91Ó°ÊÓ

For the disk.$râ\copyright ½a$in.

The moment of inertia about the diameter is calculated as:

$\mathit{I}\mathbf{=}\mathbf{∫}{\mathit{y}}^{\mathbf{2}}\mathit{d}\mathit{m}$ …(1)

The area of the disk using integral is calculated as$\mathit{d}\mathit{A}\mathbf{=}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{a}}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{2}\mathbf{π}}\mathit{r}\mathit{d}\mathit{r}\mathit{d}\mathit{θ}$:

The centroid of one quadrant of the disk is calculated as,$\stackrel{\mathbf{¯}}{\mathbf{x}}\mathbf{=}\frac{\mathbf{∫}\mathbf{x}\mathbf{d}\mathbf{m}}{\mathbf{∫}\mathbf{d}\mathbf{m}}$

The circumference of the circle is calculated as:$\mathit{C}\mathbf{=}\mathbf{∫}\mathit{d}\mathit{s}\mathbf{=}{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{2}\mathbf{π}}\mathit{r}\mathit{d}\mathit{θ}$

The centroid of the quarter arc circle is calculated as,

$\stackrel{\mathbf{¯}}{\mathbf{x}}\mathbf{=}\frac{\mathbf{∫}\mathbf{x}\mathbf{d}\mathbf{m}}{\mathbf{∫}\mathbf{d}\mathbf{m}}$ …(2)

And,

$\stackrel{\mathbf{¯}}{y}\mathbf{=}\frac{\mathbf{∫}\mathit{y}\mathbf{dm}}{\mathbf{∫}\mathbf{dm}}$ …(3)

(a)

The area of the disk using integral is calculated as,

$$\begin{gathered} dA={∫}_0^a{∫}_0^{2\mathrm{π}}rdrd\mathrm{θ} \\ ={∫}_0^{a}r[\mathrm{θ}{]}_0^{2\mathrm{π}}dr \\ =\left[\frac{r^2}{2}\right](2\mathrm{π}) \\ =\frac{a^2}{2}(2\mathrm{π}) \end{gathered}$$

Solve further,$A=\mathrm{Ï€}{a}^2$

Thus, the area of the disk is $\mathrm{Ï€}{a}^2$.

(b)

The centroid of one quadrant of the disk is calculated as,$\stackrel{¯}{x}=\frac{∫xdm}{∫dm}$

Rewrite the above expression as polar coordinates.

$$\begin{gathered} d\stackrel{¯}{x}=\frac{{∫}_0^a{∫}_0^{\mathrm{Ï€}/2}r\cos \mathrm{θ}\mathrm{ÒÏ}rdrd\mathrm{θ}}{{∫}_0^a{∫}_0^{\mathrm{Ï€}/2}\mathrm{ÒÏ}rdrd\mathrm{θ}} \\ =\frac{{\left[\frac{r^3}{3}\right]}_0^a[\sin \mathrm{θ}{]}_0^{\mathrm{Ï€}/2}}{{\left[\frac{r^2}{2}\right]}_0^a[\mathrm{θ}{]}_0^{\mathrm{Ï€}}} \\ =\frac{\left(\frac{a^3}{3}\right)\left(\sin \frac{\mathrm{Ï€}}{2}-\sin 0\right)}{\left(\frac{a^2}{2}\right)\left(\frac{\mathrm{Ï€}}{2}\right)} \\ =\frac{4a}{3\mathrm{Ï€}} \end{gathered}$$

Now, the centroid of one quadrant of the disk is calculated as,$\stackrel{¯}{y}=\frac{∫ydm}{∫dm}$

Rewrite the above expression as polar coordinates.

$$\begin{gathered} d\stackrel{¯}{y}=\frac{{∫}_0^a{∫}_0^{\mathrm{Ï€}/2}r\sin \mathrm{θ}\mathrm{ÒÏ}rdrd\mathrm{θ}}{{∫}_0^a{∫}_0^{\mathrm{Ï€}/2}\mathrm{ÒÏ}rdrd\mathrm{θ}} \\ =\frac{{\left[\frac{r^3}{3}\right]}_0^a[-\cos \mathrm{θ}{]}_0^{\mathrm{Ï€}/2}}{{\left[\frac{r^2}{2}\right]}_0^a[\mathrm{θ}{]}_0^{\mathrm{Ï€}}} \\ =\frac{\left(\frac{a^3}{3}\right)\left(-\cos \frac{\mathrm{Ï€}}{2}+\cos 0\right)}{\left(\frac{a^2}{2}\right)\left(\frac{\mathrm{Ï€}}{2}\right)} \\ =\frac{4a}{3\mathrm{Ï€}} \end{gathered}$$

Thus, the required centroid is$\left(\frac{4a}{3\mathrm{Ï€}},\frac{4a}{3\mathrm{Ï€}}\right)$ .

(c)

The moment of inertia about the diameter is calculated for:

$$\begin{gathered} \frac{dm}{dA}=\frac{M}{\mathrm{Ï€}{a}^2}dm \\ =\frac{M}{\mathrm{Ï€}{a}^2}dA \\ =\frac{M}{\mathrm{Ï€}{a}^2}dxdy \end{gathered}$$

Substitute $\frac{M}{\mathrm{Ï€}{a}^2}dxdy$for dm in equation (1).

$I=∫y^2\frac{M}{\mathrm{π}{a}^2}dxdy$

Rewrite the equation in polar coordinate form

$$\begin{gathered} I={∫}_0^a{∫}_0^{2\mathrm{π}}{r}^2{\sin }^2\mathrm{θ}\frac{M}{\mathrm{π}{a}^2}rdrd\mathrm{θ} \\ =\frac{M}{2\mathrm{π}{a}^2}{\left[\frac{r^4}{4}\right]}_0^a{\left[\mathrm{θ}-\frac{\sin 2\mathrm{θ}}{2}\right]}_0^{2\mathrm{π}} \\ =\frac{M}{2\mathrm{π}{a}^2}\left[\frac{a^4}{4}\right]\left[2\mathrm{π}-\frac{\sin 2\left(2\mathrm{π}\right)}{2}\right]-[0-0] \\ =\frac{M}{2\mathrm{π}{a}^2}\left[\frac{a^4}{4}\right]\left(2\mathrm{π}\right) \end{gathered}$$

Solve further,$I=\frac{M{a}^2}{4}$

Thus, the moment of inertia is $\frac{M{a}^2}{4}$.

(d)

The circumference of the circle is calculated as,

$$\begin{gathered} C=∫ds \\ ={∫}_0^{2\mathrm{π}}rd\mathrm{θ} \\ =r[\mathrm{θ}{]}_0^{2\mathrm{π}} \\ =2\mathrm{π}r \end{gathered}$$

Substitute for in the above expression:$C=2\mathrm{Ï€}a$

Thus, the circumference of the circle is $2\mathrm{Ï€}a$.

(e)

The centroid of the quarter arc circle is calculated as,

$\stackrel{¯}{x}=\frac{∫xdm}{∫dm}$ …â¶Ä¦(2)

And,

$\stackrel{¯}{y}=\frac{∫{\displaystyle y}dm}{∫dm}$ …â¶Ä¦(3)

Now,$dm=\mathrm{λ}rd\mathrm{θ}$

Substitute $\mathrm{λ}rd\mathrm{θ}$for dm and $r\cos \mathrm{θ}$for x in equation (2).

$$\begin{gathered} \stackrel{¯}{x}=\frac{{∫}_0^{\mathrm{π}/2}r\cos \mathrm{θ}\mathrm{λ}rd\mathrm{θ}}{{∫}_0^{\mathrm{π}/2}\mathrm{λ}rd\mathrm{θ}} \\ =\frac{r^2}{r}\frac{[\sin \mathrm{θ}{]}_0^{\mathrm{π}/2}}{[\mathrm{θ}{]}_0^{\mathrm{π}/2}} \\ =\frac{r\left(\sin \frac{\mathrm{π}}{2}\right)}{\frac{\mathrm{π}}{2}} \\ =\frac{2r}{\mathrm{π}} \end{gathered}$$

Substitute a for r in the above expression:$\stackrel{¯}{x}=\frac{2a}{\mathrm{π}}$

Substitute$\mathrm{λ}rd\mathrm{θ}$ for dm and$r\sin \mathrm{θ}$ for y in equation (3).

$$\begin{gathered} d\stackrel{¯}{y}=\frac{{∫}_0^{\mathrm{π}/2}r\sin \mathrm{θ}\mathrm{λ}rd\mathrm{θ}}{{∫}_0^{\mathrm{π}/2}\mathrm{λ}rd\mathrm{θ}} \\ =\frac{r^2}{r}\frac{[-\cos \mathrm{θ}{]}_0^{\mathrm{π}/2}}{[\mathrm{θ}{]}_0^{\mathrm{π}/2}} \\ =\frac{r\left(\cos 0\right)}{\frac{\mathrm{π}}{2}} \\ =\frac{2r}{\mathrm{π}} \end{gathered}$$

Substitute $a$for $r$in the above expression,$\stackrel{¯}{y}=\frac{2a}{\mathrm{π}}$

Thus, the required centroid is$\left(\frac{2a}{\mathrm{Ï€}},\frac{2a}{\mathrm{Ï€}}\right)$ .