# What is the moment of inertia of a pendulum with a mass of 2 kg that is 1 m from the pivot?

Jun 1, 2017

The moment of inertia is $= 2 k g {m}^{2}$

#### Explanation:

We assume that it is a point mass.

The moment of inertia is

$I = m {r}^{2}$

The mass is $m = 2 k g$

The distance is $r = 1 m$

The moment of inertia is

$I = m {r}^{2} = 2 \cdot {1}^{2} = 2 k g {m}^{2}$

Jun 1, 2017

wkt $T = 2 \pi \cdot \sqrt{\frac{l}{g}}$
$T = 2 \pi \cdot \sqrt{\frac{I}{m} g d}$

#### Explanation:

So relating both the formulas we get
$T = 2 \pi \cdot \sqrt{\frac{l}{g}} = 2 \pi \cdot \sqrt{\frac{I}{m} g d}$

$\implies \sqrt{\frac{l}{g}} = \sqrt{\frac{I}{m} g d}$ ------( on squaring)
$\implies \left(\frac{l}{g}\right) = \left(\frac{I}{m} g d\right)$
$\implies \left(l \cdot m d\right) = I$ ------------( g got cancelled)
$\implies {l}^{2} \cdot m = I$---------------( l and d are same)
$\implies I = 1 \cdot 1 \cdot 2 = 2$ ------------( hence answer)
...............................hope it helps.........................:)