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System of Particles and Rotational Motion

Centre of mass

Consider a two particle system with particles having masses m_

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Consider a two particle system with particles having masses $m_{1}$ and $m_{2}$ . If the first particle is pushed towards the centre of mass through a distance $d$ , by what distance should the second particle be moved, so as to keep the centre of mass at the same position? (a) $\frac{m _{2}}{m _{1}} d$ (b) $\frac{m _{1}}{m _{1} + m _{2}} d$ (c) $\frac{m _{1}}{m _{2}} d$ (d) $d$

Views: 6,089 students

Updated on: May 23, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2006

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Sebastian Discussed

Consider a two particle system with particles having masses

m_{1}

and

m_{2}

. If the first particle is pushed towards the centre of mass through a distance

d

, by what distance should the second particle be moved, so as to keep the centre of mass at the same position? (a)

\frac{m _{2}}{m _{1}} d

(b)

\frac{m _{1}}{m _{1} + m _{2}} d

(c)

\frac{m _{1}}{m _{2}} d

(d)

d

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Exp. (c)
To keep the centre of mass at the same position, velocity of centre of mass is zero, so

\frac{m _{1} v _{1} + m _{2} v _{2}}{m _{1} + m _{2}} = 0

where,

v_{1}

and

v_{2}

are the velocities of particles 1 and 2 , respectively.

\Rightarrow m_{1} \frac{d r _{1}}{d t} + m_{2} \frac{d r _{2}}{d t} = 0

[∵ v_{1} = \frac{d r _{1}}{d t} & v_{2} = \frac{d r _{2}}{d t}]

\Rightarrow m_{1} d r_{1} + m_{2} d r_{2} = 0

of particles. d r_{2} represent the change in displacement

Let 2 nd particle has been displaced by distance x

\Rightarrow m_{1} (d) + m_{2} (x) = 0 or x = - \frac{m _{1} d}{m _{2}}

Alternate Solution

x_{CM} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}

.. (i)

After moving

m_{1}

through a distance

d

towards

CM

and to keep the position of

CM

unchanged, let

d^{'}

be the shift of

m_{2}

x_{CM} = \frac{m _{1} ( x _{1} - d ) + m _{2} ( x _{2} + d ^{'} )}{m _{1} + m _{2}}

From Eqs. (i) and (ii), we get

\frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}} = \frac{m _{1} ( x _{1} - d ^{'} ) + m _{2} ( x _{2} + d )}{m _{1} + m _{2}}

\Rightarrow - m_{1} d + m_{2} d^{'} = 0

∴ d^{'} = \frac{m _{1}}{m _{2}} d

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Question Text	Consider a two particle system with particles having masses $m_{1}$ and $m_{2}$ . If the first particle is pushed towards the centre of mass through a distance $d$ , by what distance should the second particle be moved, so as to keep the centre of mass at the same position? (a) $\frac{m _{2}}{m _{1}} d$ (b) $\frac{m _{1}}{m _{1} + m _{2}} d$ (c) $\frac{m _{1}}{m _{2}} d$ (d) $d$
Updated On	May 23, 2023
Topic	System of Particles and Rotational Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 6
Upvotes	690
Avg. Video Duration	7 min

Text solutionVerified

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