Question

The moment of inertia of a disc of mass M and radius
R about an axis, which is tangential to circumference of
disc and parallel to its diameter is​

Answer 1

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!