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Rotational Motion question

2015 · Q138
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Rotational Motion question

2015 · Q138

NEETPhysicsRotational MotionMCQ+4 / −1
Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω\omegaω0 is minimum, is given by

AIPMT 2015 Physics - Rotational Motion Question 78 English
  1. A
    x=m2m1Lx = {{{m_2}} \over {{m_1}}}Lx=m1​m2​​L
  2. B
    x=m2Lm1+m2x = {{{m_2}L} \over {{m_1} + {m_2}}}x=m1​+m2​m2​L​
  3. C
    x=m1Lm1+m2x = {{{m_1}L} \over {{m_1} + {m_2}}}x=m1​+m2​m1​L​
  4. D
    x=m1m2Lx = {{{m_1}} \over {{m_2}}}Lx=m2​m1​​L
View written solutionFree

Correct answer: B

K=12(m1x2)ω02+12m2(L−x)2ω02K = {1 \over 2}\left( {{m_1}{x^2}} \right)\omega _0^2 + {1 \over 2}{m_2}{\left( {L - x} \right)^2}\omega _0^2K=21​(m1​x2)ω02​+21​m2​(L−x)2ω02​

dK/dx=0dK/dx = 0dK/dx=0

m1x=m2(L−x){m_1}x = {m_2}\left( {L - x} \right)m1​x=m2​(L−x)

m1x=m2L−m2x{m_1}x = {m_2}L - {m_2}xm1​x=m2​L−m2​x

⇒x=m2Lm1+m2 \Rightarrow x = {{{m_2}L} \over {{m_1} + {m_2}}}⇒x=m1​+m2​m2​L​

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