Question

Three particles each of mass 1 kg are placed at the corners of a right angled triangle AOB, O being the origin of the coordinate system (OA and OB along positive X- direction and positive Y- direction). If OA=OB=1m, the positive vector of the centre of mass (in metres) is:

• A. $\frac{\widehat{i}+\widehat{j}}{3}$
• B. $\frac{2\left( \widehat{i}+\widehat{j} \right)}{3}$
• C. $\frac{2\left( \widehat{i}-\widehat{j} \right)}{3}$
• D. $\left( \widehat{i}-\widehat{j} \right)$

Answer 1

Answer: (A)

$\frac{\widehat{i}+\widehat{j}}{3}$

Answer 2

Three particles each of mass 1 kg are placed the three corners of a right angled triangle. Here, ${{m}_{1}}={{m}_{2}}={{m}_{3}}=1$ $({{x}_{1}},{{y}_{1}})=(0,0)$ $({{x}_{3}},{{y}_{3}})=(0,1)$ $({{x}_{CM}},{{y}_{CM}})=?$ ${{x}_{CM}}=\frac{{{m}_{1}}{{x}_{1}}+{{m}_{2}}{{x}_{2}}+{{m}_{3}}{{x}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}$ ${{x}_{CM}}=\frac{1.0+1.1+1.0}{1+1+1}$ ${{x}_{CM}}=\frac{1}{3}$ ${{y}_{CM}}=\frac{{{m}_{1}}{{y}_{1}}+{{m}_{2}}{{y}_{2}}+{{m}_{3}}{{y}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}$ ${{y}_{CM}}=\frac{1.0+1.0+1.1}{1+1+1}$ ${{y}_{CM}}=\frac{1}{3}$ $\therefore$ $\vec{r}\,CM={{x}_{CM}}\hat{i}+{{y}_{CM}}\hat{j}=\frac{1}{3}\hat{i}+\frac{1}{3}\hat{j}$ ${{\vec{r}}_{CM}}=\frac{\hat{i}+\hat{j}}{3}$