Question

An engineer has an odd-shaped $\mathbf{10}\mathbf{}\mathbf{kg}$object and needs to find its rotational inertia about an axis through its center of mass. The object is supported on a wire stretched along the desired axis. The wire has a torsion constant, $\mathbf{k}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{N}\mathbf{.}\mathbf{m}$. If this torsion pendulum oscillates through$\mathbf{50}$cycles in$\mathbf{50}\mathbf{}\mathit{s}$, what is the rotational inertia of the object?

Short answer

The rotational inertia of an object about an axis through its center of mass is$0.079\mathrm{kg}{\mathrm{m}}^2$.

Step-by-step solution

The given data

• The mass of the object is,$M=10\mathrm{kg}$.
• The torsion constant of wire is,$k=0.50\mathrm{N}$.
• The pendulum oscillates through 20 cycles inrole="math" localid="1657263969416" $50\mathrm{s}$.

Understanding the concept of rotational inertia

Using the formula for the period of torsion pendulum, we can find the rotational inertia of an object about an axis through its center of mass by inserting the values of period and torsion constant.

Formula:

The period of torsion pendulum, $T=2\pi \sqrt{\frac{l}{k}}$ (i)

Calculation of rotational inertia

The pendulum oscillates through 20 cycles in 50 s. Hence, its period is given as:

$$\begin{gathered} T=\frac{50S}{20} \\ =2.5\mathrm{s} \end{gathered}$$

Now, from the equation (i), we can get the rotational inertia of the body as:

$$\begin{gathered} l=\frac{k{T}^2}{4{\mathrm{\pi }}^2} \\ =\frac{\left[\left(0.50\right)\left(2.5^2\right)\right]}{4{\left(3.142\right)}^2} \\ =0.079\mathrm{kg}·{\mathrm{m}}^2 \end{gathered}$$

Therefore, the rotational inertia of an abject about an axis through its center of mass is $0.079\mathrm{kg}·{\mathrm{m}}^2$.