Question 37: Centre of Mass and Rotational Motion

Question

Two identical bodies A and B of equal masses have initial velocities $\overrightarrow{v_1}=4 \hat{i} \mathrm{~m} / \mathrm{s}$ and $\overrightarrow{v_2}=4 \hat{j} \mathrm{~m} / \mathrm{s}$ respectively. The body A has acceleration $\overrightarrow{a_1}=6 \hat{i}+6 \hat{j} \mathrm{~m} / \mathrm{s}^2$ while the acceleration of the other body B is zero. The centre of mass of the two bodies moves in $\_\_\_\_$ path.

Accepted Answer

For two bodies of equal mass $\left(\mathrm{m}_1=\mathrm{m}_2=\mathrm{m}ight)$, the velocity of the COM is:
$$ \overrightarrow{\mathrm{v}}_{\mathrm{cm}}=\frac{\mathrm{m} \overrightarrow{\mathrm{v}}_1+\mathrm{m} \overrightarrow{\mathrm{v}}_2}{\mathrm{~m}+\mathrm{m}}=\frac{\overrightarrow{\mathrm{v}}_1+\overrightarrow{\mathrm{v}}_2}{2} $$
The velocity of body $A$ is $\vec{v}_1=4 \hat{i}$ and that of body $B$ it is $\vec{v}_2=4 \hat{j}$ :
So,
$$ \overrightarrow{\mathrm{v}}_{\mathrm{cm}}=\frac{4 \hat{\imath}+4 \hat{\jmath}}{2}=(2 \hat{\imath}+2 \hat{\jmath}) \mathrm{m} / \mathrm{s} $$
Similarly, for equal masses the acceleration of center of mass is,
$$ \overrightarrow{\mathrm{a}}_{\mathrm{cm}}=\frac{\overrightarrow{\mathrm{a}}_1+\overrightarrow{\mathrm{a}}_2}{2} $$
The acceleration of body A is $\overrightarrow{\mathrm{a}}_1=6 \hat{i}+6 \hat{\mathrm{j}}$ and that of body B is $\overrightarrow{\mathrm{a}}_2=0$ :
$$ \overrightarrow{\mathrm{a}}_{\mathrm{cm}}=\frac{(6 \hat{\imath}+6 \hat{\jmath})+0}{2}=(3 \hat{\imath}+3 \hat{\jmath}) \mathrm{m} / \mathrm{s}^2 $$
A particle moves in a straight line if its acceleration vector is parallel (or anti-parallel) to its velocity vector. If they are not parallel, the path is generally a parabola (under constant acceleration).
Direction of velocity of centre of mass is,
$$ 	an 	heta_{\mathrm{v}}=\frac{\mathrm{v}_{\mathrm{y}}}{\mathrm{v}_{\mathrm{x}}}=\frac{2}{2}=1 \Rightarrow 	heta_{\mathrm{v}}=	an ^{-1}(1)=\frac{\pi}{4} \mathrm{rad} $$
Direction of acceleration of centre of mass is,
$$ 	an 	heta_{\mathrm{a}}=\frac{\mathrm{a}_{\mathrm{y}}}{\mathrm{a}_{\mathrm{x}}}=\frac{3}{3}=1 \Rightarrow 	heta_{\mathrm{a}}=	an ^{-1}(1)=\frac{\pi}{4} \mathrm{rad} $$
Since both vectors have the same direction (they both make an angle of $45^{\circ}$ with the positive x -axis), the acceleration is acting exactly along the line of motion.
Because $\overrightarrow{\mathrm{a}}_{\mathrm{cm}}$ is parallel to $\overrightarrow{\mathrm{v}}_{\mathrm{cm}}$, the center of mass moves in a straight line. Hence, the correct option is (C).

Answer

circular

Answer

parabolic

Answer

straight line

Answer

elliptical

Centre of Mass and Rotational Motion

Choose the correct answer: