How Is the Center of Gravity Calculated for a Non-Uniform Rod?

  • Thread starter decathlonist
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In summary, the conversation discusses a problem involving the center of gravity of a rod with a given length and linear mass density. The answer is found to be 2L/3 and the mathematical definition of center of gravity is also explained. Riemann sums and integrals are utilized to solve the problem.
  • #1
decathlonist
Hello,

I was studying for the test and I ran into the problem that I can't solve.

This problem involves the use of Calculus (in particular - integration). Although I do have a solution for the problem, I don't have any explanation for it. Can someone please tell me how we arrived to the final answer?

"A rod of length L and linear mass density l(x)=Ax kg/m, where A is a constant and x is measured in meters from one end of the rod. Assuming the rod is in a uniform gravitational field, determine the location of the center of the rod."

( center of gravity = 2L/3 )

Thank you
 
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  • #2
I presume you know what the mathematical definition of the centre of gravity is?
 
  • #3
Are you sure that the answer is 2L/3 ?
Because I think it's L*sqrt(2)/2 (the same as L/sqrt(2))
sqrt -> square root;
[?]
 
  • #4
As "sir-pinski" was pointing out, you can't do a problem involving center of gravity until you know what that means! (It's remarkable how many people don't seem to notice that you can't do a problem if you don't know the definitions!)
The "center of gravity" is the point at which the rod would balance: the torque (twisting force) on one side about that point is equal to (and opposite) the torque on the other. The torque about a point is itself defined as the force applied times the distance from the point.

Imagine dividing the rod into small lengths, each of length dx, and treat the weight as constant on each small length (this technique should be familiar to you). Take the left end of the bar as x=0 and let "a" be the distance to the center of gravity (I would prefer x with a bar over it but..). For each point on the bar, at distance x from the left end, the distance from the center of gravity is |x-a|. However, if we just use x-a (rather than the absolute value) that will also allow us to treat left and right of it as positive and negative. The mass of each little "piece" of the rod is its density times its length: l(x)dx. The torque about the center of gravity of a little piece of the rod is (x-a)l(x)dx. The total torque would be [sigma](x-a)l(x)dx and in order to be balanced must be 0: [sigma](x-a)l(x)dx = 0 which is the same as [sigma]xl(x)dx= a[sigma]l(x)dx.

These are Riemann sums and in the limit become the integral:

[integral]xl(x)dx= a [integral]l(x)dx.
(The integral on the right is, of course, the total weight of the rod.)

In this particular case l(x)= Ax so [integral]l(x)dx= [integral]Axdx= (1/2)Ax2 evaluated between 0 and L= (1/2)Al2.

[integral]xl(x)dx= [integral]Ax<sup>2</sup>dx= (1/3)Ax<sup>3</sup>
evaluated between 0 and L= (1/3)AL3.

We must have (1/3)A L3= a(1/2)A L2 so

a= (2/3)L as advertised.
 

Related to How Is the Center of Gravity Calculated for a Non-Uniform Rod?

1. What is the difference between physics and calculus?

Physics is the study of the natural world and its behavior, while calculus is a branch of mathematics that focuses on rates of change and accumulation. Physics uses calculus to describe and analyze the physical world.

2. How does calculus apply to physics?

Calculus is used in physics to describe and analyze physical phenomena such as motion, forces, and energy. It allows for precise calculations and predictions of these phenomena by using mathematical models and equations.

3. What are some common applications of calculus in physics?

Some common applications of calculus in physics include kinematics, which uses derivatives to describe motion, and Newton's laws, which use integrals to calculate forces and acceleration. Calculus is also used in electromagnetism, thermodynamics, and quantum mechanics.

4. Why is it important to understand calculus in physics?

Understanding calculus is crucial in physics because it allows for precise and accurate analysis and predictions of physical phenomena. It also provides a deeper understanding of the fundamental principles and laws governing the natural world.

5. Can calculus be used to solve real-world problems in physics?

Yes, calculus is essential for solving real-world problems in physics. It allows for the precise calculation of physical quantities and the prediction of future behavior. Many technological advancements, such as space exploration and engineering, rely on the application of calculus in physics.

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