Question 51: Centre of Mass and Rotational Motion

Question

A rod of length 5 L is bent right angle keeping one side length as 2 L .
The position of the centre of mass of the system :
(Consider $\mathrm{L}=10 \mathrm{~cm}$)

Accepted Answer

$$\begin{aligned}
& \mathrm{x}_{\mathrm{com}}=\frac{2 \mathrm{~m}(10)+3 \mathrm{~m}(0)}{5 \mathrm{~m}}=4 \mathrm{~cm} \
& \mathrm{y}_{\mathrm{com}}=\frac{2 \mathrm{~m}(0)+3 \mathrm{~m}(15)}{5 \mathrm{~m}}=9 \mathrm{~cm} \
& \overrightarrow{\mathrm{r}}_{\mathrm{com}}=4 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}
\end{aligned}$$

Answer

$4 \hat{i}+9 \hat{j}$

Answer

$2 \hat{i}+3 \hat{j}$

Answer

$5 \hat{i}+8 \hat{j}$

Answer

$3 \hat{i}+7 \hat{j}$

Centre of Mass and Rotational Motion

A rod of length 5 L is bent right angle keeping one side length as 2 L . The position of the centre of mass of the system : (Consider \mathrm{L}=10 \mathrm{~cm})

Choose the correct answer: