From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a 90\(^{\circ}\) sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times ‘MR2 ’. Then the value of ‘K’ is
\(\frac{1}{8}\)
\(\frac{3}{4}\)
\(\frac{7}{8}\)
\(\frac{1}{4}\)
The Correct Option is B
Solution and Explanation
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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