The moment of inertia about an axis passing through centre of mass of disc and perpendicular to its plane is \(\text{I}_{CM}\) = \(\frac{1}{2}\)\(\text{MR}^2\)
where M is the mass of disc and R its radius. According to theorem of parallel axis, moment of inertia of circular disc about an axis touching the disc at its diameter and normal to the disc is
\(\text {I}\) = \(\text{I}_{CM}\) + \(\text{MR}^2\)
= \(\frac{1}{2}\) \(\text{MR}^2\) + \(\text{MR}^2\)
= \(\frac{3}{2}\) \(\text{MR}^2\)
Therefore, the correct option is (C): \((\frac{3}{2})\text{MR}^2\)
In a Vernier caliper, \(N+1\) divisions of vernier scale coincide with \(N\) divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:
Identify the major product C formed in the following reaction sequence:
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.
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
To calculate the moment of inertia, we use two important theorems-