Question

A triangular lamina has vertices (0,0),(0,6) and(6,0),and uniform density. Find:

(a) $\overline{\mathbf{x}}\mathbf{,}\overline{\mathbf{y}}$

(b)${\mathit{I}}_{\mathbf{x}}$,

(c)${\mathit{I}}_{\mathbf{m}}$ about an axis parallel to thex-axis. Hint: Use Problemcarefully.

Short answer

(a) The $\overline{\mathrm{x}},\overline{\mathrm{y}}$coordinates of centre of mass for a triangular lamina is 2,2.

(b) The moment if inertia $\left(I_x\right)$ about axis for a triangular lamina is 6M .

(c) The moment of inertia $\left(I_m\right)$about the axis parallel with x-axis and going through the centre of massfor a triangular lamina is 2M.

Step-by-step solution

Definition of Moment of Inertia definition

In physics, a moment of inertia is a quantitative measure of a body's rotational inertia—that is, the body's resistance to having its speed of rotation about an axis changed by the application of a torque (turning force). The axis might be internal or exterior, and it can be fixed or not.

(a) Determining x,y coordinates of centre of mass

Consider the figure, as given below –

The centre of mass is at –

$$\begin{gathered} \overline{x},=\overline{y}=\frac{1}{M}{\int }_{x=0}^6\left[{\int }_{y=0}^{6}ydy\right]\sigma dx \\ =\frac{1}{18}{\int }_0^6\frac{1}{2}{\left(6-x\right)}^{2}d\left(6-x\right)\left(-1\right)\sigma \\ =-\frac{1}{108}{\overline{){\left(6-x\right)}^3}}_0^6 \\ =\frac{216}{108} \end{gathered}$$

Where $M=18\sigma$. Thus, the centre of mass is at 2,2 .

Hence, the$\overline{x},\overline{y}$ coordinates of centre of mass is obtained as 2,2.

(b) Determining Ix moment of inertia about  x-axis

The moment of inertia about the and axis will be identical –

$$\begin{gathered} I_x={\int }_{y=0}^6\left[{\int }_{x=0}^{6-y}{x}^{2}dx\right]dy\sigma \\ =\frac{\sigma }{3}{\int }_0^6{\left(6-y\right)}^{2}d\left(6-y\right)\left(-1\right) \\ =-\frac{\sigma }{12}{\overline{){\left(6-x\right)}^4}}_0^6 \\ =108\sigma \\ =6M \end{gathered}$$

Hence, the moment if inertia $\left(I_x\right)$about axis is obtained as 6M.

(c) Determining Im about an axis parallel to the x-axis

The moment of inertia about the axis parallel withaxis and going through the centre of mass is –

$$\begin{gathered} I_x=I_m+M{D}^2 \\ {I}_m=I_x-4M \\ =2M \end{gathered}$$

Hence, the moment of inertia $\left(I_m\right)$about the axis parallel with x-axis and going through the centre of mass is obtained as 2M.