A solid cylinder of radius R and length L have moment of inertia I1 and a second solid cylinder of radius R2 and length L2 cut from it have moment of inertia I2. Find 11/I2.
64
32
128
256
I1 = M(\((\frac{R^2}{4})\)+\((\frac{L_2}{12})\))
I1 = \((\frac{M}{4})\)(R2+\((\frac{L^2}{3})\))
M = ρπR2L
M2 = ρπ\((\frac{R^2}{8})\)L = \(\frac{M}{8}\)
I2 = \((\frac{M}{8})\)x(\((\frac{1}{4})\)[\((\frac{R^2}{4})\)+\((\frac{L_2}{12})\)]
= \((\frac{M}{128})\)[R2+\((\frac{L_2}{3})\)]
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.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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-