Question

There are two masses m₁ and m₂ placed at a distance l apart, let the centre of mass of this system is at a point named C. If m₁ is displaced by l₁ towards C and m₂ is displaced by l₂ away from C, find the distance from C where the new centre of mass will be located.

Answer 1

Simple question

Steps to solve

1)First take out the original position of m1 and m2 from the centre of mass

2)After this, using the.formula of centre mass,caculate the original position of centre of mass

3) Predict the new position of masses

4)Then calculate the new position of centre ofass

5)Then simply, subtract the old position.of centre of mass new postion to get the distance

Solution:-Let the m1 is located at distance x from centre of mass

then, distance of m2 from Xcm=l-x

now ,we can say

m1x=m2(l-x)

m1x+m2x=m2l

x=m2l/(m1+m2)

distance of m1 from initial Xcm=m2l/(m1+m2)

distant of m2 from initial Xcm=l-m2l/(m1+m2)

=m1l/(m1+m2)

now

initial postion of Xcm=(m1x1+m2x2)/(m1+m2)

=(m1m2l +m2m1×l)/(m1+m2)^2

=2m1m2l/(m1+m2)^2

Now ,when the m1 and m2 is moved from their original position

then

position of m1 from initial Xcm=x-l1

={m2l-(m1+m2)I1}/(m1+m2)

position of m2 from initial Xcm=l-x+l2

={m1l+(m1+m2)l2}/(m1+m2)

Let the position of new Xcm from initial be x towards m2

Then

postion of m1 from new Xcm ={m2l-(m1+m2)l1+(m1+m2)x}/(m1+m2)

position of m2 from new Xcm={m1l+(m1+m2)l2-(m1+m2)x}/(m1+m2)

now equating The centre ofass we get

{2m1m2l+x×(m1+m2)^2}/(m1+m2)^2=[m1×{m2l-(m1+m2)l1+(m1+m2)x}+m2×{m1l+(m1+m2)l2-(m1+m2)x]/(m1+m2)^2

=>2m1m2l+x(m1+m2)^2=m1m2l-l1×(m1)^2-m1m2l1+

m1x(m1+m2)+m1m2l+m1m2l2+(m2)^2×I2-m2x(m1+m2)

cancelling out we get

(m2+m2)×x=m1l1+m2l2

x=(m1l1+m2l2)/(m1+m2)

this will be displacement of Xcm

Answer 2

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For given :-

We have :-

$\bf{{m1 \: x = m 2 \: (L−x)}}$

$\bf{{⇒x = \: \frac{m2l}{m1 + m2} }} \: \: \: \: \: \: \: \: (1)$

m1 displaced by L1 and m2 displaced by L2, with new centre of mass C

We have :-

$\bf{{m1 (L 1 +x−y) = m 2 (L−x+y+L 2 ) \: \: \: \: \: \: \: (2)}}$

from (1) and (2), we have,

$\bf{{m1 (L1 + ( \frac{m2L}{m1 + m2}) - y) = }}$

$\bf{{m2(L-( \frac{m2L}{m1 + m2} ) + y + L2)}}$

upon simplification, we get the value of "y" as,

$\bf{{y = \frac{m1L1 - m2L}{m1 + m2} }}$

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