Moment of Inertia

Practice Questions:

Question 11: moderate

The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its midpoint and perpendicular to its length is $I_{0}$. Its moment of inertia about an axis passing through one of its ends perpendicular to its length is

(2011 Mains)

1. $I_{0} + ML^{2}/2$

2. $I_{0} + ML^{2}/4$

3. $I_{0} + 2ML^{2}$

4. $I_{0} + ML^{2}$

View Answer

Using the parallel axis theorem, $I = I_{cm} + Md^{2}$. Here, the center of mass moment of inertia is $I_{cm} = I_{0}$ and the distance to the parallel axis is $d = \frac{L}{2}$. Thus, $I = I_{0} + M(\frac{L}{2})^{2} = I_{0} + \frac{ML^{2}}{4}$.

💬 Discuss this Question

Question 12: moderate

From a circular disc of radius $R$ and mass $9M$, a small disc of mass $M$ and radius $\frac{R}{3}$ is removed concentrically. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its center is:

(2010 Mains)

1. $\frac{40}{9}MR^{2}$

2. $MR^{2}$

3. $4MR^{2}$

4. $\frac{4}{9}MR^{2}$

View Answer

Initial moment of inertia $I_{original} = \frac{1}{2}(9M)R^{2} = \frac{9}{2}MR^{2}$. Moment of inertia of the removed part is $I_{removed} = \frac{1}{2}M(\frac{R}{3})^{2} = \frac{1}{18}MR^{2}$. The remaining moment of inertia is $I = I_{original} - I_{removed} = \frac{9}{2}MR^{2} - \frac{1}{18}MR^{2} = \frac{40}{9}MR^{2}$.

💬 Discuss this Question

« Previous 1 2