SATHEE: Rotational Motion Question 61

Rotational Motion Question 61

Question: The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc is:

[AIPMT (S) 2006]

Options:

A) $M R^{2}$

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

C) $\frac{3}{2} M R^{2}$

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

Show Answer

Answer:

Correct Answer: C

Solution:

The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is

$I_{C m} = \frac{1}{2} M R^{2}$

where M is the mass of the disc and R its radius.

According to theorem of parallel axis, ML of circular disc about an axis touching the disc at its diameter and normal to the disc is

$I = I_{C M} + M R^{2}$

$= \frac{1}{2} M R^{2} + M R^{2}$

$= \frac{3}{2} M R^{2}$

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Rotational Motion Question 61

Question: The moment of inertia of a uniform circular disc of radius R and mass M about an axis touching the disc at its diameter and normal to the disc is:

Options:

Answer:

Solution:

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