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6. system of particles and rotational motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics6. system of particles and rotational motion1 Mark
MCQ
Three point masses ${m_1},\,{m_2},\,{m_3}$ are located at the vertices of an equilateral triangle of length $'a'$. The moment of inertia of the system about an axis along the altitude of the triangle passing through ${m_1}$ is 
  • ✓
    $({m_2} + {m_3})\frac{{{a^2}}}{4}$
  • B
     $({m_1} + {m_2} + {m_3}){a^2}$
  • C
    $({m_1} + {m_2})\frac{{{a^2}}}{2}$
  • D
    $({m_2} + {m_3}){a^2}$
✓

Answer

Correct option: A.
$({m_2} + {m_3})\frac{{{a^2}}}{4}$
a
moment of inertia is the product of mass and square of separation between particle and axis of rotation.

e.g. $M . I=m r^{2}$

here, we see, separation of mass $\mathrm{m} 1$ and altitude $N N^{\prime} i s 0 .$

alteration between mass $m_{2}$ and $N N^{\prime}$ is $\left(\frac{a}{2}\right)$ also for $m_{3}$ separation is $\left(\frac{a}{2}\right)$

moment of inertia about altitude passing through $m_{1}=I_{1}+I_{2}+I_{3}$

where $I_{1}, I_{2},$ and $I_{3}$ are $M . I o f m_{1}, m_{2}$ and $m_{3}$ respectively.

$M . I=m_{1} \cdot(0)+m_{2}\left(\frac{a}{2}\right)^{2}+m_{3}\left(\frac{a}{2}\right)^{2}$

$=\frac{a^{2}}{4 \times\left(m_{2}+m_{3}\right)}$

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