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Question
Prove the theorem of parallel axes.
(Hint: If the centre of mass is chosen to be the origin `summ_ir_i = 0`
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Solution
Theorem of parallel axes: According to this theorem, moment of inertia of a rigid body about any axis AB is equal to moment of inertia of the body about another axis KL passing through centre of mass C of the body in a direction parallel to AB, plus the product of total mass M of the body and square of the perpendicular distance between the two parallel axes. If h is perpendicular distance between the axes AB and KL, then Suppose the rigid body is made up of n particles m1, m2, …. mn, mn at perpendicular distances r1, r2, ri…. rn. respectively from the axis KL passing through the centre of mass C of the body.
If h is the perpendicular distance of the particle of mass m{ from KL, then
The perpendicular distance of ith particular from the axis
`AB = (r_i + n)`
or `I_(AB) = sum_(i)m_i(r_i + h)^2`
`= sum_(i) m_ir_i^2 + sum_(i) m_ih^2 + 2h sum_(i) m_ir_i` ....(ii)
As the body is balanced about the centre of mass the algebraic sum of the moments of the weights of all particles about an axis passing through C must be zero
`sum_(i)(m_ig)r_i = 0 or g sum_(i) m_ir_i` or `sum_(i) m_ir_i = 0` ...(iii)
From equation (ii) we have
`I_(AB) = sum_(i) m_ir_i^2 + (summ_i)h^2 + 0`
or `I_(AB) = I_(KL) + Mh^2`
Where `I_"KL" = sum_(i) m_ir_i^2` and `M = sum m_i`
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Answer in brief:
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