# What is the moment of inertia of a pendulum with a mass of 3 kg that is 1 m from the pivot?

Jan 15, 2018

$3 K g . {m}^{2}$

#### Explanation:

Here,you will have to consider that the bob of the pendulum is a point mass,because its radius is not given.

so,the given situation comes out to be that a mass of $3 K g$ is placed$1 m$ away from a point,so its moment of inertia w.r.t that point will be $m {r}^{2}$ or $3 K g . {m}^{2}$

if,the radius of the bob was given as$r$,its moment of inertia w.r.t a diameter would be $\frac{2}{5} m {r}^{2}$ (bob is actually a sphere)

so,distance between the centre of the bob and the pivoted point would become $\left(1 + r\right)$

hence,w.r.t the pivoted point moment of inertia would have been$\left(\frac{2}{5} m {r}^{2} + m {\left(1 + r\right)}^{2}\right)$