Question

The moment of inertia of a uniform solid disc of mass M and radius R about an axis normal to the disk and passing through its center is MR^2/2 . What is the moment of inertia of the same disc about an axis lying in its plane and tangent to it (as shown in the figure)?

Answer 1

Given :

Mass of the disc = M

Radius of the disc = R

Moment of inertia normal to the disc = MR²/2

To Find :

The moment of inertia of disc lying in same plane and tangent to it

Solution :

• Parallel axis theorem :  moment of inertia of a body about an axis that is  parallel to an axis passing  through the center of mass is equal to  sum of moment of inertia of the body about the axis passing through center of  mass and product of mass and square of the distance between axes.
• By using parallel axis theorem

                     $I=\frac{MR^{2} }{2} +M\times (R^){2}$

                     $I=\frac{MR^{2} }{2} +MR^{2}$

                     $I=\frac{3MR^{2} }{2}$

The moment of inertia lying in the same plane and tangent to disc is 3MR²/2

Answer 2

Answer:

I think the correct answer is  $\frac{5MR^{2} }{4}$ . I think this question is from one of the IISER aptitude sample papers? And the correct answer is option D. I may not be right, but here's my reasoning. If you know Grade 11 Physics, you should be able to solve this.

Explanation:

First, we know that the moment of inertia of a disc perpendicular to its plane is  $\frac{MR^{2}}{2}$ . Using the perpendicular axis theorem, we can determine that the moment of inertia of a disc along one of its diameters will be half of that, which is  $\frac{MR^{2}}{4}$ .

Now, using parallel axis theorem ( $I_{r} = I_{cm} + MR^{2}$ ), we see that the moment of inertia of the same disc along a tangent (at a distance R from its centre of mass) will be:

$\frac{MR^{2}}{4} + MR^{2}$

=  $\frac{MR^{2}}{4} + \frac{4MR^{2}}{4}$

=  $\frac{5MR^{2}}{4}$

And I think that's the answer. Hope this helps!