The moment of inertia of a rod about an axis through its centre and perpendicular to it is I. The rod is bent in the middle, so that the two halves make an angle 60°. Now the new M.I about the same axis is

Question

The moment of inertia of a rod about an axis through its centre and perpendicular to it is I. The rod is bent in the middle, so that the two halves make an angle 60°. Now the new M.I about the same axis is (A) I / 2 (B) I/4 (C) I (D) 2I

Tardigrade Admin · Accepted Answer

I =(1/12) ML 2 I '=2 × (1/3)(( M /2))(( L /2))2 =(1/2) ML 2= I

Q. The moment of inertia of a rod about an axis through its centre and perpendicular to it is I. The rod is bent in the middle, so that the two halves make an angle $6 0^{\circ}$ . Now the new M.I about the same axis is

$I /2$

I/4

I

2I

$I = \frac{1}{12} M L^{2}$
$I^{^{'}} = 2 \times \frac{1}{3} (\frac{M}{2}) (\frac{L}{2})^{2}$
$= \frac{1}{2} M L^{2} = I$

Q. The moment of inertia of a rod about an axis through its centre and perpendicular to it is I. The rod is bent in the middle, so that the two halves make an angle 60∘. Now the new M.I about the same axis is

I/2

I/4

I

2I

I=121​ML2 I′=2×31​(2M​)(2L​)2 =21​ML2=I

Q. The moment of inertia of a rod about an axis through its centre and perpendicular to it is I. The rod is bent in the middle, so that the two halves make an angle $6 0^{\circ}$ . Now the new M.I about the same axis is

$I /2$

$I = \frac{1}{12} M L^{2}$
$I^{^{'}} = 2 \times \frac{1}{3} (\frac{M}{2}) (\frac{L}{2})^{2}$
$= \frac{1}{2} M L^{2} = I$