Calculate the rotational inertia of a meter stick, with mass 0.499 kg, about an axis perpendicular to the stick and located at the 35.2 cm mark. (Treat the stick as a thin rod.)

The rotational inertia of the meter stick is 0.0618 kgm².

Explanation:

Given;

mass of the meter stick, m = 0.499 kg

perpendicular distance to the rotational axis, r = 35.2 cm = 0.352 m

The rotational inertia or moment of inertia for a point mass is given by;

I = mr²

where;

m is the point mass

r is the perpendicular distance

Substitute the givens and solve for moment of inertia.

I = (0.499)(0.352)²

I = 0.0618 kgm²

Therefore, the rotational inertia of the meter stick is 0.0618 kgm².

what is the actual question though?