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Two masses m(1) and m(2) are placed at a...

Two masses `m_(1) and m_(2)` are placed at a distance r from each other. Find out the moment of inertia of system about an axis passing through their centre of mass.

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To find the moment of inertia of a system of two masses \( m_1 \) and \( m_2 \) placed at a distance \( r \) from each other about an axis passing through their center of mass, we can follow these steps: ### Step 1: Determine the position of the center of mass The center of mass \( R_{cm} \) of the two masses can be calculated using the formula: \[ R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2} \] ...
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Knowledge Check

  • Two particles of masses m_(1) and m_(2) are separated by a distance 'd'. Then moment of inertia of the system about an axis passing through centre of mass and perpendicular the line joining them is

    A
    `((m_(1)m_(2))/(m_(1)+m_(2)))(d^(2))/(2)`
    B
    `((m_(1)m_(2))/(m_(1)+m_(2)))d`
    C
    `((m_(1)m_(2))/(m_(1)+m_(2)))d^(2)`
    D
    `((2m_(1)m_(2))/(m_(1)+m_(2)))d^(2)`

  • Two point masses m and 3m are placed at distance r. The moment of inertia of the system about an axis passing through the centre of mass of system and perpendicular to the joining the point masses is

    A
    `(3)/(5)mr^(2)`
    B
    `(3)/(4)mr^(2)`
    C
    `(3)/(2)mr^(2)`
    D
    `(6)/(7)mr^(2)`

  • Two identical circular plates each of mass M and radius R are attached to each other with their planes bot^r to each other. The moment of inertia of system about an axis passing through their centres and the point of contact is

    A
    `(MR^(2))/4`
    B
    `(5MR^(2))/4`
    C
    `3/4 MR^(2)`
    D
    `MR^(2)`

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    Two rings have the same mass (m) and radius (r). They are placed in such away that their centres are at a common point and their planes are perpendicular to each other. What is the moment of inertia of the system about an axis passing through their centre and perpendicular to the plane of one of the ring ?

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