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Rotational Motion question

2016 · Q149
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Rotational Motion question

2016 · Q149

NEETPhysicsRotational MotionMCQ+4 / −1
A light rod of length lll has two masses m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is
  1. A
    m1m2m1+m2l2{{{m_1}{m_2}} \over {{m_1} + {m_2}}}{l^2}m1​+m2​m1​m2​​l2
  2. B
    m1+m2m1m2l2{{{m_1} + {m_2}} \over {{m_1}{m_2}}}{l^2}m1​m2​m1​+m2​​l2
  3. C
    (m1+m2)l2\left( {{m_1} + {m_2}} \right){l^2}(m1​+m2​)l2
  4. D
    m1m2 l2\sqrt {{m_1}{m_2}} \,{l^2}m1​m2​​l2
View written solutionFree

Correct answer: A

Here, l1+l2=l{l_1} + {l_2} = ll1​+l2​=l

Centre of mass of the system,

l1=m1×0+m2×lm1+m2=m2lm1+m2{l_1} = {{{m_1} \times 0 + {m_2} \times l} \over {{m_1} + {m_2}}} = {{{m_2}l} \over {{m_1} + {m_2}}}l1​=m1​+m2​m1​×0+m2​×l​=m1​+m2​m2​l​

l2=l−l1=m1lm1+m2{l_2} = l - {l_1} = {{{m_1}l} \over {{m_1} + {m_2}}}l2​=l−l1​=m1​+m2​m1​l​

Required moment of inertia of the system,

I=m1l12+m2l22I = {m_1}l_1^2 + {m_2}l_2^2I=m1​l12​+m2​l22​

=(m1m22+m2m12)l2(m1+m2)2 = \left( {{m_1}m_2^2 + {m_2}m_1^2} \right){{{l^2}} \over {{{\left( {{m_1} + {m_2}} \right)}^2}}}=(m1​m22​+m2​m12​)(m1​+m2​)2l2​

=m1m2(m1+m2)l2(m1+m2)2=m1m2m1+m2l2 = {{{m_1}{m_2}\left( {{m_1} + {m_2}} \right){l^2}} \over {{{\left( {{m_1} + {m_2}} \right)}^2}}} = {{{m_1}{m_2}} \over {{m_1} + {m_2}}}{l^2}=(m1​+m2​)2m1​m2​(m1​+m2​)l2​=m1​+m2​m1​m2​​l2

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