Question 39: Centre of Mass and Rotational Motion

Question

The position of center of mass of three masses $2 \mathrm{~kg}, 3 \mathrm{~kg}$ and 15 kg placed with respect to mid point $(p)$ of normal bisector, as shown in the figure is $\_\_\_\_$ .

Accepted Answer

If we have multiple masses $\mathrm{m}_1, \mathrm{~m}_2, \ldots$ at positions $\left(\mathrm{x}_1, \mathrm{y}_1ight),\left(\mathrm{x}_2, \mathrm{y}_2ight), \ldots$, the coordinates of the COM are
$$ \begin{aligned} \mathrm{x}_{\mathrm{cm}} & =\frac{\mathrm{m}_1 \mathrm{x}_1+\mathrm{m}_2 \mathrm{x}_2+\mathrm{m}_3 \mathrm{x}_3 \ldots}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3 \ldots} \ \mathrm{Y}_{\mathrm{cm}} & =\frac{\mathrm{m}_1 \mathrm{y}_1+\mathrm{m}_2 \mathrm{y}_2+\mathrm{m}_3 \mathrm{y}_3 \ldots}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3 \ldots} \end{aligned} $$
Let the midpoint p be the origin $(0,0)$. From the geometry,
In right angled triangle $\mathrm{ACD}, \cos 60^{\circ}=\frac{\mathrm{CD}}{\mathrm{AC}} \Rightarrow \mathrm{CD}=\mathrm{AC} \cos 60^{\circ}=10 	imes \frac{1}{2} \mathrm{~m}=5 \mathrm{~m}$
Point P is the midpoint of CD , so $\mathrm{PC}=\mathrm{PD}=\frac{\mathrm{CD}}{2}=\frac{5 \mathrm{~m}}{2}=2.5 \mathrm{~m}$
As P is the origin, the coordinate of 15 kg is $(0,2.5) \mathrm{m}$ and the coordinates of D is $(0,-2.5) \mathrm{m}$.
As 2 kg and 3 kg is on the same horizontal level as D, so y-coordinate of 2 kg and 3 kg is $\mathrm{y}_1=-2.5 \mathrm{~m}$
Also, $\sin 60^{\circ}=\frac{\mathrm{AD}}{10} \Rightarrow \mathrm{AD}=10 \sin 60^{\circ}=10 	imes \frac{\sqrt{3}}{2}=5 \sqrt{3} \mathrm{~m}$
Hence, the $x$-coordinate of 2 kg is $-5 \sqrt{3} \mathrm{~m}$ and that of 3 kg it is $5 \sqrt{3} \mathrm{~m}$
Hence, the coordinates of 2 kg are $(-5 \sqrt{3},-2.5)$, the coordinate of 3 kg are $(5 \sqrt{3},-2.5)$, and that of 15 kg is (0,2.5).
The total mass of system is $\mathrm{M}=2+3+15=20 \mathrm{~kg}$.
So, the x -coordinate of centre of mass is:
$$ \mathrm{x}_{\mathrm{cm}}=\frac{(15 	imes(0))+(3 	imes 5 \sqrt{3})+(2 	imes(-5 \sqrt{3}))}{20}=\frac{5 \sqrt{3}}{20}=\frac{\sqrt{3}}{4} $$
And the y -coordinate of centre of mass is:
$$ y_{c m}=\frac{(15 	imes 2.5)+(3 	imes(-2.5))+(2 	imes(-2.5))}{20}=\frac{37.5-12.5}{20} $$
$$\Rightarrow $$ $$ \mathrm{y}_{\mathrm{cm}}=\frac{25}{20}=\frac{5}{4}=1.25 $$
Therefore, the centre of mass of system of masses is $\left(\frac{\sqrt{3}}{4}, 1.25ight)$.
Hence, the correct option is (A).

Answer

$$ \left(\frac{\sqrt{3}}{4}, 1.25\right) $$

Answer

$$ \left(\frac{\sqrt{3}}{4}, 1.0\right) $$

Answer

$$ (0,0) $$

Answer

$$ (1.25,0) $$

Centre of Mass and Rotational Motion

The position of center of mass of three masses 2 \mathrm{~kg}, 3 \mathrm{~kg} and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is \_\_\_\_ .

Choose the correct answer: