SATHEE: Rotational Motion Question 12

Rotational Motion Question 12

Question: The moment of inertia of uniform semi-circular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is

[AIEEE 2005]

Options:

A) $\frac{1}{4} M r^{2}$

B) $\frac{2}{5} M r^{2}$

C) $M r^{2}$

D) $\frac{1}{2} M r^{2}$

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Answer:

Correct Answer: D

Solution:

[d] The mass of complete (circular) disc is

The moment of inertia of disc about the given axis is

$l = \frac{2 M r^{2}}{2} = M r^{2}$

The disc may be assumed as combination of two semi-circular parts.

Thus, the moment of inertia of a semi-circular part of disc,

$l_{1} - l - l_{1}$

$∴ l_{1} = \frac{l}{2} = \frac{M r^{2}}{2}$

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Rotational Motion Question 12

Question: The moment of inertia of uniform semi-circular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is

Options:

Answer:

Solution:

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