## 17Calculus Physics - Moments, Center of Mass, and Centroids of a Planar Lamina

##### 17Calculus

Moments, Center of Mass and Centroid of a Planar Lamina

Moment, Center of Mass, Center of Gravity and Centroid of a Planar Lamina

For a Region (Thin Plate, Planar Lamina)

 The center of mass a region is the point in which the region will be perfectly balanced horizontally if suspended from that point. [src-Paul's Online Notes]   The center of mass is affected by it's shape, it's density and gravity. The center of gravity is the same as the center of mass if gravity is the same everywhere in the region.   This is so common that many people will use the terms center of mass and center of gravity interchangably. The centroid is the same as center of mass if the density is the same everywhere in the region.   Again, this is extremely common, so some people just talk about the centroid.

Before we go on, let's watch a video discussing the center of gravity for a plate-like object.   We usually call this type of object a planar lamina.

### Michel vanBiezen - Mechanical Engineering: Centroids & Center of Gravity - What is Center of Gravity?

video by Michel vanBiezen

Here is a related video discussing centroid and center of gravity.   His figure actually has a thickness but since it is the same thickness everywhere, it doesn't come into play.

### Michel vanBiezen - Mechanical Engineering: Centroids & Center of Gravity - Centroids

video by Michel vanBiezen

Okay, now we are ready to discuss the moments of a plane area and a wire or line.   Here is a great video.

### Michel vanBiezen - Mechanical Engineering: Centroids & Center of Gravity - 1st Moments of Areas and Lines

video by Michel vanBiezen

Table of Equations

The moment and center of mass equations for a planar lamina are given in Table 1.   In this table, ρ is the density.

Table 1 - Planar Lamina Moments and Center of Mass Equations

$$\displaystyle{M_x=\rho\int_{a}^{b}{(1/2)[f(x)+g(x)][f(x)-g(x)]~dx}}$$

$$\displaystyle{M_y=\rho\int_{a}^{b}{x[f(x)-g(x)]~dx}}$$

total mass

$$\displaystyle{m=\rho\int_{a}^{b}{f(x)-g(x)~dx}}$$

center of mass $$\bar{x},\bar{y}$$

$$\displaystyle{\bar{x}=\frac{M_y}{m}, \bar{y}=\frac{M_x}{m}}$$

$$f(x) \geq g(x), [a,b]$$; $$\rho$$ is the planar mass density and is a constant in these equations

Here is a short video explaining how to derive these equations.

### David Lippman - Deriving center of mass equations for a lamina [6min-1sec]

video by David Lippman

Okay, with all that information you should be able to succesfully work these practice problems.

Practice

Unless otherwise instructed, solve these problems giving your answers in exact, simplified form.

Find the centroid of the region bounded by $$y = \sqrt{x}$$, $$y = 0$$ and $$x = 4$$.   Also, find the moments $$M_x$$ and $$M_y$$ for the planar laminar of uniform density ρ.

Problem Statement

Find the centroid of the region bounded by $$y = \sqrt{x}$$, $$y = 0$$ and $$x = 4$$.   Also, find the moments $$M_x$$ and $$M_y$$ for the planar laminar of uniform density ρ.

$$M_x = 4\rho$$, $$M_y = 64\rho/5$$, centroid = $$(12/5, 3/4)$$

Problem Statement

Find the centroid of the region bounded by $$y = \sqrt{x}$$, $$y = 0$$ and $$x = 4$$.   Also, find the moments $$M_x$$ and $$M_y$$ for the planar laminar of uniform density ρ.

Solution

### The Organic Chemistry Tutor - 2445 video solution

$$M_x = 4\rho$$, $$M_y = 64\rho/5$$, centroid = $$(12/5, 3/4)$$

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Find the centroid of the region bounded by $$y = \sqrt{x}$$ and $$y = x$$.

Problem Statement

Find the centroid of the region bounded by $$y = \sqrt{x}$$ and $$y = x$$.

Solution

### The Organic Chemistry Tutor - 3595 video solution

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Find the centroid of the region bounded by $$y = 4-x^2$$ and $$y = x+2$$.

Problem Statement

Find the centroid of the region bounded by $$y = 4-x^2$$ and $$y = x+2$$.

Solution

### The Organic Chemistry Tutor - 3596 video solution

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Find the centroid of the region bounded by $$y = x^2$$ and $$y = 8-x^2$$.

Problem Statement

Find the centroid of the region bounded by $$y = x^2$$ and $$y = 8-x^2$$.

Solution

This problem is solved in two consecutive videos.

### PatrickJMT - 3566 video solution

video by PatrickJMT

### PatrickJMT - 3566 video solution

video by PatrickJMT

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Find the centroid of the region bounded by $$y = 1-x^2$$ in the first quadrant.

Problem Statement

Find the centroid of the region bounded by $$y = 1-x^2$$ in the first quadrant.

Solution

### blackpenredpen - 3567 video solution

video by blackpenredpen

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Find the center of gravity of part of a circle that lies in the first quadrant with radius $$R$$.

Problem Statement

Find the center of gravity of part of a circle that lies in the first quadrant with radius $$R$$.

Hint

Because of the symmetry of the area, $$\bar{x} = \bar{y}$$.

Problem Statement

Find the center of gravity of part of a circle that lies in the first quadrant with radius $$R$$.

Hint

Because of the symmetry of the area, $$\bar{x} = \bar{y}$$.

Solution

### Michel vanBiezen - 3568 video solution

video by Michel vanBiezen

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Find the center of gravity of a semi-circle that lies above the x-axis with radius $$R$$.

Problem Statement

Find the center of gravity of a semi-circle that lies above the x-axis with radius $$R$$.

Hint

Because of the symmetry of the area, $$\bar{x} = 0$$.

Problem Statement

Find the center of gravity of a semi-circle that lies above the x-axis with radius $$R$$.

Hint

Because of the symmetry of the area, $$\bar{x} = 0$$.

Solution

### Michel vanBiezen - 3569 video solution

video by Michel vanBiezen

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Find the center of gravity of a triangle with height $$h$$ and width $$b$$. Draw the triangle so that the base is on the x-axis and y-axis cuts the triangle in half, going through the third point, which means that the other two sides are equal.

Problem Statement

Find the center of gravity of a triangle with height $$h$$ and width $$b$$. Draw the triangle so that the base is on the x-axis and y-axis cuts the triangle in half, going through the third point, which means that the other two sides are equal.

Hint

Because of the symmetry of the area, $$\bar{x} = 0$$.

Problem Statement

Find the center of gravity of a triangle with height $$h$$ and width $$b$$. Draw the triangle so that the base is on the x-axis and y-axis cuts the triangle in half, going through the third point, which means that the other two sides are equal.

Hint

Because of the symmetry of the area, $$\bar{x} = 0$$.

Solution

### Michel vanBiezen - 3570 video solution

video by Michel vanBiezen

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Find the center of gravity of the top half of an ellipse, centered at the origin. The equation of the ellipse is $$\displaystyle{ \frac{x^2}{a} + \frac{x^2}{b} = 1 }$$ with $$a$$ along the x-axis.

Problem Statement

Find the center of gravity of the top half of an ellipse, centered at the origin. The equation of the ellipse is $$\displaystyle{ \frac{x^2}{a} + \frac{x^2}{b} = 1 }$$ with $$a$$ along the x-axis.

Solution

### Michel vanBiezen - 3571 video solution

video by Michel vanBiezen

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Find the center of gravity of the parabolic area $$y = kx^2$$ with width $$2a$$ and height $$h$$.

Problem Statement

Find the center of gravity of the parabolic area $$y = kx^2$$ with width $$2a$$ and height $$h$$.

Hint

$$k = h/a^2$$

Problem Statement

Find the center of gravity of the parabolic area $$y = kx^2$$ with width $$2a$$ and height $$h$$.

Hint

$$k = h/a^2$$

Solution

### Michel vanBiezen - 3572 video solution

video by Michel vanBiezen

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Find the center of gravity of the area bounded by $$y = kx^2$$, $$y = 0$$ and $$x = a$$.

Problem Statement

Find the center of gravity of the area bounded by $$y = kx^2$$, $$y = 0$$ and $$x = a$$.

Hint

In this video solution, he labels the height of the area $$h$$ and gives his answer in terms of $$h$$. However, $$h=ka^2$$.

Problem Statement

Find the center of gravity of the area bounded by $$y = kx^2$$, $$y = 0$$ and $$x = a$$.

Hint

In this video solution, he labels the height of the area $$h$$ and gives his answer in terms of $$h$$. However, $$h=ka^2$$.

Solution

### Michel vanBiezen - 3573 video solution

video by Michel vanBiezen

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Find the center of gravity of the area bounded by $$y = kx^n$$, $$y = 0$$ and $$x = a$$.

Problem Statement

Find the center of gravity of the area bounded by $$y = kx^n$$, $$y = 0$$ and $$x = a$$.

Hint

In this video solution, he labels the height of the area $$h$$ and gives his answer in terms of $$h$$. However, $$h=ka^n$$.

Problem Statement

Find the center of gravity of the area bounded by $$y = kx^n$$, $$y = 0$$ and $$x = a$$.

Hint

In this video solution, he labels the height of the area $$h$$ and gives his answer in terms of $$h$$. However, $$h=ka^n$$.

Solution

This problem is solved in two consecutive videos.

### Michel vanBiezen - 3574 video solution

video by Michel vanBiezen

### Michel vanBiezen - 3574 video solution

video by Michel vanBiezen

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Find the center of gravity of a section of a circle with radius $$R$$ and angle $$2\alpha$$. Center the circle so that the origin is at the center and the positive x-axis cuts the sector into two equal sections, $$\alpha$$ above the x-axis and $$\alpha$$ below the x-axis.

Problem Statement

Find the center of gravity of a section of a circle with radius $$R$$ and angle $$2\alpha$$. Center the circle so that the origin is at the center and the positive x-axis cuts the sector into two equal sections, $$\alpha$$ above the x-axis and $$\alpha$$ below the x-axis.

Hint

Because of symmetry, $$\bar{y} = 0$$

Problem Statement

Find the center of gravity of a section of a circle with radius $$R$$ and angle $$2\alpha$$. Center the circle so that the origin is at the center and the positive x-axis cuts the sector into two equal sections, $$\alpha$$ above the x-axis and $$\alpha$$ below the x-axis.

Hint

Because of symmetry, $$\bar{y} = 0$$

Solution

### Michel vanBiezen - 3575 video solution

video by Michel vanBiezen

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Find the center of mass of the region bounded by the triangle in the figure. Problem Statement

Find the center of mass of the region bounded by the triangle in the figure. Solution

This problem is solved in two consecutive videos.

### patrickJMTPhysics - 3597 video solution

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