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): \((\frac{3}{2})\text{MR}^2\)
A string of length \( L \) is fixed at one end and carries a mass of \( M \) at the other end. The mass makes \( \frac{3}{\pi} \) rotations per second about the vertical axis passing through the end of the string as shown. The tension in the string is ________________ ML.
The current passing through the battery in the given circuit, is:
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
A full wave rectifier circuit with diodes (\(D_1\)) and (\(D_2\)) is shown in the figure. If input supply voltage \(V_{in} = 220 \sin(100 \pi t)\) volt, then at \(t = 15\) msec:
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-