Find Center of Mass of Homogeneous Semicircular Plate

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Homework Help Overview

The discussion revolves around finding the center of mass of a homogeneous semicircular plate and a cube box without a top. The subject area includes concepts from geometry and integration.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use integration to find the center of mass of the semicircular plate, questioning the need for multiple integrals due to changing radius. They also describe their approach to finding the center of mass of the cube box, expressing difficulty in calculating the vertical component.
  • Another participant provides a mathematical expression related to the center of mass calculation.
  • There is a request for hints regarding the second question about the cube box.

Discussion Status

The discussion is ongoing, with participants exploring different methods and raising questions about their approaches. Some guidance has been offered, but there is no explicit consensus on the solutions or methods being discussed.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the type of assistance they can receive. There are indications of confusion regarding the integration setup and the assumptions made in the calculations.

team31
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Find the center of mass of a homogeneous semicircular plate, let R be the radius of the circle.
I have think I need to use ingergration to slove this problem, I'm stuck with that R is also changing as if I'm cutting little slices, so does that mean i have to set up two intergra, one for R solve that, and put it into another intergra which is for Ycm?

another question is there is a cube box with no top lays on a coordinate system x,y,z, z is vertical. and the side is 40 cm, I can find xcm=20 and ycm=20 very easily, but i had a hard time to find z, and i used harder way to find z, which is tp set zcm=x, and calculate the weight between the top half of the box and bottom half of the box,
40(40-x)+2(40-x)2=20*20+40x+40x.
from there I got Zcm=17.5, and that is not even the right answer, can u guys suggest a better and easier way for me to slove it?
 
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[tex]X=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{xdm}{M}[/tex]
 
huh, i got it, 4r/3pi is my answer, is that right?
 
could anybody give me a hint on my question #2, please
 

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