Question

What is the moment of inertia of a pendulum with a mass of $3{\text{ Kg}}$ that $1{\text{ m}}$ from the pivot ?

Answer 1

Moment of inertia various with respect to the different axes. Use the formula of moment of inertia and also the parallel axis theorem to get the moment of inertia both along the center of mass of the bob, as well as the axis with respect to the pivoted point.

Formula used:
Moment of inertia is given as, $I = \sum {{m_i}{r_i}}$ where $i$ is the ith point of the object at ${r_i}$ units from the center of mass. Also, according to the parallel axis theorem, the moment of inertia along an axes h unit away from the center of mass is $I = {I_c} + m{h^2}$.

Complete step by step answer:
The mass of the A pendulum is a weight that is suspended from a hinge (a fixed-point pivot) and can freely swing. A restoring force acts on a pendulum as it is displaced sideways from its resting equilibrium position; as it reaches its highest point in its motion, gravity accelerates it back to the equilibrium position. When the restoring force is released, it causes the pendulum to oscillate around the equilibrium point, swinging back and forth.

Here, we consider the bob to be a point mass, so its moment of inertia with respect to the axis through its center of mass is $I = m{r^2}$[Here $r$ represents the distance of the point mass from the axis]. Since it is a point mass, its center of mass is at 0 units from the point mass. Hence, the moment of inertia with respect to the center of mass of the point bob is $I = 0$.

Now we can use the parallel axis theorem to find the moment of inertia with respect to the pivot. According to the parallel axis theorem, the moment of inertia along an axis which is $h$ units away from the axis through the center of mass is $I = {I_c} + m{h^2}$. Here the axis is along the fixed line through the pivot. And in it given that the pivot is $1m$ away from the bob. So, $h = 1m$. Also, we already found the ${I_c} = 0$. Hence, the moment of inertia along the axis through the pivot is $I = 0 + m{h^2}$
$\Rightarrow I = 3 \times 1{\text{ Kg}}{{\text{m}}^2}$
$\therefore I = 3{\text{ Kg}}{{\text{m}}^2}$

Hence, the moment of inertia of a pendulum is $3{\text{ Kg}}{{\text{m}}^2}$.

Note: Here, we assume the bob to be a point mass since, nothing about its dimensions is given. But if its radius is given, we have to use the formula of moment of inertia of a collection of points as mentioned above.