Chapter 12, Problem 42A

What happens to the rotational inertia of a simple pendulum when the mass of the bob is doubled and the length of the pendulum is halved? (See Figure 12.6.)

To determine

Resulting rotational inertia.

Answer to Problem 42A

The resulting rotational inertia is halved that is $I\text{'}=\frac{I}{2}$

Explanation of Solution

Given:

Mass of the bob is doubled.

Length of the pendulum is halved.

Formula Used:

The rotational inertia of simple pendulum is given as,

  $I=m{L}^2$

Where, $m$ is the mass of bob and $L$ is length of pendulum.

Calculation:

New mass of bob is, $m\text{'}=2m$ and

New length of pendulum is, $L\text{'}=\frac{L}{2}$

So resulting rotational inertia is,

  $\begin{array}{l}I\text{'}=m\text{'}L{\text{'}}^2\\ =2m\times {\left(\frac{L}{2}\right)}^2\\ =\frac{m{L}^2}{2}\\ \Rightarrow I\text{'}=\frac{I}{2}\end{array}$

Conclusion:

Therefore, the resulting rotational inertia of simple pendulum is halved that is $I\text{'}=\frac{I}{2}$