. (edit: ahh, which is not to imply it doesn't work for non-rigid bodies; what i meant to say is: why the center of mass concept is so useful for examining a physical pendulum)
 
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1. What is a rigid solid problem?

A rigid solid problem is a mathematical problem involving an object or system that cannot be deformed or changed in shape. This means that the distances between points on the object remain constant, regardless of any external forces acting upon it.

2. Why are rigid solid problems important?

Rigid solid problems are important because they help us understand and analyze real-world objects and systems. They allow us to predict how these objects will behave under different conditions, which is crucial for many fields such as engineering, physics, and architecture.

3. How do you approach solving a rigid solid problem?

The first step in solving a rigid solid problem is to clearly define the problem and identify all relevant variables. Then, you can use mathematical equations and principles, such as Newton's laws of motion, to analyze the problem and find a solution.

4. What are some common challenges when solving rigid solid problems?

Some common challenges when solving rigid solid problems include determining the appropriate equations to use, dealing with complex geometry, and accounting for all external forces and constraints. It is important to carefully consider all factors and double-check your calculations to ensure accuracy.

5. What are some tips for successfully solving rigid solid problems?

Some tips for successfully solving rigid solid problems include breaking the problem down into smaller, more manageable parts, using visual aids and diagrams to better understand the problem, and checking your work for errors. It may also be helpful to discuss the problem with others and seek guidance from a mentor or teacher.

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