The moment of inertia of a uniform circular disc of radius R and mass M about an axis passing from the edge of the disc and normal to the disc is:
\(\frac{5}{2}MR^2\)
\(MR^2\)
\(\frac{7}{2}MR^2\)
\(\frac{3}{2}MR^2\)
The Correct Option is D
Solution and Explanation
Moment of inertia of the disc around an axis perpendicular to its plane and passing through its centre of mass ICM = \(\frac{1}{2}\) MR2
The disc's moment of inertia about an axis perpendicular to its plane and running across its diameter may be calculated using the parallel axis theorem by using the formula I = ICM + Md2.
Therefore, I = \(\frac{1}{2}\) MR2+MR2 = \(\frac{3}{2}MR^2\).
Correct option is (D): \(\frac{3}{2}MR^2\)
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Concepts Used:
Moment of Inertia
Moment of inertia is defined as the quantity expressed by the body resisting angular acceleration which is the sum of the product of the mass of every particle with its square of a distance from the axis of rotation.
Moment of inertia mainly depends on the following three factors:
- The density of the material
- Shape and size of the body
- Axis of rotation
Formula:
In general form, the moment of inertia can be expressed as,
I = m × r²
Where,
I = Moment of inertia.
m = sum of the product of the mass.
r = distance from the axis of the rotation.
M¹ L² T° is the dimensional formula of the moment of inertia.
The equation for moment of inertia is given by,
I = I = ∑mi ri²
Methods to calculate Moment of Inertia:
To calculate the moment of inertia, we use two important theorems-
- Perpendicular axis theorem
- Parallel axis theorem