# What is the moment of inertia of a pendulum with a mass of 3 kg that is 7 m from the pivot?

Jan 12, 2016

$147 k g . {m}^{2}$

#### Explanation:

Assuming this is a simple pendulum (mass and dimensions negligible), we may use the definition of moment of inertia to obtain :

$I = {\sum}_{j = 1}^{n} {m}_{j} {r}_{p e r p . j}^{2}$

$= 3 \times {7}^{2}$

$= 147 k g . {m}^{2}$.

Note that if we consider it as a compound pendulum for more accurate results, then we would need to use the more general formula

$I = {\lim}_{n \to \infty} {\sum}_{j = 1}^{n} {m}_{j} {r}_{p e r p . j}^{2} = {\int}_{M} {r}^{2} \mathrm{dm}$.

In this case we would need more information regarding the dimensions of the pendulum, and the centre of mass together with the parallel axis theorem would be required to calculate the total moment of inertia in that case.