The position of the centre of mass of the system: (Consider $\mathrm{L}=10 \mathrm{~cm}$)
A rod of length 5 L is bent right angle keeping one side length as 2 L. The position of the centre of mass…
The position of the centre of mass of the system: (Consider $\mathrm{L}=10 \mathrm{~cm}$)
- $2 \hat{i}+3 \hat{j}$
- $3 \hat{i}+7 \hat{\mathrm{j}}$
- $5 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}$
- $4 \hat{i}+9 \hat{j}$
Solution
$\mathrm{x}_{\mathrm{com}}=\frac{2 \mathrm{~m}(10)+3 \mathrm{~m}(0)}{5 \mathrm{~m}}=4 \mathrm{~cm}$
$\mathrm{y}_{\mathrm{com}}=\frac{2 \mathrm{~m}(0)+3 \mathrm{~m}(15)}{5 \mathrm{~m}}=9 \mathrm{~cm}$
$\overrightarrow{\mathrm{r}}_{\mathrm{com}}=4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}$
Asked in: JEE Main 2025 (07 Apr Shift 1)