Question

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

Answer 1

The moment of inertia is `=2kgm^2`

Explanation:

We assume that it is a point mass.

The moment of inertia is

`I=mr^2`

The mass is `m=2kg`

The distance is `r=1m`

The moment of inertia is

`I=mr^2=2*1^2=2kgm^2`

Answer 2

wkt `T=2pi* sqrt(l/g)`
`T= 2pi* sqrt(I/mgd)`

Explanation:

So relating both the formulas we get
`T=2pi* sqrt(l/g) = 2pi* sqrt(I/mgd)`

`=> sqrt(l/g) = sqrt(I/mgd)` ------( on squaring)
`=> (l/g) = (I/mgd)`
`=>( l*md) = I` ------------( g got cancelled)
`=> l^2 *m = I`---------------( l and d are same)
`=> I= 1*1 * 2 = 2` ------------( hence answer)
...............................hope it helps.........................:)