Center of mass, moment of inertia for x = y^2 and y = x - 2 with density d = 3x

In summary, the problem involves finding the center of mass, moment of inertia, and radius of gyration of a region bounded by a parabola and line. Using the given density, the center of mass is found to be at \bar{x} = \frac{235}{112} and \bar{y} = \frac{25}{32}. The moment of inertia is calculated to be I_{y} = 110.7 and the radius of gyration R_{y} \approx 2.26. These calculations have been verified by Maple.
  • #1
VinnyCee
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Here is the problem:

A region is defined as being bounded by the parabola [tex]x = y^2[/tex] and the line [tex]y = x - 2[/tex].

The density of this region is [tex]\delta = 3x[/tex].

a) Find the center of mass.

b) Find the moment of inertia about the y-axis.

c) Find the radius of gyration about the y-axis.

Here is what I have:

[tex]M = \int_{-1}^{2}\int_{y^2}^{y + 2}\;3x\;dx\;dy = \frac{108}{5}[/tex]

[tex]M_{x} = \int_{-1}^{2}\int_{y^2}^{y + 2} 3xy\;dx\;dy = \frac{135}{8}[/tex]

[tex]M_{y} = \int_{-1}^{2}\int_{y^2}^{y + 2} 3x^2\;dx\;dy = \frac{1269}{28}[/tex]

[tex]\bar{x} = \frac{\frac{1269}{28}}{\frac{108}{5}} = \frac{235}{112}\;\;and\;\;\bar{y} = \frac{\frac{135}{8}}{\frac{108}{5}} = \frac{25}{32}[/tex]

[tex]I_{y} = \int_{-1}^{2}\int_{y^2}^{y + 2} x^2 \left(3x\right)\;dx\;dy = 110.7[/tex]

[tex]R_{y} = \sqrt{\frac{110.7}{\frac{108}{5}}} \approx 2.26[/tex]

Does this look correct?
 

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  • #2
I get what you do for M but 153/8 rather than 135/8 forMx. I get completely different answer for the others- but I may have tried to do them too fast.
 
  • #3
I don't know what Halls did,but all your calculations are perfect.I double checked them with Maple...


Daniel.
 
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FAQ: Center of mass, moment of inertia for x = y^2 and y = x - 2 with density d = 3x

What is the formula for calculating the center of mass for a system with density d = 3x?

The formula for calculating the center of mass for a system with density d = 3x is x̅ = ∫xdm / ∫dm, where x̅ represents the position of the center of mass, m represents the mass element, and the integration is taken over the entire system.

How do you calculate the moment of inertia for x = y^2 and y = x - 2 with density d = 3x?

The moment of inertia for a system with density d = 3x can be calculated using the formula I = ∫r^2dm, where r is the distance from the axis of rotation to the mass element and the integration is taken over the entire system. For the given equations x = y^2 and y = x - 2, the moment of inertia can be calculated by substituting the equations into the formula and solving the resulting integral.

What is the physical interpretation of the center of mass for a system with non-uniform density?

The center of mass for a system with non-uniform density represents the point where the entire mass of the system can be considered to be concentrated. This point is useful for analyzing the overall motion of the system and can be used to simplify complex systems by treating them as a single point mass.

How does the center of mass change as the density of a system varies?

The center of mass of a system changes as the density varies because the position of the center of mass is dependent on the distribution of mass within the system. As the density changes, the distribution of mass also changes, resulting in a different position for the center of mass.

Can the center of mass and moment of inertia be calculated for three-dimensional objects?

Yes, the center of mass and moment of inertia can be calculated for three-dimensional objects using similar formulas. The center of mass is calculated by finding the weighted average of the x, y, and z coordinates, while the moment of inertia is calculated by integrating over all three dimensions. However, the calculations may become more complex for three-dimensional objects compared to two-dimensional objects.

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