Physics, asked by MiniDoraemon, 6 months ago

Four point masses, each of the value m , are placed at the corner of square ABCD of side L . The moment of inertia of this system about an axis passing through A and parallel to BD is [AIEEE2006 ] ​

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Answered by nehaimadabathuni123
1

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Answered by Ekaro
9

Answer :

Four point masses, each of mass m are placed at the corner of square ABCD of side L.

We have to find the moment of inertia of this system about an axis passing through A and parallel to BD.

Diagram :

Refer to the attachment for better understanding.

➤ Moment of inertia of a body about the given axis of rotation,

\dag\:\underline{\boxed{\tt{I=m_1r_1^2+m_2r_2^2+\dots+m_nr_n^2}}}

\longrightarrow\sf\:I=(mr^2)_A+(mr^2)_B+(mr^2)_C+(mr^2)_D

\longrightarrow\sf\:I=0+\dfrac{mL^2}{2}+m(\sqrt{2}L)^2+\dfrac{mL^2}{2}

\longrightarrow\sf\:I=mL^2+2mL^2

\longrightarrow\bf\:I=3mL^2

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