Find the location of the CM of a hollow ice cream cone

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Homework Help Overview

The discussion revolves around finding the center of mass (CM) of a hollow ice cream cone with a specified base radius and height, as well as considering how the answer changes for a solid cone. Participants are exploring the implications of mass density and geometric properties in their calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using slices to determine the mass and center of mass, with one suggesting a formula involving density and area. Another participant presents a formula for the CM of the hollow cone and compares it to that of a solid cone. Questions arise regarding the complexity of the calculations and the meaning of variables used, such as H.

Discussion Status

The discussion is active, with participants sharing their attempts at deriving the center of mass. Some express uncertainty about the complexity of the formulas and the definitions of certain variables, indicating a need for clarification. There is no explicit consensus yet, but participants are engaging with the problem and offering insights.

Contextual Notes

There is mention of uniform mass density and the need for clarity on the variable H, which some participants question. The discussion reflects the constraints of homework rules, as participants are navigating their understanding without providing direct solutions.

Tonyt88
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Find the location of the CM of a hollow ice cream cone, with base radius R and height h, and uniform mass denisty. How does your answer change if the cone is solid, instead of hollow?

Okay, so I'm pretty sure that I need to work with slices, and that you need the mass which I believe is [where sigma is density]

σ ( (pi) r^2 + (pi) r √r2 + h2)

Though I don't know where to go from here
 
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Nm, I think I got it, I have:

[h^2]/[(3)(R+sqrt(R^2 + H^2))]

and for the filled cone:

h/4
 
\vec{R}_{CM}=\frac{\int_A\sigma\vec{r}dA}{\int_A\sigma dA}

Notice that the angle that the cone makes with its symetry axis is \tan\theta=R/h. I leave it to you to evaluate the integrals.
 
Tonyt88 said:
Nm, I think I got it, I have:

[h^2]/[(3)(R+sqrt(R^2 + H^2))]

and for the filled cone:

h/4
The first one looks far too complicated. What is H anyway? Is that where the CM is located? The filled cone looks good.
 

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