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

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Physics

Parallel axis theorem

The rotational inertia of a rigid system can change depending on the orientation of its rotational axis
- The minimum rotational inertia in any plane is always when the axis passes through the system's center of mass
A rigid system may not always rotate about an axis passing through its center of mass
- As a result, the rotational inertia will always be greater than if the axis passed through the system's center of mass
The rotational inertia about an axis that is parallel to any axis passing through the center of mass can be calculated using the parallel axis theorem

$I' = I_{c m} + M d^{2}$

Where:
- $I_{c m}$ = rotational inertia about an axis passing through the system’s center of mass
- $I'$ = rotational inertia about an axis parallel to $I_{c m}$
- $M$ = mass of the system, in $kg$
- $d$ = distance between the rotational axis and the system’s center of mass
This means that the rotational inertia of a system will be the same when rotating about any parallel axis at an equal distance from the center of mass

Rotational inertia around a parallel axis

Diagram showing an irregular shaped object with a yellow cylindrical rod which is parallel to its center of mass. Variables I', Icm, and d indicate distances and axes. — **The rotational inertia about another axis that is parallel to the axis through the center of mass, at a distance d from the object’s center of mass, can be determined using the parallel axis theorem**

Worked Example

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

$I' = I_{c m} + M d^{2}$

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

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

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

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