Two identical thin uniform rods of length ‘L’ each are joined to form ‘T’ shape as shown in figure. The distance of CM from D is

Two identical thin uniform rods of length L each are joined to form T as shown.

Let the mass of each rod per unit length = $\rho$

We know uniform metal rods obviously contain the C.M at the centre of rod.

Let the centre of mass of AB is at C

i.e. $\text{AC}=\text{CB}=\frac{\text{L}}{\text{2}}$

The C.M of CD is at E

i.e. $\text{CE}=\text{ED}=\frac{\text{L}}{\text{2}}$

consider coordinate axis taking origin (0, 0) at D as shown.

Now we can consider total mass of AB at C i.e ${\text{m}}_{\text{1}}=\rho \text{L}$ .

Similarly total mass of CD at E i.e. ${\text{m}}_{\text{2}}=\rho \text{L}$

Coordinates of C = (0, L)

Coordinates of E $=\left(0,\frac{\text{L}}{\text{2}}\right)$

We know coordinates of C.M of system of two masses.

$x=\frac{{\text{m}}_1{x}_1+{\text{m}}_2{x}_2}{{\text{m}}_1+{\text{m}}_2},y=\frac{{\text{m}}_1{y}_1+{\text{m}}_2{y}_2}{{\text{m}}_1+{\text{m}}_2}$

We get on substituting the values

$x=0$   

$y=\frac{\rho \text{L}\left(\text{L}\right)+\rho \text{L}\left(\frac{\text{L}}{\text{2}}\right)}{\rho \text{L}+\rho \text{L}}$  $\Rightarrow \frac{3\frac{L}{2}}{2}=\frac{3L}{4}$

The coordinates of C.M of system of two rods kept in T shape i.e. $P=\left(0,\frac{3L}{4}\right)$

Hence the distance of C.M from $D=\frac{3L}{4}$ .