Parallel Axis Theorem (College Board AP® Physics 1: Algebra-Based): Revision Note

The rotational inertia of a rod of mass $M$ and length $L$rotating about its center of mass is $\frac{1}{12}M{L}^2$. Which of the following describes the rotational inertia of the rod rotating about one of its ends?

A $\frac{1}{2}M{L}^2$

B $\frac{1}{3}M{L}^2$

C $\frac{1}{6}M{L}^2$

D $\frac{1}{12}M{L}^2$

The correct answer is B

Answer:

Step 1: Determine the distance between the center of mass and the parallel axis

$d=\frac{L}{2}$

Step 2: Apply the parallel axis theorem to the rod

$I\text{'}=I_{cm}+\hspace{0.17em}M{d}^2$

$I\text{'}=\frac{1}{12}M{L}^2+M{\left(\frac{L}{2}\right)}^2$

$I\text{'}=\frac{1}{12}M{L}^2+\frac{1}{4}M{L}^2$

$I\text{'}=\frac{1}{3}M{L}^2$

Two solid spheres form a dumbbell when attached to each end of a thin uniform rod. The dumbbell rotates with the center of mass of each sphere at a distance of 22 cm from the axis of rotation, as shown in the diagram.

The rod has a mass of 150 g. Each sphere has a radius of 4 cm and a mass of 750 g.

Rotational inertia of a thin rod about its center = $\frac{1}{12}m{L}^2$

Rotational inertia of a solid sphere about its center = $\frac{2}{5}M{R}^2$

(A) Calculate the total rotational inertia of the dumbbell arrangement

(B) Determine the proportion of rotational inertia the thin rod contributes to the dumbbell arrangement

Answer:

Part (A)

Step 1: Analyze the scenario

• The total rotational inertia of the dumbbell is the sum of

  • the rotational inertia of each solid sphere about the center of the rod

  • the rotational inertia of the thin rod about its center

Step 2: Calculate the rotational inertia of each sphere

• The rotational inertia of a solid sphere about its own center of mass is $I_{sphere,cm}=\frac{2}{5}M{R}^2$

• The rotational inertia of each sphere about the center of the rod can be found using the parallel axis theorem

$I_{sphere}=I_{sphere,cm}+M{d}^2$

• The mass of each sphere is $M=750\mathrm{g}=0.75\mathrm{kg}$

• The radius of each sphere is $R=4\mathrm{cm}=0.04\mathrm{m}$

• The distance between the rotational axis and each sphere is $d=22\mathrm{cm}=0.22\mathrm{m}$

$I_{sphere}=\frac{2}{5}M{R}^2+\hspace{0.17em}M{d}^2$

$I_{sphere}=\frac{2}{5}\left(0.75\right){\left(0.04\right)}^2+\left(0.75\right){\left(0.22\right)}^2=0.0368\mathrm{kg}·{\mathrm{m}}^2$

Step 3: Calculate the rotational inertia of the rod

• The mass of the rod is $m=150\mathrm{g}=0.15\mathrm{kg}$

• The length of the rod is $L=2\times \left(22-4\right)=36\mathrm{cm}=0.36\mathrm{m}$

$I_{rod}=\frac{1}{12}m{L}^2$

$I_{rod}=\frac{1}{12}\left(0.15\right){\left(0.36\right)}^2=0.00162\mathrm{kg}·{\mathrm{m}}^2$

Step 4: Calculate the total rotational inertia of the system

$I_{tot}=\sum m{r}^2$

$I_{tot}=2I_{sphere}+I_{rod}$

$I_{tot}=2\left(0.0368\right)+0.00162=0.0752\mathrm{kg}·{\mathrm{m}}^2$

Part (B)

Step 1: Analyze the scenario

• The proportion of rotational inertia contributed by the thin rod is given by

$\frac{I_{rod}}{I_{tot}}$

Step 2: Calculate the ratio of the rotational inertias

$\frac{I_{rod}}{I_{tot}}=\frac{0.00162}{0.0752}=0.0215\approx 2\%$

• This means the rod contributes about 2% of the overall rotational inertia of the dumbbell