Question 738: Centre of Mass and Rotational Motion

Question

One quarter section is cut from a uniform circular disc of radius $R$. This section has a mass $M$. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

Accepted Answer

To determine the moment of inertia of a quarter section of a uniform circular disc of radius $R$ and mass $M$ about an axis perpendicular to its plane and passing through the center of the original disc, we start with the moment of inertia of the full disc.
The moment of inertia of a full uniform circular disc about an axis perpendicular to its plane and passing through its center is given by:
$$ I_{	ext{full}} = \frac{1}{2} M_{	ext{full}} R^2 $$
Since the original disc is uniformly distributed, the mass of the full disc $M_{	ext{full}}$ can be split evenly into four equal parts, and mass $M$ corresponds to the quarter section. Hence, we have:
$$ M = \frac{M_{	ext{full}}}{4} $$
We also know that the moment of inertia of a quarter section of the disc about the same axis would be one-fourth of that of the entire disc because the mass and area are uniformly distributed. Therefore, the moment of inertia of the quarter section $I_{	ext{quarter}}$ is:
$$ I_{	ext{quarter}} = \frac{1}{4} I_{	ext{full}} $$
Substituting $$ I_{	ext{full}} $$ from above:
$$ I_{	ext{quarter}} = \frac{1}{4} \left( \frac{1}{2} M_{	ext{full}} R^2 ight) $$
Now, replacing $$ M_{	ext{full}} $$ with $$ 4M $$ (since $$ M = \frac{M_{	ext{full}}}{4} $$), we get:
$$ I_{	ext{quarter}} = \frac{1}{4} \left( \frac{1}{2} (4M) R^2 ight) $$
This simplifies to:
$$ I_{	ext{quarter}} = \frac{1}{4} \left( 2 M R^2 ight) $$
$$ I_{	ext{quarter}} = \frac{1}{2} M R^2 $$
Therefore, the moment of inertia of the quarter section of the disc about the given axis is:
$$ \boxed{ \frac{1}{2} MR^2 } $$
So, the correct answer is Option A: $$ \frac{1}{2} MR^2 $$

Answer

$\dfrac{1}{2} MR^2$

Answer

$\dfrac{1}{4} MR^2$

Answer

$\dfrac{1}{8} MR^2$

Answer

$\sqrt{2} MR^2$

Centre of Mass and Rotational Motion

One quarter section is cut from a uniform circular disc of radius R. This section has a mass M. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

Choose the correct answer: