14.1 Center of mass and rigid bodies  (Page 2/2)

$$\begin{array}{l}\gamma =\frac{m_1}{l_1}=\frac{m_2}{l_2}\end{array}$$

The COM of the L-shaped rod is :

$$\begin{array}{l}{x}_{\mathrm{COM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}\\ \Rightarrow x_{\mathrm{COM}}=\frac{\gamma l_{1}x\frac{l_1}{2}}{\gamma (l_1+l_2)}\\ \Rightarrow x_{\mathrm{COM}}=\frac{{l_1}^2}{2(l_1+l_2)}\end{array}$$

and similarly,

$$\begin{array}{l}{y}_{\mathrm{COM}}=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}\\ \Rightarrow x_{\mathrm{COM}}=\frac{\gamma l_{2}x\frac{l_2}{2}}{\gamma (l_1+l_2)}\\ \Rightarrow x_{\mathrm{COM}}=\frac{{l_2}^2}{2(l_1+l_2)}\end{array}$$

It is important to note that the expressions of COM are independent of mass term.

Semicircular uniform wire

A semicircular wire is not symmetric to all the axes of the coordinates. The wire is a planar body involving two dimensions (x and y). However, we note that the wire is symmetric about y-axis, but not about x-axis. This has two important implications :

• COM lies on y - axis (i.e on the axis of symmetry).
• We only need to evaluate "y" coordinate of COM as $x_{\mathrm{COM}}$ = 0.

Thus, we need to evalute the expression :

$$\begin{array}{l}{y}_{\mathrm{COM}}=\frac{1}{M}\int ydm\end{array}$$

Evaluation of an integral is based on identifying the differential element. The semicircular wire is composed of large numbers of particle like elements, each of mass "ⅆm". The idea of this particle like element of mass "dm" is that its center of mass i.e COM lies at "y" distance from the origin in y-direction. This element is linked to the mass "m" of the wire as :

Semicircular uniform wire

$$\begin{array}{l}dm=\gamma dl=\left(\frac{M}{\pi R}\right)dl\end{array}$$

Now,

$$\begin{array}{l}dl=Rd\theta \\ y=R\sin \theta \end{array}$$

$$\begin{array}{l}\Rightarrow dm=\left(\frac{M}{\pi R}\right)Rd\theta =\left(\frac{M}{\pi }\right)d\theta \end{array}$$

Putting in the integral,

$$\begin{array}{l}\Rightarrow y_{\mathrm{COM}}=\frac{1}{M}\int R\sin \theta \left(\frac{M}{\pi }\right)d\theta \\ \Rightarrow y_{\mathrm{COM}}=\frac{R}{\pi }\int \sin \theta d\theta \end{array}$$

Integrating between θ = 0 and θ = π,

$$\begin{array}{l}\Rightarrow y_{\mathrm{COM}}=\frac{R}{\pi }[-\cos \theta {]}_0^{\pi }\\ \Rightarrow y_{\mathrm{COM}}=\frac{R}{\pi }[-\cos \pi +\cos 0]=\frac{2R}{\pi }\end{array}$$

The COM of the semicircular wire is ( $0,\frac{2R}{\pi }$ ).

Semicircular unifrom thin disc

A semicircular disc is not symmetric to all the axes of the coordinates. The thin disc is a planar body involving two dimensions (x and y). However, we note that the disc is symmetric about y-axis, but not about x-axis. This has two important implications :

• COM lies on y - axis (i.e on the axis of symmetry).
• We only need to evaluate "y" coordinate of COM as $x_{\mathrm{COM}}$ = 0.

Thus, we need to evaluate the expression :

$$\begin{array}{l}{y}_{\mathrm{COM}}=\frac{1}{M}\int ydm\end{array}$$

Evaluation of an integral is based on identifying the differential element. The semicircular disc is composed of large numbers of semicircular thin wires. We must be careful in evaluating this integral. The elemental mass "dm" corresponds to a semicirular wire - not a particle as the case before. The term "y" in the integral denotes the distance of the COM of this elemental semicircular wire. The COM of this semi-circular thin wire element is at a distance 2r/Π (as determined earlier)in y-direction.

Semicircular uniform disc

Now,

$$\begin{array}{l}y=\frac{2r}{\pi }\\ dm=\sigma dA=(\pi rdr)\sigma \end{array}$$

where "σ" denoted the surface density. It is given by :

$$\begin{array}{l}\sigma =\frac{M}{\frac{\pi R^2}{2}}=\frac{2M}{\pi R^2}\end{array}$$

Putting in the integral,

$$\begin{array}{l}\Rightarrow y_{\mathrm{COM}}=\frac{1}{M}\int \left(\frac{2r}{\pi }\right)\sigma \pi rdr\\ \Rightarrow y_{\mathrm{COM}}=\frac{2\sigma }{M}\int r^{2}dr\end{array}$$

Integrating between r = 0 and r = R (radius of the semicircular disc),

$$\begin{array}{l}\Rightarrow y_{\mathrm{COM}}=\frac{2\sigma }{M}[\frac{r^3}{3}{]}_0^R\\ \Rightarrow y_{\mathrm{COM}}=\frac{2\sigma R^3}{3M}\\ \Rightarrow y_{\mathrm{COM}}=\frac{4M{R}^3}{3M\pi R^2}=\frac{4R}{3\pi }\end{array}$$

The COM of the semicircular disc is ( $0,\frac{4R}{3\pi }$ ).

Com rigid bodies with non-uniform density

Evaluation of integrals to determine COM of non-uniform rigid bodies may be cumbersome. There are two possibilities :

1. The variation in density is distinctly marked.
2. The variation in density is defined in mathematical form.

Variation in density is distinctly marked

If the variation in density is distinctly marked, then it is possible to determine COMs of individual uniform parts. Once COMs of the individual uniform parts are known, we can treat them as particle with their mass at their respective COMs. Thus, non-uniform rigid body is reduced to a system of particles, which renders easily for determination of its COM. This technique is illustrated in the example given here :

Problem : A thin plate (axb) is made up of two equal portions of different surface density of ${\sigma }_1$ and ${\sigma }_1$ as shown in the figure. Determine COM of the composite plate in the coordinate system shown.

Composite system of rigid bodies

Solution : COM of the first half is at $\frac{a}{4}$ , whereas COM of the second half is at $\frac{3a}{4}$ from the origin of the coordinate system in x-direction. As x-axis lies at mid point of the height of the plate, y-component of COM is zero. Now two portions are like two particles having masses :

$$\begin{array}{l}{m}_1=\frac{ab}{2}{\sigma }_1\end{array}$$

and

$$\begin{array}{l}{m}_2=\frac{ab}{2}{\sigma }_2\end{array}$$

Also, the x-coordinates of the positions of these particles are $x_1=\frac{a}{4}$ and $x_2=\frac{3a}{4}$ . The x-component of the COM of two particles system is :

$$\begin{array}{l}{x}_{\mathrm{COM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}\end{array}$$

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{\frac{ab}{2}{\sigma }_1\frac{a}{4}+\frac{ab}{2}{\sigma }_2\frac{3a}{4}}{\frac{ab}{2}{\sigma }_1+\frac{ab}{2}{\sigma }_2}\end{array}$$

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{({\sigma }_1+3{\sigma }_2)a}{4({\sigma }_1+{\sigma }_2)}\end{array}$$

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Variation in density is defined in mathematical form

If the variation in density is well defined, then it is possible to evaluate integral to determine the COM for non-uniform rigid body. The technique is illustrated in the example given here :

Problem : Linear density of a rod varies with length as γ = 2x + 3. If the length of the rod is "L", then find the COM of the rod.

Solution : The COM of the rod in one dimension is given as :

Com of a rod with varying density

$$\begin{array}{l}{x}_{\mathrm{COM}}=\frac{1}{M}\int xdm\end{array}$$

Here,

$$\begin{array}{l}dm=\gamma dx=(2x+3)dx\end{array}$$

We can use this relation to determine COM as :

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{\int xdm}{\int dm}\\ \Rightarrow x_{\mathrm{COM}}=\frac{\int x(2x+3)dx}{\int (2x+3)dx}\end{array}$$

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{2\int x^{2}dx+3\int xdx}{2\int xdx+3\int dx}\end{array}$$

Evaluating between x = 0 and x = L, we have :

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{\frac{2L^3}{3}+\frac{3L^2}{2}}{\frac{2L^2}{2}+3L}\end{array}$$

$$\begin{array}{l}\Rightarrow x_{\mathrm{COM}}=\frac{4L^2+9L}{6(L+3)}\end{array}$$

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