Question

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

Answer 1

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,

$\boxed{ \bf{I = \dfrac{Ml^2sin^2 \theta}{3}}}$

So, Moment of inertia of square frame about one of its diagonal (AB):

$\rm \leadsto I_{AB} =4I \\ \\ \rm \leadsto I_{AB} = 4 \times \dfrac{Ml^2sin^2 45 \degree}{3} \\ \\ \rm \leadsto I_{AB} = \dfrac{4}{3} Ml^2 \times \bigg( \dfrac{1}{ \sqrt{2} } \bigg) ^{2} \\ \\ \rm \leadsto I_{AB} = \dfrac{4}{3} Ml^2 \times \dfrac{1}{2} \\ \\ \rm \leadsto I_{AB} = \dfrac{2}{3} Ml^2$

$\therefore$ Momentum of inertia of square frame about one of its diagonal = $\rm \dfrac{2}{3} Ml^2$

Answer 2

Explanation:

$\underline{ \underline{\sf \: Given } } :$

• Four identical thin rods each of mass M .

• length = l from a square frame.

$\underline{ \underline{\sf \: To Find \: } } :$

• what is diagonal ?

$\underline{ \underline{\sf \: Solution \: } } :$

So the moment of inertia of each rod through the center and about an axis perpendicular to the ends.

we have formula :

$\boxed{\sf \: l= \frac{m{l}^{2} }{6} }$

Thus the moment of inertia of each rod about axis is:

$\sf \underline {From \: parallel \: axis \: theorem} :$

$\sf \leadsto \: l \: = 4 \: \times( l + m)( { \frac{l}{2} }^{2} )\: \\ \\ \sf \leadsto \: l \: = \: \frac{m {l}^{2} }{6} + \frac{m {l}^{2} }{4} \\ \\ \sf \leadsto \: l \: = \: \: \frac{2}{3} {ml}^{2}$