Question 321: Centre of Mass and Rotational Motion

Question

In the system of two particles of masses ' $\mathrm{m}_1$ ' and ' $\mathrm{m}_2$ ', the first particle is moved by a distance 'd' towards the centre of mass. To keep the centre of mass unchanged, the second particle will have to be moved by a distance

Accepted Answer

$$\begin{array}{ll} 
& m_1 x_1=m_2 x_2 \quad	ext{.... (i)}\
& m_1\left(x_1-dight)=m_2\left(x_2-d^{\prime}ight) \quad	ext{.... (ii)}\
	herefore \quad & m_1 x_1-m_1 d=m_2 x_2-m_2 d^{\prime} \
& m_1 d=m_2 d^{\prime} \quad	ext{.... [From (i)]}\
	herefore \quad & d^{\prime}=\frac{m_1}{m_2} d
\end{array}$$

Answer

$\frac{\mathrm{m}_2}{\mathrm{~m}_1} \mathrm{~d}$, towards the centre of mass.

Answer

$\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{~d}$, away from the centre of mass.

Answer

$\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{~d}$, towards the centre of mass.

Answer

$\frac{\mathrm{m}_2}{\mathrm{~m}_1} \mathrm{~d}$, away from the centre of mass.

Centre of Mass and Rotational Motion

In the system of two particles of masses ' \mathrm{m}_1 ' and ' \mathrm{m}_2 ', the first particle is moved by a distance 'd' towards the centre of mass. To keep the centre of mass unchanged, the second particle will have to be moved by a distance

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