Exercises: Centers of Mass and Centroids

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Various questions relating to centers of mass and centroids.

Find the centroid of the region bounded above by $y = x$ and below by $y = x^2$ . $\overline {x} = \answer {\frac {1}{2}} \text { and } \overline {y} = \answer {\frac {2}{5}}.$

Find the centroid of the region bounded above by $y = 4-x^2$ and below by $y = 0$ . $\overline {x} = \answer {0} \text { and } \overline {y} = \answer {\frac {8}{5}}.$

A thin plate in the plane defined by $x^2 \leq y \leq 1$ and $x \geq 0$ has density $y$ at the point $(x,y)$ . Compute the center of mass.

Use $y$ as the variable of slicing. The center of mass of a single slice $(\tilde x, \tilde y)$ is then $(\sqrt {y}/2,y)$ .

$\overline {x} = \answer {\frac {1}{2}} \text { and } \overline {y} = \answer {\frac {5}{7}}.$

A thin plate in the plane defined by $x^2 \leq y \leq 2x^2$ and $0 \leq x \leq 1$ has density $x$ at the point $(x,y)$ . Compute the center of mass.

Use $x$ as the variable of slicing. The center of mass of a single slice $(\tilde x, \tilde y)$ is then $(x,3x^2/2)$ .

$\overline {x} = \answer {\frac {4}{5}} \text { and } \overline {y} = \answer {1}.$

The same thin plate as above ( $x^2 \leq y \leq 2x^2$ and $0 \leq x \leq 1$ ) now has density $x^{-2}$ at the point $(x,y)$ . Because the density of the plate is now higher near the origin than in the previous problem, this suggests that the center of mass will shift away fromtowards the origin relative to the previous exercise.

Compute the center of mass. $\overline {x} = \answer {\frac {1}{2}} \text { and } \overline {y} = \answer {\frac {1}{2}}.$

Compute the centroid of a thin wire along the graph $y = \sqrt {1-x^2}$ between $x = 0$ and $x=1$ .

Recall that $M = \int ds$ $\overline {x} = \frac {1}{M} \int x ds$ $\overline {y} = \frac {1}{M} \int y ds$ where $ds$ is the arc length element. We also know that $\int \frac {dx}{\sqrt {1-x^2}} = \arcsin x + C.$

$\overline {x} = \answer {\frac {2}{\pi }} \text { and } \overline {y} = \answer {\frac {2}{\pi }}.$

Sample Quiz Questions

Compute the centroid of the region bounded by the inequalities $-2 \leq x \leq 0 \qquad \text { and } \qquad {3x^2-\frac {13}{2}} \leq y \leq {3x^2-\frac {11}{2}}.$ (Hints won’t be revealed until after you choose a response.)

$\displaystyle \left (-\frac {3}{2},-3 \right )$ $\displaystyle \left (-1,-3 \right )$ $\displaystyle \left (-\frac {1}{2},-3 \right )$ $\displaystyle \left (-\frac {3}{2},-2 \right )$ $\displaystyle \left (-1,-2 \right )$ $\displaystyle \left (-\frac {1}{2},-2 \right )$

The key calculations are as follows: $\begin {aligned} M & = \int _{-2}^{0} \left [ \left ({3x^2-\frac {11}{2}}\right ) - \left ({3x^2-\frac {13}{2}}\right ) \right ] dx , \\ M_y & = \int _{-2}^{0} x \left [ \left ({3x^2-\frac {11}{2}}\right ) - \left ({3x^2-\frac {13}{2}}\right ) \right ] dx, \\ M_x & = \frac {1}{2} \int _{-2}^{0} \left [ \left ({3x^2-\frac {11}{2}}\right )^2 - \left ({3x^2-\frac {13}{2}}\right )^2 \right ] dx. \end {aligned}$

$\begin {aligned} M & = \int _{-2}^{0} \left [ \left ({3x^2-\frac {11}{2}}\right ) - \left ({3x^2-\frac {13}{2}}\right ) \right ] dx = \int _{-2}^{0} \left [{1}\right ] dx = 2, \\ M_y & = \int _{-2}^{0} x \left [ \left ({3x^2-\frac {11}{2}}\right ) - \left ({3x^2-\frac {13}{2}}\right ) \right ] dx = \int _{-2}^{0} x \left [{1}\right ] dx = -2, \\ M_x & = \frac {1}{2} \int _{-2}^{0} \left [ \left ({3x^2-\frac {11}{2}}\right )^2 - \left ({3x^2-\frac {13}{2}}\right )^2 \right ] dx = \int _{-2}^{0} \left [{3x^2-6}\right ] = -4, \\ \overline {x} & = \frac {M_y}{M} = -1, \\ \overline {y} & = \frac {M_x}{M} = -2.\end {aligned}$

Sample Exam Questions

Find the $y$ -coordinate of the centroid of the region bounded by the $x$ -axis, the $y$ -axis, and the graph of $y = \cos x$ for $0 \leq x \leq \pi /2$ if the density is constant.

Use the identity $\cos ^2 x = \frac {1 + \cos 2x}{2}$ to calculate the integral of $\cos ^2 x$ .

$\displaystyle \frac {\pi }{18}$ $\displaystyle \frac {\pi }{12}$ $\displaystyle \frac {\pi }{8}$ $\displaystyle \frac {\pi }{6}$ $\displaystyle \frac {\pi }{4}$ $\displaystyle \frac {\pi }{2}$

The area of the region is given by $M = \int _0^{\frac {\pi }{2}} \cos x~dx = 1$ and $\begin {aligned} M_x & = \int _0^{\frac {\pi }{2}} \frac {0 + \cos x}{2} \cos x~ dx = \frac {1}{2} \int _0^{\frac {\pi }{2}} \cos ^2 x ~ dx \\ & \frac {1}{2} \int _0^{\frac {\pi }{2}} \frac {1 + \cos 2x}{2} dx = \frac {\pi }{8}. \end {aligned}$ Therefore $\overline {y} = M_x / M = \pi /8$ .

Find the $y$ -coordinate of the centroid of the region in the upper half-plane (i.e., for $y > 0$ ) bounded by the semicircle $y = \sqrt {1-x^2}$ . (It is easiest to use a geometric formula to find the area of the region.)

$\displaystyle \frac {4 \pi }{3}$ $\displaystyle \frac {4}{3 \pi }$ $\displaystyle \frac {7 \pi }{3}$ $\displaystyle \frac {7}{3 \pi }$ $\displaystyle \frac {28 \pi }{9}$ $\displaystyle \frac {28}{9 \pi }$