Question

Two uniform rods of mass m and length l form a
cross moment of inertia of cross about an axis
parallel to Cd passing through A is​

Answer 1

Answer:

5/12ML^2.

Explanation:

We get that the two rods are making a cross hence the moment of inertia of the two rods will pass together through the center and the equivalent moment of inertia will be of the figure of cross.

Since, for the vertical rod the moment of inertia is at the half length or M(L/2)^2 or ML^2/4. For the horizontal rod the moment of inertia is ML^2/12 + M(L/2)^2 which on solving we will get the moment of inertia to be ML^2/6.

Hence, the moment of inertia about an axis parallel to the other rod will be summation of the two inertia = ML^2/6 + ML^2/4 which on solving we will get that 5/12ML^2.

Answer 2

Answer:

$\frac{5}{12}ml^2$

Explanation:

We have been given that

The two rods are making a cross hence the Moment of Inertia of the two rods will pass together through the center and the equivalent Moment of Inertia will be of the figure of cross.

Since, for the vertical rod the moment of inertia is at the half length or $m[\frac{l}{2}]^{2}$ or $ml^{\frac{2}{4}}$. For the horizontal rod the Moment of Inertia is $ml^{\frac{2}{12}}$ + $m[\frac{l}{2}]^{2}$  which on solving we will get the Moment of Inertia to be $ml^{\frac{2}{6}}$ .

Hence, the Moment of Inertia about an axis parallel to the other rod will be summation of the two inertia = $ml^{\frac{2}{6}}$  + $ml^{\frac{2}{4}}$  which on solving we will get that $\frac{5}{12}ml^2$