Four identical thin rods each of mass M and length l, from a square frame. Moment of inertia of this frame about one of its diagonal is? answer is 2/3ml^2
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Four identical thin rods each of mass M and length l, from a square frame. Moment of inertia of this frame about one of its diagonal is? answer is 2/3ml^2
plz solve step by step
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Given:
Mass of rods = M
Length of rods = l
Type of frame formed by 4 identical rods = Square
To Find:
Momentum of inertia of square frame about one of its diagonal
Answer:
Moment of inertia of single rod inclined at an angle θ with the axis is given by,
[tex] \boxed{ \bf{I = \dfrac{Ml^2sin^2 \theta}{3}}}[/tex]
So, Moment of inertia of square frame about one of its diagonal (AB):
[tex] \rm \leadsto I_{AB} =4I \\ \\ \rm \leadsto I_{AB} = 4 \times \dfrac{Ml^2sin^2 45 \degree}{3} \\ \\ \rm \leadsto I_{AB} = \dfrac{4}{3} Ml^2 \times \bigg( \dfrac{1}{ \sqrt{2} } \bigg) ^{2} \\ \\ \rm \leadsto I_{AB} = \dfrac{4}{3} Ml^2 \times \dfrac{1}{2} \\ \\ \rm \leadsto I_{AB} = \dfrac{2}{3} Ml^2 [/tex]
[tex] \therefore [/tex] Momentum of inertia of square frame about one of its diagonal = [tex] \rm \dfrac{2}{3} Ml^2 [/tex]
Explanation:
[tex] \underline{ \underline{\sf \: Given } } : [/tex]
[tex] \underline{ \underline{\sf \: To Find \: } } : [/tex]
[tex] \underline{ \underline{\sf \: Solution \: } } : \\ [/tex]
So the moment of inertia of each rod through the center and about an axis perpendicular to the ends.
we have formula :
[tex] \boxed{\sf \: l= \frac{m{l}^{2} }{6} } \\ \\ [/tex]
Thus the moment of inertia of each rod about axis is:
[tex] \sf \underline {From \: parallel \: axis \: theorem} : [/tex]
[tex] \sf \leadsto \: l \: = 4 \: \times( l + m)( { \frac{l}{2} }^{2} )\: \\ \\ \sf \leadsto \: l \: = \: \frac{m {l}^{2} }{6} + \frac{m {l}^{2} }{4} \\ \\ \sf \leadsto \: l \: = \: \: \frac{2}{3} {ml}^{2} [/tex]