Physics 207 WebPages - Fall 1999

Problem 6 (Problem 53 in Textbook)

Calculate the rotational inertia of a meter stick, with mass 0.56 kg, about an axis perpendicular to the stick and located at the 20 cm mark. (Treat the stick as a thin rod)

Solution:

There are two ways to solve this problem. In the first way, we first treat the stick as two different parts and find the mass of each.

Calculate the rotational inertia of each part using the following equation from Table 11-2. We treat each part as they rotate about the axis through one end perpendicular to the length of the meter stick.

Summing the rotational inertia of each part gives the rotational inertia of the body.

The second method uses the parallel-axis theorem. We first calculate the rotational inertia of the meter stick with the axis of rotation through the center of mass. We obtain the following equation from Table 11-2. The radius R here should be the length of the whole meter stick, which is 1 m.

Then, we use the parallel-axis theorem to solve for the rotational inertia of the meter stick rotating about the axis perpendicular to the stick located at the 20 cm mark. The R here is the distance between the rotating axis in this case and the rotating axis through the center of mass. Thus, the distance R here is 30 cm, the rotating axis through the center of mass at 50 cm minus the rotating axis at 20 cm.

Finally, we substitute and solve.

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