Moment of inertia of a thin rod of mass M and length about an axis passing through its centre is (M L2/12). Its moment of inertia about a parallel axis at a distance of (L/4) from this axis is given by

Question

Moment of inertia of a thin rod of mass M and length about an axis passing through its centre is (M L2/12). Its moment of inertia about a parallel axis at a distance of (L/4) from this axis is given by (A) (M L2/48) (B) (M L3/48) (C) (M L2/12) (D) (7 M L2/48)

Tardigrade Admin · Accepted Answer

Apply parallel axis theorem, I=IC M+M h2, we get ⇒ =(M L2/12)+M((L/4))2 =(7 M L2/48)

Q. Moment of inertia of a thin rod of mass $M$ and length about an axis passing through its centre is $\frac{M L ^{2}}{12}$ . Its moment of inertia about a parallel axis at a distance of $\frac{L}{4}$ from this axis is given by

$\frac{M L ^{2}}{48}$

$\frac{M L ^{3}}{48}$

$\frac{M L ^{2}}{12}$

$\frac{7 M L ^{2}}{48}$

Apply parallel axis theorem,
$I = I_{CM} + M h^{2}$ , we get
$\Rightarrow= \frac{M L ^{2}}{12} + M (\frac{L}{4})^{2}$
$= \frac{7 M L ^{2}}{48}$

Q. Moment of inertia of a thin rod of mass M and length about an axis passing through its centre is 12ML2​. Its moment of inertia about a parallel axis at a distance of 4L​ from this axis is given by

48ML2​

48ML3​

12ML2​

487ML2​