Disconcerting disc

Classical Mechanics Level 5

Find the Moment of Inertia (in S.I. unit) of a uniform semicircular disc of mass $M$ and radius $R$ about an axis parallel to its plane and touching it at circumference while making an angle $\theta$ with its diameter as shown in the figure.

$\text{Given Data} : ~ M=12~kg \\ R = 5m \\ \theta = \cos ^{ -1 }{ \left( \frac {\pi }{ 4 } \right) }$

The answer is 175.

1 solution

Purushottam Abhisheikh
May 14, 2015

Let $C$ be the center of mass of disc. So $d=\dfrac{4R}{3\pi}$ .

Now, by parallel axis theorem, $I_O = I_C + M(dcos\theta)^2 \\ \implies I_C = I_O - Md^2cos^2\theta \\ \therefore I_C = M\left(\dfrac{R^2}{4}-d^2cos^2\theta\right)$ .

Again by parallel axis theorem, $I_P= I_C + M(R-dcos\theta)^2 \\ = \dfrac{MR^2}{4} - Md^2cos^2\theta + M(R^2 + d^2cos^2\theta - 2Rdcos\theta) \\ \therefore I_P = \dfrac{5MR^2}{4} - 2MRdcos\theta$

Putting the given values in the above expression we would get $I_P = 175 ~kgm^2$

Parallel Axis Theorem just didn't click !!! Had to spend quite a few pages integrating it.. :(

Ankit Kumar Jain - 3 years, 3 months ago

Dear Sir, it seems to me that you gave an incorrect final answer. As MR^2 = 300, then you will get 195, not 175.

Sergey Krotov - 5 years ago

I rechecked my answer and it is still 175.

Purushottam Abhisheikh - 5 years ago

Disconcerting disc

The answer is 175.

1 solution

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